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】星期几","tags":["分支结构","数组"],"primaryTag":"分支结构","difficulty":"入门","color":"#dc2626","url":"/problems/roj/5059","isolated":true},{"id":"roj/5058","oj":"roj","problem_id":"5058","label":"roj 5058","title":"【例3.7】最大的数","tags":["分支结构","模拟"],"primaryTag":"分支结构","difficulty":"入门","color":"#dc2626","url":"/problems/roj/5058","isolated":true},{"id":"roj/5055","oj":"roj","problem_id":"5055","label":"roj 5055","title":"【例3.3】三个数","tags":["排序","分支结构"],"primaryTag":"排序","difficulty":"入门","color":"#dc2626","url":"/problems/roj/5055","isolated":true},{"id":"roj/5054","oj":"roj","problem_id":"5054","label":"roj 5054","title":"【例3.2】范围判断","tags":["分支结构"],"primaryTag":"分支结构","difficulty":"入门","color":"#dc2626","url":"/problems/roj/5054","isolated":true},{"id":"roj/5053","oj":"roj","problem_id":"5053","label":"roj 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可互相套用。"},{"id":"common:OpenJ_Bailian/4135--luogu/P1873","source":"OpenJ_Bailian/4135","target":"luogu/P1873","type":"common","directed":false,"reason":"同为二分答案模板题，P1873 为最简入门，4135 在此基础上加入贪心分段验证。"},{"id":"pre:luogu/P3372->luogu/U420425","source":"luogu/P3372","target":"luogu/U420425","type":"pre","directed":true,"reason":"本题操作 1、2 就是 P3372 的区间加与区间和(题面也写了与线段树一一致), 懒标记线段树那部分直接复用; 本题新增的只是操作 3 的树上二分与操作 4 的单点赋值"},{"id":"pre:roj/3110->luogu/P3178","source":"roj/3110","target":"luogu/P3178","type":"pre","directed":true,"reason":"B 的解法三（main3.cpp）直接用 A 教的「双树状数组维护差分：区间加改 l、r+1 两端点，前缀和用 (x+1)Σd-Σj·d」这一步骤（RangeFenwick::range_add/prefix_sum 与 A 的 bit1/bit2 公式同构），只是在其上叠加 A 未教的 DFS 序与树链剖分，把子树加、根路径和映射成数组区间加/区间和后调用该模板。"},{"id":"pre:luogu/P8855->luogu/P3038","source":"luogu/P8855","target":"luogu/P3038","type":"pre","directed":true,"reason":"B 的路径差分步骤在 lca(u,v) 处执行 c[lca] -= 2, 直接复用 A 教的求 LCA 这一步"},{"id":"pre:luogu/P3258->luogu/P3038","source":"luogu/P3258","target":"luogu/P3038","type":"pre","directed":true,"reason":"B 的边差分就是 A 教的 LCA 路径差分这一步的变体: 点差分的 c[lca]--、c[parent(lca)]-- 在 B 里换成 c[lca]-=2, 端点标记与子树累加还原原样沿用, 再叠加 DFS 序与树状数组求子树和."},{"id":"pre:luogu/P8805->luogu/P3038","source":"luogu/P8805","target":"luogu/P3038","type":"pre","directed":true,"reason":"B 的树上差分沿用 A 的根路径容斥拆路径(路径=根路径u+根路径v−2×根路径lca, LCA 段被重复计算需修正), 只是把查询侧前缀和改造成更新侧三个单点差分"},{"id":"pre:luogu/P8855->luogu/P5588","source":"luogu/P8855","target":"luogu/P5588","type":"pre","directed":true,"reason":"解法二判断同色点共线、求外侧连通块大小都要反复算树上距离, 直接复用 A 教的倍增 LCA 距离公式 depth[u]+depth[v]-2depth[p]."},{"id":"pre:luogu/P3398->luogu/P5588","source":"luogu/P3398","target":"luogu/P5588","type":"pre","directed":true,"reason":"解法二的共线检查原样搬用 A 的距离拆分路径判定(dist(a,x)+dist(x,b)=dist(a,b) 当且仅当 x 在路径上), 在其上叠加直径端点与两侧计数."},{"id":"pre:luogu/P3379->luogu/P3398","source":"luogu/P3379","target":"luogu/P3398","type":"pre","directed":true,"reason":"本题三种解法的查询步骤都是倍增 LCA 的直接调用，先掌握「先同深、再同步上跳」的查询结构，再叠加路径相交判定。"},{"id":"pre:roj/1258->luogu/P1544","source":"roj/1258","target":"luogu/P1544","type":"pre","directed":true,"reason":"B 把 A 教的数塔“两个方向取 max 再加格点权值”转移原样嵌进自己的 dp[i][j][c]，只在其上叠加三倍次数维度 c 与 3*a[i][j] 分支"},{"id":"pre:acwing/1015->luogu/P1544","source":"acwing/1015","target":"luogu/P1544","type":"pre","directed":true,"reason":"B 的数字三角形 DP 直接复用了 A 教的“两个前驱取 max 再加当前格权值”转移（B 代码里就是 max(pre[j],pre[j-1])+a[i][j]），只是叠加了三倍次数维度 c、k=min(k,n) 截断与滚动数组，A 的纯路径 DP 是它的前置模板。"},{"id":"pre:luogu/U663298->luogu/P14360","source":"luogu/U663298","target":"luogu/P14360","type":"pre","directed":true,"reason":"B 的前缀子集和 DP 直接沿用 A 教的计数型 0/1 背包：一维 dp[s] 表示子集和恰好为 s 的方案数，dp[0]=1 初始化、容量倒序枚举 for(s=MAXV;s>=a[i];s--) dp[s]+=dp[s-a[i]]，只把 A 的输出目标从『恰好装满 V』换成『子集和 > a[i] 的方案数』（用 2^(i-1) 减前缀和统一统计超上限部分）。"},{"id":"pre:luogu/P2918->luogu/P14360","source":"luogu/P2918","target":"luogu/P14360","type":"pre","directed":true,"reason":"把超过目标重量的状态统一截断到阈值，是本题只精确维护不超过值域上限的子集和、超过部分合并统计的直接前置思想。"},{"id":"pre:luogu/P1115->luogu/P1122","source":"luogu/P1115","target":"luogu/P1122","type":"pre","directed":true,"reason":"A 教的「以 i 结尾/必须包含当前元素」状态设计与「负贡献就不接」取舍，被 B 逐孩子复用成 dp[u]=beauty[u]+sum(max(0,dp[v]))（main.cpp 中 if (dp[v] > 0) dp[u] += dp[v]），只是把一行序列换成了树上后序遍历"},{"id":"pre:luogu/P3009->luogu/P1122","source":"luogu/P3009","target":"luogu/P1122","type":"pre","directed":true,"reason":"B 的树形 DP 直接复用 A 教的「状态必须包含当前元素 + 前面的负贡献不接」这一步：A 在序列上写成 dp[i]=max(a[i],dp[i-1]+a[i]) 并在 dp[i-1] 是负贡献时重新起头，B 把它逐孩子套成 dp[u]=beauty[u]+sum(max(0,dp[v]))（main.cpp 的 if (dp[v] > 0) dp[u] += dp[v]），再用同一句「最终答案是所有 dp 中的最大值」收尾，额外叠加的只是树上后序 DFS 与父节点去重，难度只提高一级。"},{"id":"pre:roj/3148->luogu/P2015","source":"roj/3148","target":"luogu/P2015","type":"pre","directed":true,"reason":"B 的 dp[u][j] 直接复用 A 教的「子树彼此独立、与外界唯一联系是根，故可压成局部状态」这一步：B 原文把状态定义为「在 u 子树里且这些边都能通过 u 连回根」，并在显式栈后序下逐儿子合并；A 只教了 0/1 两档状态相加，B 额外叠加「取儿子子树须先保留父边 take+1」的容量维树上分组背包，构成普及+/提高- 到 普及+/提高 的一级台阶，属模板级复用。"},{"id":"pre:luogu/U661995->luogu/P2015","source":"luogu/U661995","target":"luogu/P2015","type":"pre","directed":true,"reason":"B 的 main.cpp 完整复用了 A 教的「冻结旧状态、组内候选项统一从旧状态转移」这一步：先 new_dp[t]=dp[u][t]（不选该组），再用 new_dp[used+take+1]=max(..., dp[u][used]+dp[v][take]+w) 从冻结的 dp[u] 转移，只是把 A 的一维容量组背包扩成「每个儿子子树为一组、take 为组内选项」的树上分组背包，并额外叠加父边必留的 +1 与后序合并流程。"},{"id":"pre:roj/3549->luogu/P2014","source":"roj/3549","target":"luogu/P2014","type":"pre","directed":true,"reason":"A 教的「依赖先在家族内枚举压成组内选项、再用一维倒序容量在互斥组间做分组背包（每组至多选一个）」正是 B 树形背包合并每棵孩子子树时实际使用的步骤：B 把每个孩子子树当成一组、名额 j 倒序转移 dp[u][j-k]+dp[v][k]（main.cpp 的 for (j=m+1;j>=1;j--)），只是在其上叠加虚拟根 0 把森林变树与子树状态 dp[u][j] 的定义。"},{"id":"pre:luogu/U661995->luogu/P2014","source":"luogu/U661995","target":"luogu/P2014","type":"pre","directed":true,"reason":"B 的 main.cpp:33-41 复用了 A 的分组背包步骤——把每个孩子子树当成一组物品、按组分配名额转移 dp[u][j-k]+dp[v][k]，A 教的“一组当整体决策单元、组间用独立状态隔离”正是这一步的基础；B 在其上叠加虚拟根把森林变树、树形 DP 与子树状态定义。"},{"id":"pre:luogu/P1775->luogu/P3205","source":"luogu/P1775","target":"luogu/P3205","type":"pre","directed":true,"reason":"B 复用 A 的「dp[l][r] 连续区间子问题、按区间长度递增递推」这一步区间 DP 框架，再叠加最后插入者在左端/右端的 0/1 维度与向两端扩张的计数转移"},{"id":"pre:luogu/P8218->luogu/P1775","source":"luogu/P8218","target":"luogu/P1775","type":"pre","directed":true,"reason":"B 的区间 DP 转移 dp[l][r]=min(...+sum(l,r)) 反复求区间质量，直接复用 A 教的前缀和数组 prefix[r]-prefix[l-1] 这一步，B 在此之上叠加枚举最后一次合并断点的区间 DP。"},{"id":"pre:luogu/P7113->luogu/P7077","source":"luogu/P7113","target":"luogu/P7077","type":"pre","directed":true,"reason":"B 复用 A 的「在调用 DAG 上按拓扑序沿边传播数值」这一步递推流程，只是把污水平均分换成 mul_all 乘积与 coef 系数的双向拓扑传播，并叠加乘法放大加法的线性合成"},{"id":"pre:luogu/P2863->luogu/P2515","source":"luogu/P2863","target":"luogu/P2515","type":"pre","directed":true,"reason":"B 的缩点步骤复用 A 教的 Tarjan 求 SCC 并按分量聚合规模，把每个 SCC 聚成超点并求环内重量价值之和，再叠加入度为 0 连虚拟根与树形背包"},{"id":"pre:luogu/P2014->luogu/P2515","source":"luogu/P2014","target":"luogu/P2515","type":"pre","directed":true,"reason":"树形背包原型题，P2515 缩点后转化为该模型。"},{"id":"pre:luogu/P3387->luogu/P2515","source":"luogu/P3387","target":"luogu/P2515","type":"pre","directed":true,"reason":"需要先掌握 Tarjan 缩点，把循环依赖压成 DAG 或树结构。"},{"id":"pre:luogu/P1775->luogu/P1063","source":"luogu/P1775","target":"luogu/P1063","type":"pre","directed":true,"reason":"B 原样复用 A 教的「只关心最后一次合并断在哪里、dp[l][k]+dp[k+1][r] 按区间长度递增递推」这一步（同一断点分解与长度序），只是把代价换成三标签乘积取 max 并叠加 A 未教的环断链复制。"},{"id":"pre:roj/1571->luogu/P1063","source":"roj/1571","target":"luogu/P1063","type":"pre","directed":true,"reason":"B 的聚合区间 DP 正解在选劈分顶点 k 劈成两个连续子区间再合并（dp[l][k]+dp[k+1][r]+a[l]*a[k+1]*a[r+1]）这一步上完整复用 A 教的区间劈分+三端点乘积合并骨架，只把 A 未涉及的外层破环成链与长度 N 取最值叠加成「最后一次聚合断点」转移，故 A 是 B 的台阶而非同难度孪生。"},{"id":"pre:luogu/P1775->luogu/P1880","source":"luogu/P1775","target":"luogu/P1880","type":"pre","directed":true,"reason":"B 的链上区间 DP 直接复用 A 教的枚举最后一次合并断点 dp[l][r]=min(dp[l][k]+dp[k+1][r]+sum) 与按长度递增递推，再叠加破环成链与 min/max 双答案"},{"id":"pre:roj/3526->luogu/P1880","source":"roj/3526","target":"luogu/P1880","type":"pre","directed":true,"reason":"A 教的关键步骤是把「连续区间 [l,r] 作为状态、枚举断点 k 劈成左右两段独立子问题、按区间长度从小到大递推」，B 的链上区间 DP（main.cpp 中 dp[l][k]+dp[k+1][r]+seg_sum）正是同一步，只是在此之上叠加复制数组破环成链与 min/max 双表。"},{"id":"pre:roj/3145->luogu/P1880","source":"roj/3145","target":"luogu/P1880","type":"pre","directed":true,"reason":"B 的链上区间 DP（main.cpp 中 dp[l][k]+dp[k+1][r]+seg_sum）就是 A 教的「枚举最后一次合并的断点 k」转移式，B 只是在其上叠加复制数组的环形处理并同时维护最大最小"},{"id":"pre:luogu/P1656->luogu/P2860","source":"luogu/P1656","target":"luogu/P2860","type":"pre","directed":true,"reason":"先把 Tarjan 判桥（low[v] > dfn[u]，重边时跳过进入边的反向边）练熟，本题第一步就是同一套判桥模板"},{"id":"pre:luogu/P8436->luogu/P2860","source":"luogu/P8436","target":"luogu/P2860","type":"pre","directed":true,"reason":"先会用「删掉所有桥再染色」求边双连通分量，本题在此基础上把边双缩成桥树并数叶子"},{"id":"pre:luogu/P3388->luogu/P3469","source":"luogu/P3388","target":"luogu/P3469","type":"pre","directed":true,"reason":"B 明确复用 A 教的「low[v]>=dfn[u] 判割点」这一步来确定删 u 后哪些子树独立成块（B 原文“这正是 Tarjan 割点里 low[v]>=dfn[u] 的含义”），再叠加 A 未教的答案拆两部分（2*(n-1) 直接访问 + 跨连通块访问对计数）。"},{"id":"pre:POJ/1144->luogu/P3469","source":"POJ/1144","target":"luogu/P3469","type":"pre","directed":true,"reason":"B 的核心步骤复用 A 教的 Tarjan 割点判定 low[v]>=dfn[u]（A 的 1.cpp 即 if(low[v]>=dfn[u] && u!=root)），回溯时按该判据取出被切下的子树块并计算作废访问贡献"},{"id":"pre:POJ/1523->luogu/P3469","source":"POJ/1523","target":"luogu/P3469","type":"pre","directed":true,"reason":"B 明确复用 A 的「low[v]>=dfn[u] 判定割点、每个满足条件的子节点切出一块子网」这一步，回溯时拿 sub_size[v] 累加块间作废访问数，再叠加 2*(n-1) 直接损失与剩余大块贡献"},{"id":"pre:luogu/P1194->luogu/P4180","source":"luogu/P1194","target":"luogu/P4180","type":"pre","directed":true,"reason":"A 教的 Kruskal（并查集按边权升序选边）正是 B 的第一步；B 在其上叠加「枚举非树边 + 树上换边」与倍增 LCA 维护路径最值。"},{"id":"pre:luogu/P1396->luogu/P4180","source":"luogu/P1396","target":"luogu/P4180","type":"pre","directed":true,"reason":"A 教的「树上/加边过程中的路径瓶颈最值」思想在 B 中升级为倍增表同时维护最大与严格次大边权，用于换边时找「严格小于 w 的最大树边」。"},{"id":"pre:luogu/P1546->luogu/P1194","source":"luogu/P1546","target":"luogu/P1194","type":"pre","directed":true,"reason":"B 的 Prim 主循环原样复用 A 教的「维护最小连边代价、每次选代价最小者加入、再用新点更新其它点」这一步（B 自述“关键不是 Prim 模板本身，而是建模”），再叠加 A 未教的虚拟源点把直接买/优惠买建成 MST 的建模。"},{"id":"pre:roj/1350->luogu/P1194","source":"roj/1350","target":"luogu/P1194","type":"pre","directed":true,"reason":"B 的 Prim 主循环与 A 逐句同构：min_cost 即 dis，接入新礼物 u 后用 K[u] 行做一遍 min 松弛；A 教的这一步正是 B 主解法的骨架，B 仅叠加虚拟源点建模"},{"id":"pre:roj/1349->luogu/P1194","source":"roj/1349","target":"luogu/P1194","type":"pre","directed":true,"reason":"B 的 Prim 直接复用 A 教的「维护到已接入点集的最便宜接入边 + 用新接入点那一行松弛」这一步，只把初值改成虚拟源点的 A 并跳过 0 边，再在其上叠加虚拟源点建模"},{"id":"pre:luogu/P1396->luogu/P1967","source":"luogu/P1396","target":"luogu/P1967","type":"pre","directed":true,"reason":"A 是单次询问的瓶颈路（最小化路径最大边权），把「阈值连通性」的论证讲透了；B 是它的镜像问题（最大化路径最小边权）且升级为多次询问，正好把 Kruskal 的阈值论证从「单对点首次连通」推广到「整张图的最大生成森林」"},{"id":"pre:luogu/P2504->luogu/P1967","source":"luogu/P2504","target":"luogu/P1967","type":"pre","directed":true,"reason":"B 复用 A 教的并查集按边判连通再合并的 Kruskal 加边机制，A 的关键性质「最小生成树最大边 = 全局最小可行跳跃上限」是 B 的「最大生成树路径最小边权 = 点对最大瓶颈值」的全局版，两题论证完全同源"},{"id":"pre:luogu/P5903->luogu/P1967","source":"luogu/P5903","target":"luogu/P1967","type":"pre","directed":true,"reason":"B 的倍增查询直接复用 A 教的 up[x][j] 倍增表与按二进制拆分上跳，只是上跳时沿途对 min_edge 取最小值，以回答路径瓶颈查询"},{"id":"pre:luogu/P4779->luogu/P3403","source":"luogu/P4779","target":"luogu/P3403","type":"pre","directed":true,"reason":"B 的 main.py 原样保留 A 教的堆优化 Dijkstra 步骤（含 current!=distance 时跳过过期条目）用于定出每个余数的最小楼层，B 在此之上叠加取模建余数图与 (h-dist[r])//base+1 计数。"},{"id":"pre:roj/1382->luogu/P3403","source":"roj/1382","target":"luogu/P3403","type":"pre","directed":true,"reason":"B 的余数图里每个余数的最早可达楼层 dist[r] 正是由 A 教的「弹出即定型 + 惰性删除」Dijkstra 一步步定出的（B 的 main.py 也原样保留了 `current != distance[residue]` 的过期条目跳过），B 只是在此之上叠加取模建图与 (h-dist[r])//base+1 计数的新流程。"},{"id":"pre:shumeng/CSP201509D->luogu/P3275","source":"shumeng/CSP201509D","target":"luogu/P3275","type":"pre","directed":true,"reason":"B 沿用 A 的把互相可达关系归结为强连通分量并求出分量这一步，用来判定分量内权 1 边构成正环，再叠加差分约束与缩点 DAG 最长路"},{"id":"pre:HDU/3836->luogu/P3275","source":"HDU/3836","target":"luogu/P3275","type":"pre","directed":true,"reason":"B 的正环判定与最长路都建立在 A 的「求 SCC 并缩点成 DAG」这一步上（同一 SCC 内出现权 1 边即正环无解，缩点 DAG 上再做 DP），再叠加 A 未教的差分约束 x_v>=x_u+w 建图与分量大小乘最长路求总糖果。"},{"id":"pre:luogu/P2910->luogu/P6464","source":"luogu/P2910","target":"luogu/P6464","type":"pre","directed":true,"reason":"B 直接复用 A 的 Floyd 全源最短路矩阵作基底，再叠加枚举传送门端点对点对距离取 min 的新步骤"},{"id":"pre:roj/1382->luogu/P1875","source":"roj/1382","target":"luogu/P1875","type":"pre","directed":true,"reason":"B 复用 A 教的堆优化 Dijkstra「按距离从小到大定型」单调序：A 证明候选距离最小者一次定型，B 照此先按最小成本弹出定型药水（main.py 同样用小根堆加 current!=distance 跳过过期条目），再按成本从小到大统计配方超边的松弛与最优方案数，额外叠加的只是 A+B->C 超边建模与方案计数这一层 A 未讲的流程。"},{"id":"pre:luogu/P4779->luogu/P1875","source":"luogu/P4779","target":"luogu/P1875","type":"pre","directed":true,"reason":"B 复用 A 的堆取当前最小距离并松弛的 Dijkstra 主循环，把单前驱出边松弛扩成两个前驱到齐才松弛的配方超边"},{"id":"pre:roj/3067->luogu/P1144","source":"roj/3067","target":"luogu/P1144","type":"pre","directed":true,"reason":"B 直接复用 A 教的那一步「BFS 首次访问即最短距离、dist 兼作访问标记」（B 代码里 dist_node[v]==-1 时置 dist_node[u]+1 并 ways[v]=ways[u]），只是在这次分层求距离之上再叠加一条额外流程：对 dist[v]==dist[u]+1 的同层边累加 ways[v]+=ways[u] 做最短路计数。"},{"id":"pre:luogu/P4779->luogu/P4568","source":"luogu/P4779","target":"luogu/P4568","type":"pre","directed":true,"reason":"B 的 main.py 把 A 教的「堆优化 Dijkstra + 弹出时跳过过期条目」整体搬到分层状态 (city,used) 上（current != distance[used][node] 即 A 的惰性删除），再叠加 A 未教的免费次数进状态分层与 0 代价跨层转移。"},{"id":"pre:roj/1382->luogu/P4568","source":"roj/1382","target":"luogu/P4568","type":"pre","directed":true,"reason":"B 把 A 教的堆优化 Dijkstra 骨架整体搬来做起点（文章明说“直接运行 Dijkstra”，main.py 里 current != distance[used][node] 跳过过期条目正是 A 教的惰性删除），只是把状态从一维 dist 扩成 (city,used) 分层图并叠加“花一次机会以 0 代价进入下一层”的新流程。"},{"id":"pre:roj/1416->roj/1415","source":"roj/1416","target":"roj/1415","type":"pre","directed":true,"reason":"1416 教的是「矩形覆盖点」的扫描线做法：把矩形沿 x 拆成进入/离开事件，y 方向用差分 + 一维树状数组做区间加、单点查。本题只是把加矩形变成了在线操作，x 无法提前扫完，于是把这套「差分 + 前缀和」整体从一维升到二维，主解正是 1416 的二维版本。"},{"id":"pre:luogu/P1443->luogu/P3371","source":"luogu/P1443","target":"luogu/P3371","type":"pre","directed":true,"reason":"无权图上的 BFS 最短路，先建立「按边松弛求最短距离」的直觉"},{"id":"common:luogu/P3371--luogu/P4779","source":"luogu/P3371","target":"luogu/P4779","type":"common","directed":false,"reason":"同一单源最短路模型，非负权改用堆优化 Dijkstra 的标准做法"},{"id":"common:luogu/P3371--luogu/P3385","source":"luogu/P3371","target":"luogu/P3385","type":"common","directed":false,"reason":"同一份 SPFA 代码，用最短路边数判可达负环"},{"id":"pre:roj/1485->luogu/P14363","source":"roj/1485","target":"luogu/P14363","type":"pre","directed":true,"reason":"本文解法二直接复用了 A 教的关键观察：把「当前状态 fail 链上的终止节点」当成本次读入所新增的全部匹配，A 用它在 BFS 里下传词尾命中标记判定「已命中」，B 用它在扫描询问时枚举所有以位置 i 结尾的规则匹配，再叠加字符对建模、覆盖 [L,R] 拆末尾子问题、fail 树 DFS 序 + 树状数组按深度阈值离线计数这些新流程；A（提高）是简洁的 AC 自动机判定/状态基础题，B（省选/NOI-）在其上叠加多阶段流程，Δrank=2，方向与台阶性均成立。"},{"id":"pre:luogu/P1536->roj/1389","source":"luogu/P1536","target":"roj/1389","type":"pre","directed":true,"reason":"B 的 M 操作直接复用 A 教的「合并两个不同代表元时同步维护集合大小」这一步（A 用 size 累加、blocks-=1，B 把计数搬到根上做 sz[新根]+=sz[旧根]），只把 A 的输出「blocks-1」换成 Q 查询 sz[find(a)] 并额外叠加按大小合并与坏行 scanf 语义复刻。"},{"id":"pre:luogu/P3366->roj/1386","source":"luogu/P3366","target":"roj/1386","type":"pre","directed":true,"reason":"B 在倒序加点时把 A 教的「用并查集判断两端是否属于不同连通块、不同才合并」直接用作加边判据（main.py 中调用 dsu.unite(i, j)，由并查集内部在两端根不同时才合并并按大小累加 sz），并沿用 A 的 union-by-size 与 sz 累加，只在其上叠加 A 未教的「前缀删点=只看后缀」倒序等价与 f(k) 单调不增的首触发即答案判定。"},{"id":"pre:luogu/P2078->roj/1346","source":"luogu/P2078","target":"roj/1346","type":"pre","directed":true,"reason":"B 复用 A 教的「并查集维护传递性并合并关系」这一步，只是 A 合并后只按根统计集合大小取 min，B 在同样合并的基础上叠加 Q 次 find(代表元) 比较来回答询问"},{"id":"pre:luogu/P3367->roj/1346","source":"luogu/P3367","target":"roj/1346","type":"pre","directed":true,"reason":"B 直接把 A 教的代表元相同即同集合用于每次询问的两次 find 比较，只是把 A 的单次判定改成 Q 次两两询问"},{"id":"pre:HDU/1213->roj/1346","source":"HDU/1213","target":"roj/1346","type":"pre","directed":true,"reason":"B 把 A 教的 find 找根(代表元)与按根相同判断同集合直接用于 Q 次询问，只额外叠加按大小合并与百万级 IO，A 只统计根节点个数而 B 改为多次两两比较"},{"id":"pre:luogu/P1551->roj/1347","source":"luogu/P1551","target":"roj/1347","type":"pre","directed":true,"reason":"B 把 A 教的并查集合并后比较代表元的判定直接用作 unite(u, v) 返回 False（两端已同根）的封圈条件，只是把「是否同组」这一问改挂到「加边是否成圈」上"},{"id":"pre:luogu/P3367->roj/1347","source":"luogu/P3367","target":"roj/1347","type":"pre","directed":true,"reason":"A 作为并查集模板教的 find 与「比较两个根是否相同」，在 B 里直接充当封圈判据：`unite(u, v)` 返回 `False`（两端已同根）即连通性已存在，于是输出步数，否则合并两棵树。"},{"id":"pre:luogu/P8654->roj/1347","source":"luogu/P8654","target":"roj/1347","type":"pre","directed":true,"reason":"A 教的并查集把两点合并到同一集合并判定是否原本已同集合，B 把同一判定直接用作“两端同根即成圈”的封圈条件，只是把 blocks-- 换成输出步数"},{"id":"pre:luogu/P5903->roj/3192","source":"luogu/P5903","target":"roj/3192","type":"pre","directed":true,"reason":"B 的圆方树做法在方点 LCA 分支里用 up_k 沿倍增表上跳，取 x、y 向上跳到方点的两个儿子 ax、ay，正是 A 教的「2^j 级祖先 + 二进制拆分逐位上跳」这一步；A 只做单点 K 级祖先，B 把它嵌进仙人掌缩环建圆方树之后再叠加环内取短弧。"},{"id":"pre:luogu/P8604->roj/3192","source":"luogu/P8604","target":"roj/3192","type":"pre","directed":true,"reason":"B 沿用了 A 教的「点双/环缩成方点、原图点留作圆点」的圆方树建模，只把环加上权值并叠上 LCA 与环内取短"},{"id":"pre:luogu/P8805->roj/3192","source":"luogu/P8805","target":"roj/3192","type":"pre","directed":true,"reason":"A 教的「根到点前缀和 + LCA 拆两点路径容斥」是 B 中 LCA 为圆点情形的实际使用步骤（代码里就是 sd[x]+sd[y]-2*sd[a]），B 只是把它嵌进圆方树并在 LCA 为方点时额外叠加环内取短弧。"},{"id":"pre:luogu/P1656->roj/3195","source":"luogu/P1656","target":"roj/3195","type":"pre","directed":true,"reason":"B 的第一阶段完整复用 A 教的 low[v] > dfn[u] 桥判定（main.py 第 50 行 low[top[0]] > tin[p]）求初始桥，之后才叠加桥树与并查集删路径的新流程"},{"id":"pre:roj/1553->roj/3189","source":"roj/1553","target":"roj/3189","type":"pre","directed":true,"reason":"B 处理每条路径的第一步直接复用 A 教的「求 LCA 后对路径端点打差分、在 LCA 处抵消再子树汇总」这一树上差分步骤，只是把 A 的边差分 diff[w]-=2 换成点差分 lca 与 father(lca) 各 -1，再叠加权值线段树合并与众数查询。"},{"id":"pre:luogu/P3258->roj/3189","source":"luogu/P3258","target":"roj/3189","type":"pre","directed":true,"reason":"B 解法第一步直接复用 A 教的树上点差分四点标记（端点 +1、LCA 与其父亲 -1 后子树求和），再在其上叠加权值线段树合并与众数查询"},{"id":"pre:luogu/P1129->roj/3204","source":"luogu/P1129","target":"roj/3204","type":"pre","directed":true,"reason":"B 在 A 教的匈牙利增广步骤上叠了「极大泥段建模 + 最小点覆盖」这条新流程，但增广时「目标列/横段被占就递归让它原主人换列/换段」这一步是直接复用 A 的模板，A 的复杂度 O(nE) 也与 B 的匈牙利部分同阶"},{"id":"pre:roj/3203->roj/3204","source":"roj/3203","target":"roj/3204","type":"pre","directed":true,"reason":"A 由「每个任务至少被一台机器覆盖 = 每条边至少一端被选中」把答案化为最小点覆盖，B 由「盖住每个泥格 = 每条边至少有一个端点被选中」走同一条路，二者都直接复用「二分图最小点覆盖 = 最大匹配」这条 König 定理把问题转成求最大匹配，B 只是额外叠加了横向/纵向极大泥段建图与手写匈牙利增广。"},{"id":"pre:luogu/P1129->roj/3205","source":"luogu/P1129","target":"roj/3205","type":"pre","directed":true,"reason":"B 求攻击图最大匹配时直接复用了 A 教的匈牙利增广步骤（该右部点被占则沿匹配边回退、让原左部点换搭档），再叠加二染色与 König 定理把最大独立集转成 N*M-T 减最大匹配"},{"id":"pre:roj/3200->roj/3205","source":"roj/3200","target":"roj/3205","type":"pre","directed":true,"reason":"B 直接复用了 A 的 (r+c) 奇偶染色成二分图这一步（代码同为 not (r+c)&1 取左部），在同一网格建图套路上再叠加 König 定理把最大独立集转成最大匹配，故 A 是 B 的前置台阶"},{"id":"pre:roj/2066->roj/3206","source":"roj/2066","target":"roj/3206","type":"pre","directed":true,"reason":"二分图最大匹配（匈牙利算法）的纯模板题；先掌握增广路，再看本题如何拆点建图并落到 N-最大匹配。"},{"id":"pre:roj/3201->roj/3202","source":"roj/3201","target":"roj/3202","type":"pre","directed":true,"reason":"B 的第四步判定复用了 A 教的「把资源分配翻译成二分图匹配」这一具体观察：A 把行/列作两侧、每个可放格作为一条边并指出两两不共行共列即为匹配定义，B 沿用同一等价（左部入侵者各分到一个发射位、右部发射位至多被用一次）把可行性写成是否存在大小为 M 的匹配，再在此外层叠加二分答案与发射位离散化；不过 B 的右部建模与匈牙利实现是新的，属模板级复用。"},{"id":"pre:luogu/P1129->roj/3202","source":"luogu/P1129","target":"roj/3202","type":"pre","directed":true,"reason":"B 的二分可行性判定直接复用 A 教的匈牙利算法与「匹配数是否达到要求规模」这一判定，只是把 A 的 n 行 n 列换成了 M 个入侵者与 NM 个发射位"},{"id":"pre:luogu/P1129->roj/3200","source":"luogu/P1129","target":"roj/3200","type":"pre","directed":true,"reason":"A 教的“列被占就递归让那一行换列”正是沿增广路翻转匹配边这一具体步骤，B 在网格奇偶染色建模后沿分层增广路整体翻转匹配边（Hopcroft-Karp）批量复用该步。"},{"id":"pre:HDU/1116->roj/3196","source":"HDU/1116","target":"roj/3196","type":"pre","directed":true,"reason":"A 教的「有向图欧拉回路判据（所有点入度等于出度且弱连通）」被 B 第二步直接引用，据此断定拆弧后必有回路、故不写可行性判断；B 在此判据之上再叠加链式前向星建图与迭代 Hierholzer 输出回路，属于同一欧拉判据之上的额外构造流程。"},{"id":"pre:shumeng/CSP201509D->roj/3197","source":"shumeng/CSP201509D","target":"roj/3197","type":"pre","directed":true,"reason":"B 直接复用 A 教的「互相可达 ⟺ 同属一个强连通分量、可缩点成 DAG」这一步：A 据此按分量大小累加 C(s,2) 算城市对，B 则在同一缩点 DAG 上统计源分量数 S 与汇分量数 T，再叠加 max(S,T) 加边构造与单分量特判，主要难度在度数结论而不在 SCC 本身。"},{"id":"pre:luogu/P3387->roj/3197","source":"luogu/P3387","target":"roj/3197","type":"pre","directed":true,"reason":"缩点模板题，先掌握如何把强连通分量缩成一个点、在原图和反图上各跑一遍 DFS，再来看本题的度数结论。"},{"id":"common:roj/1515--roj/3197","source":"roj/1515","target":"roj/3197","type":"common","directed":false,"reason":"同模型同公式：两问都是源分量数与 max(源分量数, 汇分量数)，可以直接对照实现。"},{"id":"common:luogu/P2746--roj/3197","source":"luogu/P2746","target":"roj/3197","type":"common","directed":false,"reason":"同源题目（IOI 1996 校园网 / Network of Schools），问题设置与答案公式完全一致。"},{"id":"pre:roj/3514->roj/3182","source":"roj/3514","target":"roj/3182","type":"pre","directed":true,"reason":"B 的 closure() 直接复用 A 教的位集 Floyd 传递闭包（按位或整行并边），只把它改造成逐条加边的增量维护并叠加判环与全序判定"},{"id":"pre:luogu/P2419->roj/3182","source":"luogu/P2419","target":"roj/3182","type":"pre","directed":true,"reason":"B 的增量补闭包直接复用 A 教的位图行位或传递闭包（bits[i] |= bits[k]），只在其上叠加自环判环与可达数互异判全序，A 更基础。"},{"id":"pre:luogu/B3611->roj/3182","source":"luogu/B3611","target":"roj/3182","type":"pre","directed":true,"reason":"B 的闭包补全与判环完全照搬 A 教的整数位集 Warshall 位或（bits[lo] |= 1<<hi 后跑同一循环），只是把该步放进逐条加边循环并额外用 bit_count 判全序"},{"id":"pre:luogu/P1119->roj/3183","source":"luogu/P1119","target":"roj/3183","type":"pre","directed":true,"reason":"B 的「先探测、再并入」直接把 A 教的「逐个加入新中间点做一轮 Floyd 增量松弛」当作第二阶段，并依赖其阶段语义（并入 k 之前 dist 只经过更早的中间点）来定义拆环所需的受限最短路，在此之上再叠加最小环建模、候选环探测与中转点表回溯"},{"id":"pre:roj/3211->roj/3183","source":"roj/3211","target":"roj/3183","type":"pre","directed":true,"reason":"A 拆解的 Floyd 轮次语义（第 k 轮前 dist 只走编号更小的中转点）正是 B「先探测后并入」所依赖的一步，用它把环长写成受限最短路加两条到 k 的边"},{"id":"pre:roj/1197->roj/3178","source":"roj/1197","target":"roj/3178","type":"pre","directed":true,"reason":"B 直接复用 A 的核心建模步骤：先去最近点等价于连续分段，再用中位村 (l+r)//2 作单段代价，B 的 segment_cost 与 dp 转移正是 A 的 cost 与 dp[j][i] 的同一形式，只在上面叠加四边形不等式分治优化，A 的 O(m^3) 转移构成 B 的朴素层，台阶关系明确。"},{"id":"pre:roj/3171->roj/3175","source":"roj/3171","target":"roj/3175","type":"pre","directed":true,"reason":"B 正解第一步（main.py 里 cost = f[i][k] + f[k+1][j] 的转移）直接复用了 A 的关键观察「最后一次聚合必然把连续段劈成左右两半、两半互不影响」；B 在此之上才叠加四边形不等式把断点范围收缩到 [p(i,j-1),p(i+1,j)]，并用 numpy 分段取最小把 O(N^2) 次转移向量化。"},{"id":"pre:roj/1569->roj/3175","source":"roj/1569","target":"roj/3175","type":"pre","directed":true,"reason":"B 正解「第一步：换成枚举断点」与其代码 f[base[rows]+(ks-rows)] + f[base[ks+1]+(end-ks-1)] 用的正是 A 教的同一状态设计：区间 [i,j] 的最后一次合并必在断点 k 拆成 [i,k] 与 [k+1,j] 两段、末次代价是与 k 无关的区间和，B 直接沿用这条断点递推后再叠加四边形不等式把断点范围收窄并向量化。"},{"id":"pre:roj/3145->roj/3175","source":"roj/3145","target":"roj/3175","type":"pre","directed":true,"reason":"本题的前置：同一套区间 DP 递推式，3145 只需 O(N^3)，本题要把它优化到 O(N^2)"},{"id":"pre:luogu/P1775->roj/3171","source":"luogu/P1775","target":"roj/3171","type":"pre","directed":true,"reason":"B 的「能量项链」主解正是复用 A 教的「枚举最后一步切点 k、左右子区间各自最优再合并」这一步区间 DP，在 B 中把断点代价从区间石子质量之和换成 a_i·a_{k+1}·a_{j+1}，再叠加环拆链复制一倍取 N 个窗口最大值，所以 A 是 B 的直接前置台阶。"},{"id":"pre:luogu/P2663->roj/3169","source":"luogu/P2663","target":"roj/3169","type":"pre","directed":true,"reason":"B 把 A 教的「布尔 dp 记录恰好凑出的和、倒序枚举保证每件只用一次」这一步原样搬进朴素可达性 0/1 背包层（每块价值 w 做 dp[s]|=dp[s-w]、s 从大到小扫），再整体换成 Python 大整数位集 reach|=reach<<w，并叠加逐块多重背包拆分与全奇/总价值奇偶剪枝；A 特有的「恰好选 n/2 人」第二维 B 并未使用，故只算模板级复用。"},{"id":"pre:roj/3141->roj/3169","source":"roj/3141","target":"roj/3169","type":"pre","directed":true,"reason":"B 的可性 DP 层就是 A 教的 0/1 背包状态与倒序避免重复选取：A 用 f[j] += f[j-A_i] 数恰好和为 j 的方案，B 把同一转移改成可达性（或运算）并把数组换成位集，再叠加上多重背包拆分与奇偶剪枝。"},{"id":"pre:luogu/P2347->roj/3169","source":"luogu/P2347","target":"roj/3169","type":"pre","directed":true,"reason":"B 的 main.py 逐块执行 A 教的位集可行性转移 dp|=dp<<w（reach|=reach<<w），只是把 A 的“统计可达重量个数”换成“查询第 total/2 位是否可达”并叠加全奇/总价值奇偶剪枝。"},{"id":"pre:luogu/P2627->roj/3157","source":"luogu/P2627","target":"roj/3157","type":"pre","directed":true,"reason":"B 第三类转移把独立项 P_i·j 提到 max 外后，决策量化为 f[i-1][m]-P_i·m，正是复用了 A 教的「窗口单调右移、用单调队列维护窗口最值」这一步，只是把枚举末尾连续长度换成枚举工匠段左端。"},{"id":"pre:roj/3151->roj/3157","source":"roj/3151","target":"roj/3157","type":"pre","directed":true,"reason":"B 复用 A 教的「把两点耦合项拆成单下标表达式、再用单调队列维护滑动窗口最大值」这一步：A 拆出 f(i)=B_i-i 取窗口最大，B 同样拆出 f[i-1][m]-P_i m 交给单调队列。"},{"id":"pre:roj/1599->roj/3157","source":"roj/1599","target":"roj/3157","type":"pre","directed":true,"reason":"B 的第三类转移原样复用 A 教的「拆项把与当前位置无关的部分提出、只留滑动窗口最值」这一步：A 提出 h(j)=dp[j-1]-S_j 后用单调队列维护窗口最大值，B 提出 f[i-1][m]-P_i*m 后用单调队列维护 m∈[j-L_i,S_i-1] 的最大值，过期从队首弹、劣质候选从队尾弹的机制完全一致，B 只在此外叠加按 S 排序与三维状态设计。"},{"id":"pre:leetcodecn/perfect-squares->roj/3659","source":"leetcodecn/perfect-squares","target":"roj/3659","type":"pre","directed":true,"reason":"B 判定面额冗余的那一步正是 A 教过的「物品可无限使用」设定：A 用它写 dp[i]=min(dp[i-j*j]+1) 求凑数最少件数，B 用它做可达集合的无限张闭包（main.py 里按 coin 平移取并），机制相同；B 只是在此之上叠加升序删冗余贪心与位集实现，A 更基础、B 更综合，难度差 1 级符合台阶方向。"},{"id":"pre:luogu/P2347->roj/3659","source":"luogu/P2347","target":"roj/3659","type":"pre","directed":true,"reason":"B 判定面额 v 是否冗余的那一步复用了 A 教的「可达集合编码成位集 + 左移取并」：A 用 dp|=(dp<<value) 加入一枚砝码，B 用同一平移取并把可达位集按 coin 闭包后查 (reachable>>v)&1，只把单枚 0/1 物品换成无限张、并在其上叠加升序删冗余贪心。"},{"id":"pre:roj/3142->roj/3659","source":"roj/3142","target":"roj/3659","type":"pre","directed":true,"reason":"B 判定面额冗余的那一步正是 A 教过的「无限件物品→完全背包正序转移」：A 用它把加数取无限次算进方案数，B 用它做可达性闭包，两者机制相同；B 只是在此之上叠加升序删冗余贪心与位集实现。"},{"id":"pre:luogu/P1419->roj/3647","source":"luogu/P1419","target":"roj/3647","type":"pre","directed":true,"reason":"P1419 的「二分出参数后，把区间和判定转成随 i 单调右移的滑窗最值、再用单调队列取队首」这一步被 3647 的 check(g) 原样复用，只是把判「窗内最小前缀和」换成判「窗内最大 f」以完成可达性转移。"},{"id":"pre:luogu/P2627->roj/3647","source":"luogu/P2627","target":"roj/3647","type":"pre","directed":true,"reason":"B 的 check 判定层把 A 教的「窗口单调右移就用单调队列维护最大值」直接套到 f_i 的转移窗口上，外面再叠一层对金币数 g 的二分"},{"id":"pre:luogu/P1725->roj/3647","source":"luogu/P1725","target":"roj/3647","type":"pre","directed":true,"reason":"B 的 check(g) 线性判定直接复用 A 教的『转移即滑动窗口最大值、用单调队列让队首恒为窗内最大 dp』这一步，只是把 A 的下标窗口 [i-R,i-L] 换成坐标窗口 [x_i-hi,x_i-lo]；A 未教的二分答案与无解判据才是 B 新增的难度。"},{"id":"pre:luogu/P1294->roj/3643","source":"luogu/P1294","target":"roj/3643","type":"pre","directed":true,"reason":"B 的状压 DP 沿用了 A 教的「把集合压成二进制 mask 当状态维度、用按位或并入新元素」这一步：A 用 mask 记已访问点集并做 mask|(1<<v) 转移，B 用 dp[S] 记已消灭小猪集合并按 S|LINE[low][j] 并入；B 另叠加的抛物线预处理与最小活猪定序属模板之上的新流程。"},{"id":"pre:roj/3003->roj/3643","source":"roj/3003","target":"roj/3643","type":"pre","directed":true,"reason":"B 的 DP 直接沿用 A 教的「把离散集合压成整数二进制位、作为状压状态维度」这一步：A 把已走点集压成整数位做 f(s,j)，B 把已消灭小猪集合压成 bitmask 做 dp[S]，并用 S|LINE 并入新集合转移，只在这层模板之上叠加过原点抛物线的 O(n²) 预处理与最小活猪定序。"},{"id":"pre:luogu/P3258->roj/3633","source":"luogu/P3258","target":"roj/3633","type":"pre","directed":true,"reason":"B 的 feasible 判定直接复用 A 教的「用 LCA 定出路径差分标记、再自底向上子树求和」这一步，只把点差分改成 LCA 处 -2 的边差分，以统计每条边被多少条超标计划覆盖"},{"id":"pre:luogu/P3128->roj/3633","source":"luogu/P3128","target":"roj/3633","type":"pre","directed":true,"reason":"B 的可行判定直接复用 A 教的树上差分端点标记加自底向上子树汇总，只是 A 汇总点覆盖次数、B 汇总每条边的覆盖数来找出超标计划的公共边"},{"id":"pre:luogu/P9246->roj/3633","source":"luogu/P9246","target":"roj/3633","type":"pre","directed":true,"reason":"P9246 教会的“端点+1、LCA−2 再自底向上求和得到每条边被路径覆盖的次数”被 3633 原样用于统计超标计划的公共边"},{"id":"pre:luogu/P1103->roj/3617","source":"luogu/P1103","target":"roj/3617","type":"pre","directed":true,"reason":"B 固定行组合后把 A 的“已选个数+末位、枚举更小前驱”选位顺序 DP 原样套到列维上（dp[k][j]=v[j]+min_{i<j}(dp[k-1][i]+h[i][j])，代码 cur[j]=v[j]+min(dp[i]+h[..])），只是把单步代价 abs(w[j]-w[p]) 换成 v[j]+h[i][j]。"},{"id":"pre:luogu/P5149->roj/3609","source":"luogu/P5149","target":"roj/3609","type":"pre","directed":true,"reason":"B 第三步直接复用 A 教的「把待处理的每个元素逐项换成它在参考序列中的位置编号，再对这个位置排列数逆序对得到答案」这一步：A 把名字换成原座位下标，B 把第2列按排名换成第1列同排名伙伴的位置 p（代码 pos_of_rank 查表得 order），只是把归并排序换成树状数组，并在其前叠加排序不等式定目标配对、三角不等式说明只重排一列等额外推导。"},{"id":"pre:luogu/P4779->roj/3597","source":"luogu/P4779","target":"roj/3597","type":"pre","directed":true,"reason":"B 直接复用 A 教的堆优化 Dijkstra 作核心引擎，只把状态从单点扩成 (国家, 已学文化集合)，额外叠加状压建模与 A* 剪枝。"},{"id":"pre:roj/1382->roj/3597","source":"roj/1382","target":"roj/3597","type":"pre","directed":true,"reason":"B 的核心引擎复用了 A 教的堆优化 Dijkstra 与「首次弹出即定型、命中终点提前返回」这一步：A 用来定型单个点，B 把它搬到 (国家,已学文化集合) 状态图上，首次弹出 u=T 的状态即得答案，再叠加状压建模、A* 启发与 -1 预检。"},{"id":"pre:luogu/P4822->roj/3597","source":"luogu/P4822","target":"roj/3597","type":"pre","directed":true,"reason":"B 复用 A 教的“把影响后续的历史维度加进状态、在状态图上跑 Dijkstra”这一步：A 把已用卡数写进 dist[u][used]，B 把已学文化集合写进 (u,mask)，只是把计数维度换成指数位集，再叠加 A 未教的 A* 启发与 -1 预检。"},{"id":"pre:luogu/P5903->roj/3600","source":"luogu/P5903","target":"roj/3600","type":"pre","directed":true,"reason":"B 的按轮倍增表与求 k 级祖先完全同构：把 A 的 up[x][j]=up[up[x][j-1]][j-1] 前半段落点拼后半段换成 g[k][i]=g[k-1][g[k-1][i]]，同一步再用于二进制拆分跳跃；差异只是 B 的跳跃代价带权并需从高到低贪心。"},{"id":"pre:luogu/P7167->roj/3600","source":"luogu/P7167","target":"roj/3600","type":"pre","directed":true,"reason":"B 的 drive 把 A 教的自大到小 2^k 段跳判定搬来贪心跳整轮，界从点权容量和换成整轮里程和，递推与单调性前提一致"},{"id":"pre:luogu/P2234->roj/3600","source":"luogu/P2234","target":"roj/3600","type":"pre","directed":true,"reason":"B 的 nearest_two 直接复用 A 教的「坐标压缩 + Fenwick 维护排名计数 + kth 二进制倍增查前驱/后继」：A 用它求最近值，B 把两侧各扩到次前驱、次后继凑出 4 个候选，再叠加倍增表与预算贪心跳跃作为新流程。"},{"id":"pre:luogu/P1873->roj/3592","source":"luogu/P1873","target":"roj/3592","type":"pre","directed":true,"reason":"B 复用了 A 教的「先证阈值越大可行量越少的单调性、再在单调谓词上二分找边界」这一步：A 用它找最后一个 true 得到最大锯片高度，B 把它套到 y(W) 上找第一个 P(W)=(y(W)≤S) 的分界点 W，只额外叠加前缀和把单次求值从 O(n) 区间统计压到 O(n+m)、以及在分界点两侧取最小差值这两个 A 未教的流程。"},{"id":"pre:leetcodecn/n-queens->roj/3575","source":"leetcodecn/n-queens","target":"roj/3575","type":"pre","directed":true,"reason":"B 的 DFS 骨架直接沿用 A 教的「用标记数组实时记录已占用约束、放置前检查、递归后恢复」这一步：把 A 的 col/diag1/diag2 换成行/列/宫三个位掩码，放置 |= 、撤销 ^=，只是在其上叠加 MRV 选格与逐层真实上界分支定界来最大化靶形总分。"},{"id":"common:roj/3076--roj/3575","source":"roj/3076","target":"roj/3575","type":"common","directed":false,"reason":"同一张盘面：先求任意一个合法填法，本题再升级为在解集合上最大化靶形加权总分"},{"id":"pre:luogu/P2627->roj/3571","source":"luogu/P2627","target":"roj/3571","type":"pre","directed":true,"reason":"B 的转移 dp[j]=P_c[j]+max(dp[k]-P_c[k]-cost) 正是 A 教的「常数项 + 窗口最大值、用单调队列维护断点」这一步的直接复用，只是把单条窗口扩成 n 条对角线各配一个队列，并额外叠加了 A 未涉及的对角线前缀和观察。"},{"id":"pre:luogu/P1725->roj/3571","source":"luogu/P1725","target":"roj/3571","type":"pre","directed":true,"reason":"A 教的「区间最值转移 = 滑动窗口最大值 → 单调队列存 (下标, dp 值)、弹队尾更差者、队头过期出队、队头即窗口最值」被 B 逐字复用：B 的 main.py 对每条对角线 c 的队列存 (k, dp[k]-P_c[k]-cost)、while q[-1][1]<=value 弹尾、q[0][0]<j-p 弹头，队头参与 dp[j]=P_c[j]+q[0][1]，再在此之上叠加时间轴分段模型与对角线前缀和这两个 A 没有的新步骤。"},{"id":"pre:roj/1266->roj/3549","source":"roj/1266","target":"roj/3549","type":"pre","directed":true,"reason":"B 把 A 教的「组内枚举选项、组间互斥」的分组背包转移整套搬到家族上：A 的第 i 组枚举 k 台对应 B 的组内枚举买法，B 的 dp[j]=max(dp[j],dp[j-cost]+gain) 就是这条转移的一维倒序版，B 额外叠加的流程（先在家族内部枚举买法消掉主件-附件依赖）是 A 没教的、也是它更难的部分。"},{"id":"pre:luogu/P1734->roj/3549","source":"luogu/P1734","target":"roj/3549","type":"pre","directed":true,"reason":"B 的一维 knapsack 沿用 A 教的「倒序枚举容量保证每个数字只参与一轮转移」这一具体判定，只是把单个物品换成家族买法选项，使同组方案不被叠加、每组至多选一个；B 在此台阶上再叠加家族内枚举消依赖的新流程。"},{"id":"pre:roj/3141->roj/3549","source":"roj/3141","target":"roj/3549","type":"pre","directed":true,"reason":"A 讲透的一维滚动倒序枚举（使 dp[j-cost] 仍是上一轮旧值、每件物品至多选一次）被 B 的分组背包原样复用：B 用同一个倒序容量循环保证同一家族的两个选项不会被叠加、每组至多选一种买法。"},{"id":"pre:luogu/P1775->roj/3548","source":"luogu/P1775","target":"roj/3548","type":"pre","directed":true,"reason":"A 教的『最后一次合并必在断点 k 处把区间劈成两半』正是 B 的 f(i,j)=max{f(i,k)+f(k+1,j)+代价} 这一步，B 只是把代价换成 a_i*a_(k+1)*a_(j+1)，再叠加环拆链复制一倍、对 N 个长度 N 窗口取最大。"},{"id":"pre:luogu/P2858->roj/3526","source":"luogu/P2858","target":"roj/3526","type":"pre","directed":true,"reason":"B 的加分二叉树直接沿用 A 教的「连续区间即完整状态 + 按区间长度从小到大递推」这一步（main.py 的 best(l,r) 递归先算短区间），只是把 A 的两端取数转移换成枚举根 k 拼左右子段，并叠加记录最优根还原前序这一新流程，属模板级复用。"},{"id":"pre:luogu/P1775->roj/3526","source":"luogu/P1775","target":"roj/3526","type":"pre","directed":true,"reason":"B 的前序还原解法直接复用 A 教的区间 DP 步骤：同样把连续区间 [l,r] 当作状态、枚举断点 k 把区间劈成左右两段独立子问题，代码 `best(l,r)` 里 `for k in range(l, r+1)` 分成 [l,k-1]/[k+1,r] 与 A 的 dp[l][k]+dp[k+1][r] 是同一状态设计与断点枚举，B 只是在其上叠加“记录最优根并栈式还原前序遍历”这一层新流程。"},{"id":"pre:luogu/P2935->roj/3511","source":"luogu/P2935","target":"roj/3511","type":"pre","directed":true,"reason":"A 教的关键换法是「不要枚举候选点，而是把一组特定点（喜欢牧场）当作最短路源点来统计距离」；B 第三步按同一换法把「出发机场任选」写成把 A 城 4 个机场距离初值置 0 一起入堆的多源 Dijkstra（首次弹出 B 城机场即答案），并沿用 A 的堆优化 Dijkstra 骨架，只在其外叠加补矩形第 4 点与机场粒度建图。"},{"id":"pre:luogu/P1775->roj/3501","source":"luogu/P1775","target":"roj/3501","type":"pre","directed":true,"reason":"B 的转移 (dp[j][i]=max(dp[j-1][t]*num(t+1,i))) 直接复用 A 教的关键观察——枚举最后一次操作的位置、把序列切成前缀子问题加末段：A 枚举最后一次合并的断点 k 得区间 dp，B 枚举最后一个乘号的位置 t 得分层 dp[j][i]，只把区间状态换成(乘号数,前缀长度)并把区间和改成大整数乘法。"},{"id":"pre:roj/2066->roj/3221","source":"roj/2066","target":"roj/3221","type":"pre","directed":true,"reason":"二分图最大匹配（匈牙利算法）的模板题；先掌握增广路，再回到本题看它如何被套进拆点建图。"},{"id":"pre:luogu/P1129->roj/3219","source":"luogu/P1129","target":"roj/3219","type":"pre","directed":true,"reason":"B 沿用 A 的行列二分图建模，只是先用墙壁把行列切成横向段与纵向段，再对同一「格连边求最大匹配」判定套用 Hopcroft-Karp"},{"id":"pre:roj/3201->roj/3219","source":"roj/3201","target":"roj/3219","type":"pre","directed":true,"reason":"B 直接复用 A 的建模步骤：把「行/列」换成被墙切出的横向段/纵向段后，仍然是两侧顶点—格为边—不冲突等价匹配、答案即最大匹配，新增的只是切段（草地不切断）这一前置处理。"},{"id":"pre:roj/3200->roj/3219","source":"roj/3200","target":"roj/3219","type":"pre","directed":true,"reason":"A 教的 BFS 分层加分层方向批量增广直接成为 B 求段-段二分图最大匹配的求解步骤，B 只替换了建图方式"},{"id":"pre:HDU/1878->roj/3217","source":"HDU/1878","target":"roj/3217","type":"pre","directed":true,"reason":"A 教的欧拉回路存在判定（图连通+无奇度点）被 B 第三步原样拿来判 de Bruijn 图，改写成有向图版本「每点两条出边两条入边且任意点可达」后得出欧拉回路存在，再叠加 A 未教的窗口建模与 Hierholzer 构造。"},{"id":"pre:POJ/2976->roj/3215","source":"POJ/2976","target":"roj/3215","type":"pre","directed":true,"reason":"B 的判定层直接复用 A 的分数规划变形：都是把「比值 ≥ x」乘分母整理为 Σ(权 - x·计数) ≥ 0 的线性判定，只是 A 把选项求和后用贪心取最大的 n-k 项验证、B 在缩点图上用 SPFA 判正环验证，并额外叠加四舍五入整数化与 ≥0 到严格正环的放大等价变形。"},{"id":"pre:luogu/P3385->roj/3215","source":"luogu/P3385","target":"roj/3215","type":"pre","directed":true,"reason":"B 的 has_cycle_avg_ge 直接复用 A 教的关键判定步骤——松弛成功时记录前驱链边数、达到 n 即认定存在环（B 只把负环对偶成正环，初值全 0 当超级源点），并在此之上叠加 01 分数规划二分与 ≥0→严格正环的放大变换。"},{"id":"pre:luogu/P4779->roj/3209","source":"luogu/P4779","target":"roj/3209","type":"pre","directed":true,"reason":"B 的两遍最短路（第一遍求 base、第二遍分层计数）都直接复用了 A 教的堆优化 Dijkstra：同样是堆里存“(权值, 结点)”按升序弹出最小状态并沿边松弛，A 用于单次单源最短路，B 把它套在拆点后的 2N 状态图上分发 ways，再叠加超额 0/1 建模与答案相加。"},{"id":"pre:roj/1382->roj/3209","source":"roj/1382","target":"roj/3209","type":"pre","directed":true,"reason":"B 的 main.py 先跑一遍与 A 同款的堆优化 Dijkstra（shortest_from 里保留 `if d > dist[u]: continue` 的惰性删除）求出 base 数组，再用「首次弹出/到达即定型」这一条把每个状态 (v,k) 的总长与条数一次定死并向后继分发；A 只教单源最短路模板，B 在其上叠加超额只有 0/1 的证明、两层状态建模与 ways 条数累加。"},{"id":"pre:luogu/P5903->roj/3155","source":"luogu/P5903","target":"roj/3155","type":"pre","directed":true,"reason":"B 的确定性转移表沿用 A 的倍增表自拼方式（up[k]=up[k-1] 在状态上自拼），再用二进制拆分挑选，只是把「跳 k 步」换成「走 2^k 份 s2」，新增难度在贪心压缩与占用份数判据"},{"id":"pre:luogu/P7167->roj/3155","source":"luogu/P7167","target":"roj/3155","type":"pre","directed":true,"reason":"A 教的倍增表自拼（up[k]=up[k-1]∘up[k-1]）加「从最大 k 往小试、能跳就跳」的二进制挑选，被 B 原样搬到「走 2^k 份 s2」的确定性转移表上，只是把跳跃状态换成（消耗份数,落点偏移）。"},{"id":"pre:luogu/P1896->roj/3153","source":"luogu/P1896","target":"roj/3153","type":"pre","directed":true,"reason":"B 的炮兵阵地直接复用 A 教的「逐行状压 + 位运算判行间冲突」这一步：先像 A 一样预计算出行内合法掩码，再把 A 的 cur & prev 扩写成 z & (x|y)，只是把状态记忆从一行扩到两行并叠加地形过滤与取最大值。"},{"id":"pre:roj/1593->roj/3153","source":"roj/1593","target":"roj/3153","type":"pre","directed":true,"reason":"B 直接沿用 A 的「按行推进 + 位运算筛行内合法掩码 + 用该行地形掩码过滤可选方案」这一步，只把行内条件加上列距 2，并把跨行状态从一行扩到两行。"},{"id":"pre:luogu/P1879->roj/3153","source":"luogu/P1879","target":"roj/3153","type":"pre","directed":true,"reason":"B 逐行转移时把行内合法性提前枚举掉，用的正是 A 教的同一列掩码位与左移判相邻技巧（s & (s<<1)），只是把左右各 1 格扩成左右各 2 格，并叠加了地形过滤与最近两行的状态设计。"},{"id":"pre:luogu/P1725->roj/3151","source":"luogu/P1725","target":"roj/3151","type":"pre","directed":true,"reason":"B 在断环成链、把距离拆成 f(i)+g(j) 之后，直接沿用 A 教的单调队列：队尾弹出不更优者、队首弹出过期者，队首即窗口内最大值的左端点，只是窗口从 [i-R,i-L] 换成 [j-K,j-1]。"},{"id":"pre:luogu/P2627->roj/3151","source":"luogu/P2627","target":"roj/3151","type":"pre","directed":true,"reason":"B 的第三步直接复用 A 教的“窗口两端单调右移→单调队列取窗口最大值”，只是把窗口值的下标范围从 i-k 换成 j-K"},{"id":"pre:roj/1593->roj/3152","source":"roj/1593","target":"roj/3152","type":"pre","directed":true,"reason":"A 教的是「按行推进 + 行掩码状态、行间只靠两掩码的按位关系传递约束」，B 的阶段与状态设计直接沿用这一骨架（f_{j,s} 与代码 g(j,s) 都是行编号加 N 位掩码，并用 s & b 做列冲突判定），只是把掩码从「本行种植」改成「竖牌伸出列」，再叠加行内逐格 dfs 铺牌这一额外流程。"},{"id":"pre:luogu/P1879->roj/3152","source":"luogu/P1879","target":"roj/3152","type":"pre","directed":true,"reason":"B 的轮廓线 DP 直接复用 A 立的「按行划分阶段 + 把列信息压成一个二进制 mask 作状态」这一步（main.py 里 g(j,s) 用 `s & b` 判位、`cur | b` 记伸出列），只是把 mask 的语义从「本行种草格子」换成「下一行被竖牌占住的列」，再叠加行内逐格 dfs 的竖牌登记与横牌右邻判定这个新流程。"},{"id":"pre:roj/3141->roj/3144","source":"roj/3141","target":"roj/3144","type":"pre","directed":true,"reason":"A 教的「一维 01 背包倒序枚举保证每件物品至多取一次」正是 B 暴力层可行性 0/1 背包的原样写法（正解对二进制拆出的每一组仍沿用同一 0/1 语义），B 只在它之上叠加二进制拆分与大整数 bitset 推进。"},{"id":"pre:leetcodecn/longest-common-subsequence->roj/3135","source":"leetcodecn/longest-common-subsequence","target":"roj/3135","type":"pre","directed":true,"reason":"B 的朴素做法从 A 教的 LCS 状态 f(i,j) 起手，把「a[i-1]==b[j-1] 命中配对就取前驱加一」这一步保留为 $1+H(i-1,j-1,a_i)$，并在其上加「上升」维度；化简后用 best 收集可接前驱，命中时执行 dp[j]=best+1。"},{"id":"pre:roj/3015->roj/3130","source":"roj/3015","target":"roj/3130","type":"pre","directed":true,"reason":"B 的 CDQ 分治在递归处理完左右两半后，正是复用 A 教的「先让左右两半各自有序，再归并合并」这一归并骨架（solve 末尾 ev[l:r]=tmp 处按 x 归并），只是把 A 在跨半归并中数逆序对换成树状数组按 y 前缀和加容斥统计左半修改对右半询问的贡献，额外叠加的是 CDQ 时间维与三维偏序这一 A 未教的新流程。"},{"id":"pre:luogu/P1884->roj/3125","source":"luogu/P1884","target":"roj/3125","type":"pre","directed":true,"reason":"B 的水平边积分直接复用 A 教的关键观察「相邻事件之间活跃集合不变」：两者都靠条带内截面/覆盖状态不变而无需关心同坐标事件顺序，B 只是把它从面积增量改写成 2×极大区间个数×条带宽，再叠加线段树维护极大区间个数这一新数据结构的流程。"},{"id":"pre:HDU/1199->roj/3125","source":"HDU/1199","target":"roj/3125","type":"pre","directed":true,"reason":"B 的建树直接复用 A 教的「区间转点离散化」：把端点坐标排序去重、相邻两点构成基本/元区间，并把闭区间 [lo,hi] 映射为叶子下标 [idx(lo), idx(hi)-1]（A 是 [idx(a), idx(b+1)-1]，B 是 [index(y1), index(y2)-1]），只是 A 用于单个球号的染色数轴、B 用于扫描线的 y 截面，再叠加 cover/seg/left_on/right_on 维护极大区间个数与条带积分求周长。"},{"id":"pre:luogu/P5250->roj/3119","source":"luogu/P5250","target":"roj/3119","type":"pre","directed":true,"reason":"B 直接复用 A 的“前驱=严格小于 x 的个数对应的第 k 小”，把它写成 value_at(count_below(bisect_left(coords,x)))，再由该 kth 原语拼出排名、第 k 小与后继。"},{"id":"pre:luogu/P2234->roj/3119","source":"luogu/P2234","target":"roj/3119","type":"pre","directed":true,"reason":"B 的普通平衡树前驱/后继直接复用 A 教的「前缀和得 less，再用 Fenwick kth 取前驱后继」，只是在其上补齐删除、排名与第 k 小。"},{"id":"pre:luogu/P2307->roj/3122","source":"luogu/P2307","target":"roj/3122","type":"pre","directed":true,"reason":"B 的扩展域并查集在合并 a~b+n、b~a+n 之前，仍用 A 教的那一步「先 find 两端点比较根，相同即冲突」来检查 find(a)=find(b)，把「同集合=有环」换成「同集合=被迫同狱」，从而无需二分、一次降序扫描即可定位答案。"},{"id":"pre:luogu/P1892->roj/3122","source":"luogu/P1892","target":"roj/3122","type":"pre","directed":true,"reason":"B 直接沿用 A 教的对立侧合并 a~b+n 与 b~a+n 来维护分居约束，只是额外按怨气值降序扫描并用 find(a)=find(b) 判定第一条被迫同狱的边"},{"id":"pre:luogu/P4552->roj/3113","source":"luogu/P4552","target":"roj/3113","type":"pre","directed":true,"reason":"B 的修改操作直接复用 A 教的「区间加在差分数组上只影响两个边界点」这一步（代码 update(l,d)、update(r+1,-d)），先把区间加塌缩成两次单点改，再叠加 A 未教的 gcd(x,y)=gcd(x,y-x) 与线段树/树状数组维护。"},{"id":"pre:roj/3011->roj/3113","source":"roj/3011","target":"roj/3113","type":"pre","directed":true,"reason":"B 的突破口把 A 教过的「区间加塌缩成 b_l 与 b_{r+1} 两个差分单点改」直接用作区间加的处理手段，再由线段树+树状数组维护其中的单点改与前缀和。"},{"id":"pre:roj/3110->roj/3113","source":"roj/3110","target":"roj/3113","type":"pre","directed":true,"reason":"B 的 update(l,d)/update(r+1,-d) 与树状数组前缀和求 A_l 正是复用 A 教的「区间加在差分上只改 d_l、d_{r+1} 两个点 + 树状数组单点改/前缀和」，再在其上叠加 A 未教的 gcd(x,y)=gcd(x,y-x) 与线段树维护区间 gcd。"},{"id":"pre:HDU/1199->roj/3114","source":"HDU/1199","target":"roj/3114","type":"pre","directed":true,"reason":"B 的扫描线线段树直接复用 A 教的「叶子节点代表两个离散化点之间的基本段、右端点映射后减 1」这一关键建树步骤（代码里 cover(1,0,seg_cnt-1) 与 y_rank[y2]-1），再在其上叠加竖边事件、cover_cnt 不下传与面积增量累加。"},{"id":"pre:HDU/3634->roj/3114","source":"HDU/3634","target":"roj/3114","type":"pre","directed":true,"reason":"A 用矩形边界切分平面得出块内不存在部分覆盖的结论，B 的竖条分解直接复用该结论，再在其上叠加扫描线与线段树增量维护覆盖长度。"},{"id":"pre:luogu/P1884->roj/3114","source":"luogu/P1884","target":"roj/3114","type":"pre","directed":true,"reason":"A 的『把 y 离散化成段、每段桶 ±1 计数求覆盖长度』正是 B 中『在段上做区间加 ±1』的朴素版，B 只是把逐桶遍历升级为线段树维护。"},{"id":"pre:luogu/P1637->roj/3109","source":"luogu/P1637","target":"roj/3109","type":"pre","directed":true,"reason":"A 教的固定中间点拆分与树状数组统计左侧比它小的个数，被 B 正解原样复用为 query(y_j-1) 求 L<，B 只是再叠上互补公式与反转序列凑出 V/∧ 两组计数。"},{"id":"pre:roj/1539->roj/3110","source":"roj/1539","target":"roj/3110","type":"pre","directed":true,"reason":"B 直接复用 A 的差分观察，把区间翻转换成区间加后同样只在 d[l] 与 d[r+1] 两个单点改动，再叠加树状数组维护前缀和"},{"id":"pre:luogu/P8687->roj/20011","source":"luogu/P8687","target":"roj/20011","type":"pre","directed":true,"reason":"A 教的“围绕已覆盖口味而不是选哪些包设计状态、用位或合并掩码”被 B 原样搬到“不记录选了哪些人，改记录菜的覆盖情况”（代码里就是 a | mask），B 只是在此之上叠加覆盖数封顶到 2 与双 bitmask 的 0/1 背包流程。"},{"id":"pre:luogu/U661986->roj/20008","source":"luogu/U661986","target":"roj/20008","type":"pre","directed":true,"reason":"B 先把唱K化成容量 t-1 的 0/1 背包，其第三步「每首歌要么选要么不选、从大到小逆序更新」正是 A 教的「倒序枚举容量保证每件物品只选一次」这一步，只是把 dp 的最大价值换成 f 的最多首数并叠加首数优先的字典序比较。"},{"id":"pre:roj/1294->roj/20008","source":"roj/1294","target":"roj/20008","type":"pre","directed":true,"reason":"B 的选歌环节直接套用 A 教的倒序枚举容量的一维 0-1 背包骨架，把「每件物品至多选一次」迁移成「每首歌至多唱一次」，只是在同一转移上把状态由最大价值换成恰好总长 s 的最多首数，再叠加首数优先的字典序比较。"},{"id":"pre:roj/1319->roj/20005","source":"roj/1319","target":"roj/20005","type":"pre","directed":true,"reason":"B 复用了 A 的相邻逆序对交换论证：先把交换差值归到邻对本身、证明交换严格变优，再据此确定排序键，只是把 A 的『按接水时间升序』换成按比值 t_i/c_i 升序。"},{"id":"pre:roj/1151->roj/3668","source":"roj/1151","target":"roj/3668","type":"pre","directed":true,"reason":"A 教的埃氏筛「枚举筛子、标记其全部倍数」这一步被 B 直接搬用：把筛子从素数换成含 7 的数，只是把倍数改标为禁报，B 原文也点明与埃氏筛结构一模一样"},{"id":"pre:roj/1067->roj/3665","source":"roj/1067","target":"roj/3665","type":"pre","directed":true,"reason":"B 的「入桶」步骤正是复用 A 教的桶计数：当值域极小时按值维护计数、扫一遍完成统计，B 把它从三个固定计数器推广为 601 档数组"},{"id":"pre:roj/1115->roj/3665","source":"roj/1115","target":"roj/3665","type":"pre","directed":true,"reason":"B 的优化第一步直接复用 A 教的「值当下标开桶、读入时桶加一」计数手法，再在其上叠加从高分档向下累加求第 k 名的新流程"},{"id":"pre:shumeng/CSP202104A->roj/3665","source":"shumeng/CSP202104A","target":"roj/3665","type":"pre","directed":true,"reason":"B 的入桶步骤直接复用了 A 教的按值分桶 count[x]++：把灰度值换成 0..600 的分数，累计频次后再从高分档向下累加定位第 k 名，从而在线求出分数线。"},{"id":"common:luogu/P7072--roj/3665","source":"luogu/P7072","target":"roj/3665","type":"common","directed":false,"reason":"洛谷上的同一道题（CSP-J 2020），其题解改用树状数组做第 k 小查询，可与本文的桶扫描对照"},{"id":"pre:luogu/P1614->roj/3636","source":"luogu/P1614","target":"roj/3636","type":"pre","directed":true,"reason":"B 复用 A 教的“相邻窗口只差新进旧出”观察，把窗口和的滚动更新迁移成 deque 整船进出与国籍计数的两端增量维护，从而避免每轮重建窗口"},{"id":"common:luogu/P2952--roj/3636","source":"luogu/P2952","target":"roj/3636","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：A 教的「操作只发生在序列两端 ⇒ 用 deque 做头尾增减（含队首 pop_front 出队）」正是 B 窗口滑动的实际一步：main.py 把窗口内的船 append 进 deque，过期船从队首 popleft 整船移出；B 在此之上叠加了 24 小时开区间出窗判定、字典"},{"id":"pre:roj/1083->roj/3608","source":"roj/1083","target":"roj/3608","type":"pre","directed":true,"reason":"B 直接复用 A 教的取模快速幂：A 用边乘边取模求 a^b mod 7，B 用同一算法求 10^k mod n 再算 (x+m·10^k) mod n，只是多加了环上位移可叠加的观察"},{"id":"pre:noi_openjudge/ch0107-09->roj/3598","source":"noi_openjudge/ch0107-09","target":"roj/3598","type":"pre","directed":true,"reason":"A 教的「字母平移后字母表回绕」这一步被 B 的解密公式逐位复用，只是把固定 +1 换成密钥位偏移并补上 i % len(key) 密钥循环与大小写跟随密文"},{"id":"pre:luogu/P1914->roj/3598","source":"luogu/P1914","target":"roj/3598","type":"pre","directed":true,"reason":"B 把 A 教的字母数值化后取模 26 平移，直接扩成逐位 (c − k) % 26 的逆运算，并叠加 i % len(key) 密钥循环与大小写保持"},{"id":"pre:noi_openjudge/ch0107-10->roj/3598","source":"noi_openjudge/ch0107-10","target":"roj/3598","type":"pre","directed":true,"reason":"B 把 A 教的“字母转下标后减去偏移、%26 再转回字符”直接用作逐位解密，只是把固定偏移 5 换成密钥字母并叠加 i%len(key) 循环。"},{"id":"pre:luogu/P8218->roj/3595","source":"luogu/P8218","target":"roj/3595","type":"pre","directed":true,"reason":"B 的优化二复用 A 教的前缀和区间计数：pre 给出 [0,j) 的楼梯数，before = S - pre[start] 正是一次区间相减，由此 O(1) 定位第 k 个楼梯；B 在此基础上叠加取周期与两段回绕分支。"},{"id":"pre:roj/20018->roj/3595","source":"roj/20018","target":"roj/3595","type":"pre","directed":true,"reason":"A 教的把超大下标对总长取模折回单周期，被 B 迁移为把每层第 x 个楼梯对楼梯总数 S 取模消去整圈，两者都是先消周期再在单周期内定位。"},{"id":"pre:roj/8005->roj/3596","source":"roj/8005","target":"roj/3596","type":"pre","directed":true,"reason":"B 的窗口和转移整行套用了 A 的前缀和消除求和：先对上一阶段整行做前缀和，再用 $S[j]-S[j-a_i-1]$ 一次减法代替逐项枚举，把 $O(a_i)$ 压成 $O(1)$"},{"id":"pre:luogu/P1449->roj/3587","source":"luogu/P1449","target":"roj/3587","type":"pre","directed":true,"reason":"A 教的后缀求值「遇到运算符弹出栈顶两个操作数、算完把结果压回」被 B 的调度场在结算运算符时照搬为 reduce_top 的弹两操作数合并压回（先弹右再弹左），只是把数值换成 (c0,c1) 计数状态，再叠加 A 未教的优先级/括号解析。"},{"id":"pre:roj/1358->roj/3587","source":"roj/1358","target":"roj/3587","type":"pre","directed":true,"reason":"B 的调度场双栈直接复用 A 教的运算符栈按优先级弹栈与括号边界隔离，只把弹出后的求值换成 (c0,c1) 计数合并。"},{"id":"pre:roj/1356->roj/3587","source":"roj/1356","target":"roj/3587","type":"pre","directed":true,"reason":"B 的 parsing 层直接复用 A 教的调度场优先级弹栈规则（栈顶优先级不低于当前就先弹出、括号作栈边界），把 A 的操作数求值栈换成 (c0,c1) 计数栈自底向上合并，额外叠加的是计数 DP 合并公式与隐式横线补位。"},{"id":"pre:luogu/P9244->roj/3589","source":"luogu/P9244","target":"roj/3589","type":"pre","directed":true,"reason":"B 沿用 A 教的「固定右端点后按右端点归类计数」，把它从单变量 cnt 换成 valid/total 同色计数，只是把合法左端点的单调边界由 j-K+1 改为最近低价店 L(j)，并叠加 pending 摊还补记。"},{"id":"pre:noi_openjudge/ch0103-13->roj/3584","source":"noi_openjudge/ch0103-13","target":"roj/3584","type":"pre","directed":true,"reason":"A 教的关键观察是把数字留在字符串域、用切片 [::-1] 倒序而不是拆位取模；B 的正解正是在这一步上叠加摘符号、lstrip('0') 去前导零与空串兜底，把范围从固定三位数扩到含负数与末尾零的任意整数。"},{"id":"pre:roj/20001->roj/3584","source":"roj/20001","target":"roj/3584","type":"pre","directed":true,"reason":"B 的字符串域解法直接复用 A 教的关键观察与表达式——反转后前导 0 就是原串末尾 0、用切片反转 + lstrip('0') 删净（另有空串兜底），只是在外面叠加了 abs 摘符号与 sign 拼回负号这一层。"},{"id":"pre:noi_openjudge/ch0110-03->roj/3569","source":"noi_openjudge/ch0110-03","target":"roj/3569","type":"pre","directed":true,"reason":"A 教的「排序键写成 (-分数, 姓名)，负号把降序变升序」被 B 逐个迁移为按 (-成绩, 报名号) 排序，再用这一次排序同时供取分数线与输出"},{"id":"pre:roj/1176->roj/3569","source":"roj/1176","target":"roj/3569","type":"pre","directed":true,"reason":"B 复用了 A 教的「按成绩降序排序后按下标取第 k 名」这一步，把第 k 名换成第 ⌊1.5m⌋ 名的分数当分数线，再叠加 s>=L 的前缀过滤处理重分扩招。"},{"id":"pre:noi_openjudge/ch0110-01->roj/3569","source":"noi_openjudge/ch0110-01","target":"roj/3569","type":"pre","directed":true,"reason":"B 在取分数线时原样套用 A 教的「排序后名次转下标取元素」，代码里即 ranked[r-1][1]，再叠加同分报名号排序键与 >=L 过滤"},{"id":"pre:leetcodecn/climbing-stairs->roj/3562","source":"leetcodecn/climbing-stairs","target":"roj/3562","type":"pre","directed":true,"reason":"A 教的“某格方法数等于其前驱邻格方法数之和对并定初值”被 B 直接用作传球 DP：把 i-1、i-2 换成环上 (j-1)%n、(j+1)%n 即可"},{"id":"pre:luogu/P1255->roj/3562","source":"luogu/P1255","target":"roj/3562","type":"pre","directed":true,"reason":"B 直接复用 A 教的「最后一步只有两种来源就相加」计数模板：把 dp[i]=dp[i-1]+dp[i-2] 换成环上左右邻居两个前驱相加，仅额外套一层 (j±1)%n 绕回，难度落在 A 之上两档。"},{"id":"pre:leetcodecn/pascals-triangle->roj/3562","source":"leetcodecn/pascals-triangle","target":"roj/3562","type":"pre","directed":true,"reason":"B 的 DP 逐层计数直接沿用了 A 教的「本层每格 = 上一层两个相邻格子相加」这一步二维逐层递推，只把上一层下标加取模改成左右邻居并补了边界与环的处理"},{"id":"pre:noi_openjudge/ch0107-02->roj/3556","source":"noi_openjudge/ch0107-02","target":"roj/3556","type":"pre","directed":true,"reason":"B 正解的第一步就是 A 教的 Counter 频率统计：用 Counter(data) 一遍扫描把序列压成值到次数，再在此外层叠加 sorted(counter) 对键升序并逐键输出，A 的入门题只到「统计频率」为止，排序输出才是 B 的新增台阶。"},{"id":"pre:noi_openjudge/ch0111-07->roj/3553","source":"noi_openjudge/ch0111-07","target":"roj/3553","type":"pre","directed":true,"reason":"B 的纪念品分组正解直接复用 A 教的「升序排序 + 两端指针比较两数和决定指针移动」这一具体双指针步骤（main.py 中 prices[lo]+prices[hi]<=limit 时 lo+=1、每轮 hi-=1），只是在末端加了交换论证把双指针贪心的最优性补上，属于 A→B 的难度台阶。"},{"id":"pre:roj/1244->roj/3553","source":"roj/1244","target":"roj/3553","type":"pre","directed":true,"reason":"A 教的是排序后首尾双指针比较两数之和与阈值（和偏大则右指针左移）的模板，B 把同一伸缩比较直接用于最贵件：价格和 ≤ w 时让最便宜的陪它成组、否则它单独成组，再叠加交换论证把这一步升级为求最少组数的贪心。"},{"id":"pre:luogu/P2676->roj/3529","source":"luogu/P2676","target":"roj/3529","type":"pre","directed":true,"reason":"B 把 A 教的降序排序优先取最大复用为按花生数从大到小排序，再叠加时间模拟与返程判定"},{"id":"pre:roj/1083->roj/3523","source":"roj/1083","target":"roj/3523","type":"pre","directed":true,"reason":"B 的子问题二直接复用 A 教的模意义快速幂：都是把 a^b 的指数二进制拆解、每步乘法后取模以避免构造巨大整数，A 用它求 a^b mod 7 再查星期表，B 用 pow(2,P,10^500) 求 2^P mod 10^500 得到末 500 位，再叠加 A 未涉及的对数闭式位数与补零切行流程。"},{"id":"pre:roj/1326->roj/3523","source":"roj/1326","target":"roj/3523","type":"pre","directed":true,"reason":"B 的末 500 位子问题直接复用 A 教的二进制快速幂逐步取模这一关键观察，把 b^p mod k 的模板套到 (2^P-1) mod 10^500 上，再叠加 log10 位数闭式"},{"id":"pre:roj/1616->roj/3523","source":"roj/1616","target":"roj/3523","type":"pre","directed":true,"reason":"B 的末 500 位子问题直接复用 A 教的三参数 pow 模快速幂，只是把底数换成 2、模换成 10**500"},{"id":"pre:roj/1317->roj/3513","source":"roj/1317","target":"roj/3513","type":"pre","directed":true,"reason":"B 枚举组合时直接复用 A 的「递增序列表示组合 ⇒ 不重不漏」这一观察，A 用手指的递增性排除重复，B 靠同一性质给出组合去重口径，只是 B 把生成交给 itertools 并叠加了判素与记忆化"},{"id":"pre:luogu/P1157->roj/3513","source":"luogu/P1157","target":"roj/3513","type":"pre","directed":true,"reason":"B 正解第一步复用 A 教的 combinations 免手写递归生成不重不漏的组合，再叠加求和、判素与 @cache 去重"},{"id":"pre:luogu/P1877->roj/3167","source":"luogu/P1877","target":"roj/3167","type":"pre","directed":true,"reason":"B 的符号求解直接沿用 A 教的加减双向可达性 DP：把音量下标换成平移后的和值，用同一 f[i][v]=f[i-1][v-a_i]∨f[i-1][v+a_i] 转移判可达，再在 A 没有的最后一步上回溯符号并还原操作位置。"},{"id":"pre:luogu/U663295->roj/3167","source":"luogu/U663295","target":"roj/3167","type":"pre","directed":true,"reason":"B 直接把 A 教的「布尔可达性 dp、由已有可达和加/减一项推出新和」搬到符号模型：main.py 的 f[i][s] 就是 A 的 dp[c]，转移只在 ±a_i 两个方向做同一件事，A 是无价值的纯可行性 01 背包，B 在此之上叠加符号建模与操作位置还原。"},{"id":"pre:roj/1195->roj/3167","source":"roj/1195","target":"roj/3167","type":"pre","directed":true,"reason":"B 的可行性线性 DP 直接复用 A 教的「每个 a_i 沿 +a_i/-a_i 两支平移可达集合」这一步（代码里 f[i][j+ai]/f[i][j-ai] 赋符号），在其上再叠加把符号回译为减操作位置的新流程。"},{"id":"pre:luogu/P3366->roj/3106","source":"luogu/P3366","target":"roj/3106","type":"pre","directed":true,"reason":"B 把 A 教的「用 find 求根、比较根是否相同来判定两元素是否同一连通块」直接用作同列判定（main.py 的 rx, ry = find(x), find(y) 与 rx != ry 分支，M 合并仍沿用 A 的按根挂树与根上规模累加），只在其上叠加 A 未教的 front 偏移量与从根向叶累加的路径压缩，把并查集从平凡连通块升级为带权并查集。"},{"id":"pre:roj/3211->roj/3103","source":"roj/3211","target":"roj/3103","type":"pre","directed":true,"reason":"B 的第一阶段「一次 Floyd 求出所有教室对的距离」直接复用 A 教的「点少就用一趟 Floyd 求任意两点全源最短路」这一步（B 的 all_pairs() 即同一 Floyd 三重松弛），把结果从取直径改成 O(1) 查表喂给期望 DP，再叠加期望线性性与申请数 DP。"},{"id":"common:roj/1635--roj/3100","source":"roj/1635","target":"roj/3100","type":"common","directed":false,"reason":"把互质条件去掉的一般中国剩余定理：合并时要先看 gcd 是否整除差值，再约去公因子求逆元。"},{"id":"pre:luogu/P3074->roj/3095","source":"luogu/P3074","target":"roj/3095","type":"pre","directed":true,"reason":"B 复用 A 教的「Kahn 求拓扑序、在该序上对每条出边做一遍线性 DP」这一步：main.py 同样先 Kahn 得 order 再沿出边递推，只是把 A 的正拓扑序取 max 最长路换成逆拓扑序上按全期望公式求 (Σ边长+Σf[后继])/出度。"},{"id":"pre:luogu/P3183->roj/3095","source":"luogu/P3183","target":"roj/3095","type":"pre","directed":true,"reason":"A 教的「DAG 路径问题不枚举、先求拓扑序再沿边一遍 DP」正是 B 实际使用的那一步：B 的 main.py 先做 Kahn 拓扑排序再 reversed(order) 沿出边递推，只是把 A 的源点 dp=1、沿边累加路径条数改成由后继倒推并按出度取平均的期望 f[u]=(Σc+Σf[v])/K，额外叠加全期望公式与马尔可夫性。"},{"id":"pre:luogu/P1379->roj/3082","source":"luogu/P1379","target":"roj/3082","type":"pre","directed":true,"reason":"A 教的状态当点、单位边 BFS 与 dist 字典兼判重记距离，正是 B 双向 BFS 每侧扩展所用的步骤，B 只是把它铺到两端并用逆规则接上"},{"id":"pre:luogu/P1032->roj/3082","source":"luogu/P1032","target":"roj/3082","type":"pre","directed":true,"reason":"B 的双向 BFS 仍直接沿用 A 的模型（串是状态、替换是边）并把 A 教的逐位 find 枚举重叠出现原样用作 expand，只是额外叠加了 B 端按逆规则扩展的流程。"},{"id":"pre:luogu/P1747->roj/3080","source":"luogu/P1747","target":"roj/3080","type":"pre","directed":true,"reason":"A 教的无权图最短路 BFS 正是 B 的解法骨架：B 的 main.py 把非障碍格建成无权图后用队列逐层扩展、首次到达 H 即返回层数，只额外叠加障碍过滤、一维下标还原与提前退出，并未用上 A 特有的 12 种走法与反向 BFS，故属模板级复用。"},{"id":"pre:roj/1329->roj/3080","source":"roj/1329","target":"roj/3080","type":"pre","directed":true,"reason":"B 的网格 BFS 直接沿用 A 教的「入队即标记、标记兼作访问、每格至多入队一次」这一状态设计（main.py 里先写 dist[nxt] = dist[pos] + 1 再 append），只是把标记从原地抹 '0' 换成 dist 初值 -1，并叠加马的 8 种跳跃位移与首次够到 H 即最短距离这一判定。"},{"id":"pre:roj/1255->roj/3080","source":"roj/1255","target":"roj/3080","type":"pre","directed":true,"reason":"B 直接复用 A 的「首次到达即最短距离、标记与判重合一份」这一 BFS 核心步骤，只是换成马的 8 位移并把前驱标记换成 dist 数组"},{"id":"pre:roj/1335->roj/3081","source":"roj/1335","target":"roj/3081","type":"pre","directed":true,"reason":"B 的八连通 BFS 直接用 A 教的「扩展条件里一次挡掉越界与障碍、入队即标记」这一网格搜索骨架（B 代码同样先判 in_map 再把 dist 置为 week+1 后 append），A 还点明八连通只需扩 DIRS；B 只把栈换成队列并叠加层数记录与取全图最大值，得到最晚占领周数。"},{"id":"pre:roj/1255->roj/3081","source":"roj/1255","target":"roj/3081","type":"pre","directed":true,"reason":"B 直接复用 A 教的网格 BFS 中首次入队即最短步数这一层数判定，只把四连通换成八连通并把单点到点距离改为全图最大层数"},{"id":"pre:roj/1329->roj/3081","source":"roj/1329","target":"roj/3081","type":"pre","directed":true,"reason":"B 的八连通 BFS 直接复用 A 教的「入队即标记、每格至多入队一次」这一 BFS 不变式来替代逐周全图重扫，只在其上叠加八方向扩散与取最大层数作为答案，属同为 BFS 模板但强度提升一级的台阶关系。"},{"id":"pre:roj/3071->roj/3068","source":"roj/3071","target":"roj/3068","type":"pre","directed":true,"reason":"A 用反图 Dijkstra 算出到终点的最短路（h），B 把同一手法搬到状态图上作「反向最短路算剩余代价」的第一步，给出紧边贪心所需的 g(v)。"},{"id":"pre:luogu/P1948->roj/3069","source":"luogu/P1948","target":"roj/3069","type":"pre","directed":true,"reason":"A 教的 0/1 边权 + 双端队列 0-1 BFS（0 权进队头、1 权进队尾）正是 B 求最少旋转次数所用的核心最短路步骤，B 只是把它套在格点建图模型上。"},{"id":"pre:roj/1382->roj/3070","source":"roj/1382","target":"roj/3070","type":"pre","directed":true,"reason":"B 的 Dial 桶队列沿用 A 的「首次弹出即定型」关键观察：B 证明油钱单调不减、故首次取出 (E,·) 即最优，只是把 A 的二叉堆换成桶队列并把惰性删除搬到 dist[u][f]!=idx"},{"id":"pre:luogu/P4779->roj/3070","source":"luogu/P4779","target":"roj/3070","type":"pre","directed":true,"reason":"B 在分层图 (城市,剩油量) 的 Dial 桶队列最短路里原样复用 A 教的那一步过期判断——A 是 `current != distance[node]` 跳过堆中旧状态，B 换成 `dist[u][f] != idx` 跳过桶里过期条目，只在此模板上叠加 A 未教的分层状态设计与桶队列。"},{"id":"pre:luogu/P4822->roj/3070","source":"luogu/P4822","target":"roj/3070","type":"pre","directed":true,"reason":"B 直接复用 A 教的把资源消耗量升为第二维状态的做法：A 用 (城市,已用卡数)，B 用 (城市,剩余油量) 建分层图跑最短路，只在选择表示上加了权值结构与 Dial 桶优化"},{"id":"pre:roj/1382->roj/3071","source":"roj/1382","target":"roj/3071","type":"pre","directed":true,"reason":"B 的 h_values 直接在反图上复用 A 教的堆优化 Dijkstra（含惰性删除过期堆条目）求到终点最短路，再外叠 A* 才是新增难度。"},{"id":"pre:luogu/P4779->roj/3071","source":"luogu/P4779","target":"roj/3071","type":"pre","directed":true,"reason":"B 正解里 h_values 就是 A 教的堆优化 Dijkstra 原样搬到反图上（弹出最小距离、跳过过期条目、沿边松弛）作为 A* 的启发函数预处理；A 是纯模板，B 在此之上叠加 f=g+h 的 A* 检索与第 K 次弹出终点计数。"},{"id":"pre:roj/3040->roj/3053","source":"roj/3040","target":"roj/3053","type":"pre","directed":true,"reason":"B 直接复用 A 教的「周期 ⟺ border、最小周期 = n − 最长 border」这一步，只是把单串换成多串公共 border 再按位与"},{"id":"pre:luogu/P3375->roj/3053","source":"luogu/P3375","target":"roj/3053","type":"pre","directed":true,"reason":"B 的 border_mask 函数直接复用 A 教的「前缀函数记录最长 border、沿 border 链回退」这一步骤（nxt[n], nxt[nxt[n]], … 沿链回溯），把它从求单个最长 border 扩展成枚举全部 border 并压成位掩码，再叠加「周期⟺border」和行、列掩码按位与求公共 border 两步，才把一维模板题升到二维最小覆盖子矩阵。"},{"id":"pre:roj/1467->roj/3053","source":"roj/1467","target":"roj/3053","type":"pre","directed":true,"reason":"B 第一步拆出「每行周期 W、每列周期 H」后，把 A 教的「最小周期 = n − 最长 border」连同 KMP 前缀函数原样搬到每个一维串上，只用位掩码把单个串的最长 border 升级为多串的最长公共 border"},{"id":"pre:luogu/P1901->roj/3048","source":"luogu/P1901","target":"roj/3048","type":"pre","directed":true,"reason":"A 教的「入栈时弹掉违规元素、再看栈顶/新栈顶得到左侧最近合法元素」这一步被 B 搬进直方图单调栈：仅把比较方向从更高换成更矮，用弹出后的新栈顶当矩形左边界，再叠加 A 未教的按底边分组与悬垂高度递推。"},{"id":"pre:luogu/P2866->roj/3048","source":"luogu/P2866","target":"roj/3048","type":"pre","directed":true,"reason":"B 复用 A 教的『新元素到来先弹出违反单调性的栈顶』这一单调栈维护动作，把它从统计可见牛数迁移为直方图扫描时弹栈夹出左右边界结算面积。"},{"id":"pre:luogu/P3467->roj/3048","source":"luogu/P3467","target":"roj/3048","type":"pre","directed":true,"reason":"B 第四步求直方图最大矩形时，用的正是 A 教的「栈内高度严格递增、遇到更矮就弹栈」这一具体判定（main.py 中 while stack and heights[stack[-1]] >= h），只是把 A 的弹栈丢弃换成用当前下标与新栈顶夹出宽度结算面积，并在此前叠加按底边分组与悬垂高度递推这些 A 未涉及的新流程。"},{"id":"pre:roj/3038->roj/3050","source":"roj/3038","target":"roj/3050","type":"pre","directed":true,"reason":"A 教的“乘 B^len 把高位前缀对齐后相减得到子串指纹”被 B 原样搬进竖向折叠：减去顶行指纹乘 (2^B)^{A-1} 再乘 2^B 滑入底行，实现 O(1) 滑动维护。"},{"id":"pre:luogu/P8306->roj/3042","source":"luogu/P8306","target":"roj/3042","type":"pre","directed":true,"reason":"B 的 build_trie 直接复用 A 教过的“插入时沿着路径逐字符前进、没有边就新建子节点”：把模式串换成 31 位二进制串，每位在 01-Trie 上走 0/1 儿子，再叠加 A 未教的高位优先贪心与相反位查询，模板复用明确、难度只高一档。"},{"id":"pre:roj/3041->roj/3042","source":"roj/3041","target":"roj/3042","type":"pre","directed":true,"reason":"B 的 build_trie 逐位插入 01-Trie 时直接复用 A 教的「边存在就复用、不存在就新建儿子」这一建树步骤（代码里 nxt == NO_CHILD 才新建），把字符路径换成二进制位路径；A 原文也点明 01-Trie 建立在这套骨架上，B 另叠加 A 未教的高位优先相反位贪心。"},{"id":"pre:leetcodecn/kth-largest-element-in-an-array->roj/3043","source":"leetcodecn/kth-largest-element-in-an-array","target":"roj/3043","type":"pre","directed":true,"reason":"B 直接搬用 A 教的「堆大小超上限就弹出堆顶最小值」这一小根堆维护前 k 大步骤，只是把固定 k 换成随扫描递增的容量 d，再叠加按 d 扫描与反悔贪心的交换论证。"},{"id":"pre:roj/2042->roj/3044","source":"roj/2042","target":"roj/3044","type":"pre","directed":true,"reason":"B 单轮合并直接复用 A 的「把每个生产者放进小根堆、堆顶就是全局最小候选」这一步，把 A 的 K 条素数生产者队列换成分行分列单调的加法表格子后仍用同一个带下标的小顶堆弹出最小值取前 n 小，A 之外 B 只需再叠加每轮只保留前 n 个和的截断与换行生成约定。"},{"id":"pre:luogu/P1631->roj/3044","source":"luogu/P1631","target":"roj/3044","type":"pre","directed":true,"reason":"A 教的两表加法表堆多路归并（弹堆顶并补入同流下一元素）正是 B 单轮归并的实际步骤，B 只是把它加到 m 个序列上逐轮迭代并补去重约定"},{"id":"pre:luogu/P2866->roj/3032","source":"luogu/P2866","target":"roj/3032","type":"pre","directed":true,"reason":"B 的直方图单调栈复用 A 教的具体弹栈判定：扫描到 i 时若 h_i <= h_top 就弹出栈顶（与 A「弹出所有 <=current 的栈顶」同一比较），只是把 A 弹栈后累加 len(stack) 换成用新栈顶和 i 结算 h_top×(i-L-1)；在此之上再叠加「最优矩形高等于某根柱子」的转化、边界 L/R 公式与末尾哨兵 0 这些 A 未教的新流程。"},{"id":"pre:luogu/P2947->roj/3032","source":"luogu/P2947","target":"roj/3032","type":"pre","directed":true,"reason":"B 的单调栈弹出条件（带等号的 <= 才弹）与「弹出后栈内剩余元素的高度单调、新栈顶就是往左最近边界」这一步正是 A 教的「弹出所有小于等于当前值的栈顶、栈内保持单调、栈顶即最近边界」，A 用它定右侧第一个更高、B 用它定左右第一个更矮，再叠加哨兵与面积结算。"},{"id":"pre:luogu/P8218->roj/3029","source":"luogu/P8218","target":"roj/3029","type":"pre","directed":true,"reason":"B 正解把 A 教的「前缀和相减得到一段区间和」直接搬成列前缀和差分 O(1) 取竖条和，再叠加二维降维与 Kadane。"},{"id":"pre:roj/1354->roj/3030","source":"roj/1354","target":"roj/3030","type":"pre","directed":true,"reason":"A 教的“栈空弹栈即非法”这一括号合法性失败点，正是 B 第一步把火车进出站归约为合法括号序列时所用的“栈不能空着出”约束（代码用 if stack 保证不空弹），B 在此合法性骨架上再叠加字典序 DFS 枚举与前 20 整树剪枝。"},{"id":"pre:leetcodecn/valid-parentheses->roj/3030","source":"leetcodecn/valid-parentheses","target":"roj/3030","type":"pre","directed":true,"reason":"B 的第一步直接沿用 A 教的括号合法性判定（栈空不能出栈），把火车进出站序列归约为合法括号序列结构，再在其上叠加字典序 DFS 枚举与剪枝。"},{"id":"pre:luogu/P1739->roj/3030","source":"luogu/P1739","target":"roj/3030","type":"pre","directed":true,"reason":"B 的第一步把火车进出站操作序列翻译成合法括号结构，直接复用 A 教的「扫描中 balance 不能为负」这一前缀非负判定，再叠加 A 未教的 DFS 按「先出栈后进栈」字典序枚举与收满 20 个整树剪枝。"},{"id":"pre:roj/3599->roj/3021","source":"roj/3599","target":"roj/3021","type":"pre","directed":true,"reason":"B 的两条关键观察都直接沿用 A 教的『交换相邻两人只影响这两人』这一步：先由 u 与前驱 w 交换只差 A[u]-A[w]、与轮数无关推出同在父块连续，再对相邻块做同样的交换推出按平均值降序，之后才叠加树约束下的块合并与按对计费增量。"},{"id":"pre:luogu/P1293->roj/3026","source":"luogu/P1293","target":"roj/3026","type":"pre","directed":true,"reason":"B 在环流建模得到代价 Σ|c_i+x_1| 后，直接用 A 教的“中位数使到各点距离和最小”这一关键观察收尾，A 的带权中位数只是 B 中位数步骤的加权版，B 在此之上叠加了环上前缀和与守恒方程的新流程。"},{"id":"common:luogu/P2952--roj/3022","source":"luogu/P2952","target":"roj/3022","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：B 复用 A 教的「操作只在两端、故选双端队列并映射头尾操作」这一步，只是从单条牛队扩到 13 个牌堆，并额外叠加朝向计数器 up[v] 的压缩统计。"},{"id":"pre:leetcodecn/merge-intervals->roj/3019","source":"leetcodecn/merge-intervals","target":"roj/3019","type":"pre","directed":true,"reason":"B 把几何转成区间后，在区间选点贪心里复用了 A 教的『排序后单次扫描、用当前区间左端点与已维护右端点比较来决定是否新开一组』，判据同为 l > right，只是排序键换成右端点、把新开区间换成新放雷达。"},{"id":"pre:leetcodecn/sort-list->roj/3015","source":"leetcodecn/sort-list","target":"roj/3015","type":"pre","directed":true,"reason":"B 的归并统计逆序对正是把 A 教的「递归排序两半后合并两个有序段」当作归并排序的合并骨架，只在取右半元素时加一次 mid-i 计数"},{"id":"pre:luogu/P1593->roj/3010","source":"luogu/P1593","target":"roj/3010","type":"pre","directed":true,"reason":"B 的 divsum 把 A 教的「项数折半时用 (power,sum) 双量递推、偶数项合并成 total*(1+power)」直接用作求每个质因子等比和的步骤，只是把递归改成按 terms 二进制从高位翻倍的迭代；余下难度（A=0 特判、m 最大到 eB+1≈1.3e9、6k±1 试除分解）属叠加。"},{"id":"pre:roj/1326->roj/3001","source":"roj/1326","target":"roj/3001","type":"pre","directed":true,"reason":"B 完整复用了 A 的二进制拆分与逐位倍增骨架：A 用底数平方升级+位为 1 才累乘求幂，B 把同一循环里的平方换成加数自加翻倍、累乘换成累加，从而在避免中间乘积溢出的同时保持 O(log b)。"},{"id":"pre:roj/1616->roj/3001","source":"roj/1616","target":"roj/3001","type":"pre","directed":true,"reason":"B 整体套用 A 教的『二进制拆分+相邻项只差一次递推+只在位为 1 时累乘』骨架，把底数平方换成加数自加、累乘换成累加，据此在 64 位溢出场景下仍保持 O(log b)，B 另叠加中间量不超过 2p 的取模论证。"},{"id":"pre:luogu/P8687->roj/3003","source":"luogu/P8687","target":"roj/3003","type":"pre","directed":true,"reason":"A 教的「把元素集合压成二进制整数 mask、用位表示已覆盖哪些口味」是 B 状压状态的核心一步：B 直接沿用为 f(s,j) 的集合维度，只在其上叠加「最后停在哪个点」维度与边权代价。"},{"id":"pre:roj/1083->roj/3000","source":"roj/1083","target":"roj/3000","type":"pre","directed":true,"reason":"A 教的「边乘边取模」快速幂（指数逐位右移、底数自平方、最低位为 1 时累乘）正是 B 的 mod_pow 原样用来求 a^b mod p 的一步，只是把固定模 7 换成任意模 p 并补上 p=1/b=0/a>=p 的边界。"},{"id":"pre:roj/1231->roj/2095","source":"roj/1231","target":"roj/2095","type":"pre","directed":true,"reason":"B 在逐行悬垂高度直方图上复用了 A 教的「单调栈弹掉破坏单调性的栈顶」这一步，只是把单次扫描的贪心删除换成逐柱结算最大矩形，并在其上叠加悬垂高度递推与面积结算。"},{"id":"pre:luogu/P1106->roj/2095","source":"luogu/P1106","target":"roj/2095","type":"pre","directed":true,"reason":"B 的正解把 A 教的「栈顶大于当前值就弹出、栈内保持递增」这一具体判定搬进每行直方图（main.py 的 while stack and h[stack[-1]] > hj），只把删掉的数字换成用当前下标与新栈顶夹出宽度来结算矩形，B 新增的只是悬垂高度递推与面积计算"},{"id":"pre:roj/3041->roj/2096","source":"roj/3041","target":"roj/2096","type":"pre","directed":true,"reason":"B 的 01-Trie 直接复用 A 教的「按需建儿子、从根逐层下走」的 Trie 建树方式，只把字符边换成二进位边，再叠加反向位贪心与平局规则"},{"id":"pre:roj/3207->roj/2090","source":"roj/3207","target":"roj/2090","type":"pre","directed":true,"reason":"B 直接复用 A 教的拆点建网（点弧容量1、连接弧INF）跑最大流得到最小点割大小 k，再在其上叠加字典序逐点判定"},{"id":"pre:luogu/P1884->roj/2091","source":"luogu/P1884","target":"roj/2091","type":"pre","directed":true,"reason":"A 教的『离散化网格 + 差分标记 + 前缀和求覆盖次数』被 B 迁移成 x 方向差分数组 + cumsum 求本行 cover，B 再把二维压成逐行流式并叠加状态翻转求周长。"},{"id":"pre:HDU/3634->roj/2091","source":"HDU/3634","target":"roj/2091","type":"pre","directed":true,"reason":"B 直接复用 A 的坐标离散化与「小块整体被覆盖」结论，把矩形边界排序去重压成条带后才逐行比较覆盖状态"},{"id":"pre:luogu/P2341->roj/2084","source":"luogu/P2341","target":"roj/2084","type":"pre","directed":true,"reason":"B 的第三步直接把 A 教的「缩点 DAG 后统计出度为 0 的分量个数」当作 sink（代码里 has_out / total - sum(has_out)），只是再补上入度为 0 的源分量计数、取 max 并加单分量特判，A 只做唯一汇点判定。"},{"id":"pre:luogu/P3387->roj/2084","source":"luogu/P3387","target":"roj/2084","type":"pre","directed":true,"reason":"B 直接复用 A 教的「SCC 缩点、分量内边被消掉」这一步，再在缩点 DAG 上数入度 0 与出度 0 分量个数，得到基础模板题到加边变强连通综合题的台阶。"},{"id":"pre:HDU/3836->roj/2084","source":"HDU/3836","target":"roj/2084","type":"pre","directed":true,"reason":"A 教的「缩点后统计入度为 0 的 P 与出度为 0 的 Q、答案取 max(P,Q)、单分量特判 0」被 B 的 main.py 原样复用：has_in/has_out 标记后 sources/sinks 计数、0 if total==1 else max(sources,sinks)；B 只是在此之上叠加子任务 A 的入度为 0 分量计数，并把 A 的直观构造升级为源汇最大匹配的证明。"},{"id":"pre:roj/1439->roj/2080","source":"roj/1439","target":"roj/2080","type":"pre","directed":true,"reason":"B 直接复用 A 教的「部分最小化保凸」这一步：先内层对 y 三分、再据此断言外层对 x 三分成立，把 A 的两层嵌套三分模板套到点到线段距离之和上"},{"id":"pre:luogu/P4779->roj/2063","source":"luogu/P4779","target":"roj/2063","type":"pre","directed":true,"reason":"B 把 A 教的堆优化 Dijkstra 原样当作子程序（detour 里同样是弹出最小距离状态、沿边松弛、以 dictionary 距离表判重），只是额外叠加了把区域周长化为最小环、再逐条禁边求 w+dist 的最小值这一层流程。"},{"id":"pre:roj/1382->roj/2063","source":"roj/1382","target":"roj/2063","type":"pre","directed":true,"reason":"B 的最小环枚举把 A 教的堆优化 Dijkstra 当子程序原样复用：main.py 的 detour 里同样是「松弛成功就压堆、弹出时 if d > dist[x] 跳过过期条目、命中目标点立即返回」，只是加了 banned 禁边参数，A 只跑一次单源最短路，B 把它对每条篱笆各跑一次再取 w+dist 最小值，并另行叠加「相接集合当端点编号」与「区域周长=最小环」两步。"},{"id":"pre:luogu/P1144->roj/2054","source":"luogu/P1144","target":"roj/2054","type":"pre","directed":true,"reason":"B 的骑士距离表 D(k,·) 正是靠 A 教的「BFS 首次访问即最短距离」这一分层性质算出的，只是把无权图从普通图换成骑士走日字的棋盘网格，之后 B 再把这份 BFS 距离表叠加到枚举集合点与携带增量的流程里。"},{"id":"pre:roj/2036->roj/2054","source":"roj/2036","target":"roj/2054","type":"pre","directed":true,"reason":"A 教的关键一步是「不为每个格子各做一次单源最短路，而是把一批源点整体当超级源点、一次多源 BFS 同时算出所有格子的距离」，B 的 carry_min 正是把种子 D(k,u)+K(king,u) 一次灌入骑士图做多源（桶）最短路同时算出所有 t 的 E_k，把枚举接送点 u 的 O(n^2) 摊成一次 O(n)；区别只在 A 的源点是迷宫最外圈的均匀空格、B 的种子带非零权值。"},{"id":"pre:luogu/P2911->roj/2049","source":"luogu/P2911","target":"roj/2049","type":"pre","directed":true,"reason":"B 复用 A 教的「把枚举/扫描顺序定成从小到大的目标键，则首次命中的最优解天然带并列取最小的保证」这一步：A 用从小到大扫描和让次数相同时保留最小和（代码只在 times > best_count 时更新），B 把它改成按 a+b+c 递增枚举、第一组命中直接返回，省去保存候选与比较；B 另叠加的「第一个非零位定倍数 k 再回代校验」判定则是 A 未教的整数倍比例判定。"},{"id":"pre:roj/3067->roj/2050","source":"roj/3067","target":"roj/2050","type":"pre","directed":true,"reason":"A 教的分层 BFS「第一次到达的层数就是最小层数，首次入队即锁定」正是 B 状态图 BFS 的纪律二：`if nxt in parent: continue` 就是第一次到达就锁定，命中目标即回溯出最短操作串，B 只是把网格四邻换成三种置换、再额外叠加按 A/B/C 生成顺序保证字典序，所以 A 是 B 的 BFS 骨架前置。"},{"id":"pre:roj/1381->roj/2051","source":"roj/1381","target":"roj/2051","type":"pre","directed":true,"reason":"B 每枚举一个候选放糖点就跑一次与 A 完全同款的堆优化 Dijkstra（含 `if d > dist[u]: continue` 跳过过期堆记录这一步），A 只求 1 到 n 单条最短路，B 在此之上叠加无向图对称换向、按牛数加权求和与枚举全部牧场取 min。"},{"id":"pre:luogu/P1747->roj/2051","source":"luogu/P1747","target":"roj/2051","type":"pre","directed":true,"reason":"A 教的『无向图最短路可反向』被 B 用作关键的换向：把每头牛到放糖点 x 反转成从 x 单源 Dijkstra 一次，从而把 N 次单终点计算压成 P 次单源计算。"},{"id":"pre:roj/2037->roj/2051","source":"roj/2037","target":"roj/2051","type":"pre","directed":true,"reason":"B 复用 A 教的「无向图最短路对称换向」这一步：把每头牛到糖点反转成从糖点出发，从而每个候选点只需一次单源 Dijkstra，再改为加权求和取最小"},{"id":"pre:leetcodecn/perfect-squares->roj/2045","source":"leetcodecn/perfect-squares","target":"roj/2045","type":"pre","directed":true,"reason":"A 教的 min 型完全背包 dp[i]=min(dp[i-j*j]+1) 正是 B 正解「第一步：把问题换成最少张数」后照搬的转移 f(v)=1+min f(v-s)，只把物品从平方数换成邮票面值；B 的主要难度来自第一位表示可达性的位并行实现与 K·max A 上界，故属模板级复用。"},{"id":"pre:luogu/B3611->roj/2045","source":"luogu/B3611","target":"roj/2045","type":"pre","directed":true,"reason":"B 的第三步把邮资可达性编码成整数位集，用 `reachable << s` 表示集合整体加 s、按位或表示求并，直接沿用 A 教的「可达集合编码成整数位集并用大整数 OR 合并」这一步，只是把 Warshall 的行并入换成按面值平移的 K 轮集合迭代，并额外叠加最少张数建模与上界 K·maxA。"},{"id":"pre:luogu/P2347->roj/2045","source":"luogu/P2347","target":"roj/2045","type":"pre","directed":true,"reason":"B 的正解直接复用 A 教的「可达性编码成位集合 + 左移或转移」这一状态设计与关键实现，只在外面叠加 K 轮分层、K*maxA 上界与连续可达前缀计数。"},{"id":"pre:roj/1294->roj/2041","source":"roj/1294","target":"roj/2041","type":"pre","directed":true,"reason":"A 教的一维背包 f[v] 状态与转移被 B 完整沿用，B 只把 A 的倒序改成容量维正序，从而由 0-1 选取变为可重复选取的完全背包。"},{"id":"pre:luogu/U661986->roj/2041","source":"luogu/U661986","target":"roj/2041","type":"pre","directed":true,"reason":"B 的完全背包完全沿用 A 教的一维转移式 dp[t]=max(dp[t],dp[t-t_i]+p_i) 与「扫描方向决定件数」这一步，只是把 A 的容量倒序（每件只选一次）换成正序（种类可重复选），并在此之上叠加按耗时升序处理与被支配种类剪枝。"},{"id":"pre:luogu/U661988->roj/2041","source":"luogu/U661988","target":"roj/2041","type":"pre","directed":true,"reason":"A 教的容量正序枚举=物品可重复选，正是 B 完全背包解法中 range(minutes, limit+1) 这一步的直接复用，B 只是在其上叠加被支配种类剪枝"},{"id":"pre:luogu/P2085->roj/2042","source":"luogu/P2085","target":"roj/2042","type":"pre","directed":true,"reason":"B 沿用 A 教的多路归并骨架（每路只留队首、弹出堆顶后压入该路下一项），再叠加 A 未涉及的「打平一起推进」去重步骤，难度从普及/提高- 升到普及+/提高-"},{"id":"pre:luogu/P1135->roj/2036","source":"luogu/P1135","target":"roj/2036","type":"pre","directed":true,"reason":"B 的多源 BFS 灌水复用了 A 教的“BFS 分层扩展、第一次访问即为最短距离”：把两出口外整片自由空间当源点层 0，某格第一次被淹到的层号就是它到最近出口的步数"},{"id":"pre:roj/3067->roj/2036","source":"roj/3067","target":"roj/2036","type":"pre","directed":true,"reason":"B 直接把 A 教的多源 BFS 源点初始化（源点距离置 0 一起入队）搬到迷宫外圈，只是把源点从 01 矩阵里的所有 1 换成补边后最外圈整片自由空间，再叠加字符位折半换算"},{"id":"pre:luogu/P1529->roj/2038","source":"luogu/P1529","target":"roj/2038","type":"pre","directed":true,"reason":"B 的第 1 步「用 INF 哨兵建距离矩阵、对全图跑一遍 Floyd 得到全源最短路」就是 A 教的「点少就建距离矩阵、一趟 Floyd 出任意两点最短路」这一步，B 只是在其上叠加偏心距/直径、连通块染色与枚举跨块点对；A 是纯 Floyd 模板题、B 难在后续直径流程，属台阶式前置。"},{"id":"pre:roj/1342->roj/2038","source":"roj/1342","target":"roj/2038","type":"pre","directed":true,"reason":"B 的第 1 步直接复用 A 教的「坐标建带权无向邻接矩阵后 Floyd 三重松弛求全源最短路」，只是把结果的用途从查 d(s,t) 换成算偏心距与直径，再叠加连通块染色与跨块点对枚举。"},{"id":"pre:roj/1082->roj/2039","source":"roj/1082","target":"roj/2039","type":"pre","directed":true,"reason":"B 的循环体直接复用 A 教的长除法取位步骤（余数乘 10 取商、余数取模），只是额外登记余数首次出现的位数以定位循环节"},{"id":"pre:luogu/P1747->roj/2037","source":"luogu/P1747","target":"roj/2037","type":"pre","directed":true,"reason":"B 的第三步正是 A 教的无向图交换源点：A 从共同终点 (1,1) 反向 BFS 一次回答两匹马，B 把只跑一次的起点换成谷仓 Z，再由一次 Dijkstra 取 A..Y 的最小值"},{"id":"pre:luogu/U661988->roj/2033","source":"luogu/U661988","target":"roj/2033","type":"pre","directed":true,"reason":"B 把 A 教的「一维数组容量/金额正序枚举，让 dp[c-v] 已含本物品贡献」这一完全背包关键步骤，直接用到计数版上（main.py 的 for amount in range(coin, n+1): ways[amount] += ways[amount-coin]），只在 A 的空间压缩基础上叠加二维递推降维与组合计数（外层按面值分组）的额外流程。"},{"id":"pre:roj/1294->roj/2033","source":"roj/1294","target":"roj/2033","type":"pre","directed":true,"reason":"B 的组合计数沿用 A 教的一维滚动数组状态设计与「枚举方向决定物品能否重复取用」这一步：A 用倒序枚举容量保证每件至多选一次，B 把同一 ways 数组改为按面值外层、金额内层正序枚举（main.py 的 ways[amount] += ways[amount - coin]），使 ways[j-c] 含本轮值从而面值可重复取用，B 原文也直接以 0/1 背包倒序扫描作对照；在此之上 B 才叠加按面值分组去重、max 换成方案数累加与 ways[0]=1 边界。"},{"id":"pre:luogu/U661986->roj/2033","source":"luogu/U661986","target":"roj/2033","type":"pre","directed":true,"reason":"B 的完全背包计数沿用 A 教的一维滚动数组状态设计并把扫描方向规则翻转使用：A 用倒序让 dp[c-v] 保持上轮值以保证每件物品只选一次，B 在 main.py 里用正序让 ways[j-c] 含本轮值，从而同一面值可重复取用，B 原文也直接以倒序 0/1 背包作对照说明这一分界；在此之上 B 才叠加按面值分组去重、max 换成方案数累加与 ways[0]=1 边界。"},{"id":"pre:luogu/P10839->roj/2032","source":"luogu/P10839","target":"roj/2032","type":"pre","directed":true,"reason":"B 的枚举层直接复用 A 的关键观察——搜索空间是小乘积空间时「完整枚举即正解、不需要优化」，A 的 n 层每层 m 选择枚举模板正好对应 B 在 N-1 个间隙每处 3 种符号的枚举；B 只是在此基础上叠加表达式字符串生成、空格拼接求值与 ASCII 顺序输出。"},{"id":"pre:luogu/P2089->roj/2032","source":"luogu/P2089","target":"roj/2032","type":"pre","directed":true,"reason":"B 直接复用 A 教的 itertools.product 枚举固定候选并利用 product 字典序免排序这一点，只是把每位置候选从 1..3 换成符号 ' '、'+'、'-'，并叠加了拼接求值与筛选步骤"},{"id":"pre:luogu/P2919->roj/2021","source":"luogu/P2919","target":"roj/2021","type":"pre","directed":true,"reason":"B 的房间判定第一步正是 A 教的「扫描未访问格、flood fill 整块连通分量并只统计一次」，把同高度 8 连通块换成只走无墙边；B 在此基础上再叠加房间号不同必有墙的引理、合并大小取 max 与先西后南的元组平局规则。"},{"id":"pre:roj/5000->roj/1676","source":"roj/5000","target":"roj/1676","type":"pre","directed":true,"reason":"B 直接沿用 A 的单调约定去重（非增/非降一一对应）并把边界量收进记忆状态，只是把上限收紧改成下界随 head 传递，再叠加恰好 k 段的新约束"},{"id":"common:roj/1675--roj/1676","source":"roj/1675","target":"roj/1676","type":"common","directed":false,"reason":"数与数据完全相同的重题（1675「数的划分」），同样求 n 拆成恰好 k 个正整数之和的方案数，可作为同一模型的对读。"},{"id":"pre:roj/1194->roj/1660","source":"roj/1194","target":"roj/1660","type":"pre","directed":true,"reason":"B 在 A 教的「把整条路线数写成组合数」这一步之上叠加反射法，再用组合数差得广义卡特兰数"},{"id":"pre:luogu/P3390->roj/1642","source":"luogu/P3390","target":"roj/1642","type":"pre","directed":true,"reason":"B 的 power() 逐位复用 A 教的矩阵快速幂：最低位为 1 就把 base 乘进 result，再让 base 平方并右移指数，只是改用 2×2 的 Fibonacci 递推矩阵"},{"id":"pre:roj/3000->roj/1643","source":"roj/3000","target":"roj/1643","type":"pre","directed":true,"reason":"B 的 mat_pow 直接复用 A 教的「最低位为 1 就把 base 乘进 result、base 自平方、指数右移」这一快速幂迭代骨架（B 代码里 if n&1: result=mat_mul(result,base); base=mat_mul(base,base); n>>=1），只把标量乘法换成 2×2 转移矩阵乘法来求 T^{n+1}，再叠加 A 未教的 S_n=f_{n+2}-1 恒等式与矩阵表示层。"},{"id":"pre:roj/1326->roj/1643","source":"roj/1326","target":"roj/1643","type":"pre","directed":true,"reason":"B 的 mat_pow 直接复用 A 的迭代快速幂「底数每轮平方升级、答案按位乘入」骨架，只是把标量底数换成 2×2 转移矩阵"},{"id":"pre:roj/1616->roj/1643","source":"roj/1616","target":"roj/1643","type":"pre","directed":true,"reason":"B 的 mat_pow 逐行照搬 A 教的二进制快速幂骨架（重复平方、位1乘入、边乘边取模），只是把标量换成2×2矩阵求 T^(n+1)"},{"id":"pre:luogu/P1292->roj/1635","source":"luogu/P1292","target":"roj/1635","type":"pre","directed":true,"reason":"B 每次 merge 求解 m1k≡diff (mod m2) 时，正是复用 A 教过的「gcd 约去公因子后求模逆元」这一步，只是从单条同余式扩到逐条折叠并新增整除判无解与 lcm 合并"},{"id":"pre:luogu/P1516->roj/1635","source":"luogu/P1516","target":"roj/1635","type":"pre","directed":true,"reason":"B 把 A 的 gcd 整除判据当作每次合并同余方程的有解判定，再叠加逐条折叠流程"},{"id":"pre:luogu/P1075->roj/1626","source":"luogu/P1075","target":"roj/1626","type":"pre","directed":true,"reason":"B 的 divisors 函数沿用 A 教过的试除上界 sqrt(n) 来分解 b1，再在其上叠加 x|b1 收缩与因数表扩展"},{"id":"pre:roj/1151->roj/1623","source":"roj/1151","target":"roj/1623","type":"pre","directed":true,"reason":"B 的 prime_table 直接复用 A 教的埃氏筛「p 从 p^2 起整片划掉倍数」，只是筛区间换成价值上界 n+1，再叠加 A 未教的「冲突边必是质数—合数、二分染色数退化为 1 或 2」这一图论观察，属在 A 的筛法台阶上叠加额外构造流程。"},{"id":"pre:luogu/P3383->roj/1619","source":"luogu/P3383","target":"roj/1619","type":"pre","directed":true,"reason":"B 的区间筛直接复用 A 教的：先取不超过 √n 的基础素数，再对区间按 p 步长切片批量置零，只是起点换成任意 L 并叠加相邻差最值"},{"id":"pre:luogu/P1835->roj/1619","source":"luogu/P1835","target":"roj/1619","type":"pre","directed":true,"reason":"B 的区间筛完全复用 A 的 max(p*p,ceil(L/p)*p) 起点与整段划除标记，只是筛完后追加了相邻质数差最值的取对流程"},{"id":"pre:roj/1151->roj/1621","source":"roj/1151","target":"roj/1621","type":"pre","directed":true,"reason":"B 复用 A 教的埃氏筛倍数枚举：把 A 中「枚举 p 的倍数并标记」迁移为「对每个出现数字枚举其倍数并累加 cnt」，再叠加值域统计与约数和转化。"},{"id":"pre:roj/1616->roj/1618","source":"roj/1616","target":"roj/1618","type":"pre","directed":true,"reason":"B 在正难则反得到 m^n − m(m−1)^(n−1) 后，直接复用 A 教的三参数 pow 模快速幂算这两个幂，B 的额外台阶是取补集与逐位乘法原理计数。"},{"id":"pre:roj/1326->roj/1618","source":"roj/1326","target":"roj/1618","type":"pre","directed":true,"reason":"B 用正难则反得到 m^n−m(m−1)^(n−1) 后，复用 A 教的把指数按二进制拆分求模幂，直接调用 pow 算出两个模幂，只是把 A 的单一幂扩成两个幂；额外台阶是补集计数与乘法原理。"},{"id":"pre:luogu/P1226->roj/1618","source":"luogu/P1226","target":"roj/1618","type":"pre","directed":true,"reason":"B 在得出 m^n − m(m−1)^(n−1) 后，直接照 A 教的 Python 三参数 pow 写 pow(m,n,MOD) 与 pow(m-1,n-1,MOD) 求两个模幂，属模板级复用；B 的额外台阶是正难则反取补集加逐位乘法原理的计数。"},{"id":"pre:roj/3158->roj/1614","source":"roj/3158","target":"roj/1614","type":"pre","directed":true,"reason":"B 的 main.py 用 hull_query/hull_add 照搬 A 教的「下凸壳 + 单调队列、队头队尾指针只前进」这一步，把 f[i]=min_j 的直线族查询压到均摊 O(1)"},{"id":"pre:roj/1610->roj/1614","source":"roj/1610","target":"roj/1614","type":"pre","directed":true,"reason":"B 直接复用 A 的下凸壳单调队列这一步：同样把决策 j 化成直线、同样靠斜率与查询双单调让队头队尾只移动指针，只是查询点从 T[i]-1-L 换成 pos_i"},{"id":"pre:roj/3158->roj/1611","source":"roj/3158","target":"roj/1611","type":"pre","directed":true,"reason":"B 复用 A 教的「先核对决策点横坐标与查询斜率都单调、再用单调队列维护下凸壳」这一判定步骤，在仓库建设的一维分段 DP 上落地"},{"id":"pre:roj/1610->roj/1611","source":"roj/1610","target":"roj/1611","type":"pre","directed":true,"reason":"A 教的「下凸壳可以用双端队列维护」这一具体步骤被 B 直接复用：B 的仓库建设转移同样化成直线族，靠查询点 X_i 与斜率 -W_j 双单调，用同一套队首弹线、队尾弹死线的单调队列凸壳把 O(N^2) 降到 O(N)。"},{"id":"pre:roj/3158->roj/1612","source":"roj/3158","target":"roj/1612","type":"pre","directed":true,"reason":"A 教的「核对决策点 x_j 与查询斜率 k_i 都单调后用下凸壳+单调队列摊还 O(1)」正是 B 主解实际走的第五步：B 把 (S_i-S_j)² 展开成直线 l_j(u)=S_j·u+Y_j 后，同样靠 x_i≥1 使斜率单调增、a<0 使查询点单调增，再用双端队列弹队头弹队尾维护包络（main.py 的 hull 两个 while），只是把最小截距下凸壳换成最大上包络并把一维序列扩到 n=10^6，难度仅升一级。"},{"id":"pre:roj/1610->roj/1612","source":"roj/1610","target":"roj/1612","type":"pre","directed":true,"reason":"B 的斜率优化解法完整复用 A 教的这一步：同样把 dp 转移展开成 l_j(x)=m_j·x+b_j 直线族、同样利用斜率与查询双单调在双端队列上维护凸壳，队尾判死线也同样用交叉相乘（B 代码 slope_pq>=slope_qi 即 A 的 is_shadow），只是把取 min 的下凸壳镜像成取 max 的上包络，并叠加 a<0 的查询点单调与 n=10^6 的 array 存储；A 是提高级模板题，B 是提高+/省选-的加强应用，方向正确。"},{"id":"pre:roj/1606->roj/1608","source":"roj/1606","target":"roj/1608","type":"pre","directed":true,"reason":"B 的「关键一步」正是复用 A 教的交换求和次序记账：启动费由「批数 m 乘以批内 C 和」改写为「该批起点之后所有任务的 C 之和」，跨段耦合消失、状态只需前缀，B 再在其上叠加下凸壳二分优化"},{"id":"pre:roj/3158->roj/1609","source":"roj/3158","target":"roj/1609","type":"pre","directed":true,"reason":"B 的 dp_layer 直接复用 A 教的「把 j 相关项拆成 (x_j,y_j)、用下凸壳+单调队列取最小截距」这一套实现，A 以点集形式给出、B 以直线族形式套用，B 另需叠加的是将猫压成 a_i=T_i-s[H_i] 并排序划分连续段。"},{"id":"pre:roj/1610->roj/1609","source":"roj/1610","target":"roj/1609","type":"pre","directed":true,"reason":"A 教的「转移展开成直线族 + 斜率与查询双单调 ⇒ 双端队列维护下凸壳、交叉相乘判死线」正是 B 第四步每层 DP 用来替代枚举切点的同一具体步骤，B 只是在其上叠加排序压参与 P 层分段 DP。"},{"id":"pre:luogu/P1725->roj/1601","source":"luogu/P1725","target":"roj/1601","type":"pre","directed":true,"reason":"A 教的「转移只依赖区间内 dp 最值 → 滑动窗口 → 单调队列：入队时弹掉队尾更差者、队首先过期先出队」正是 B 的 relax() 对同余类窗口 [m-c,m] 求最小键 old[r+uw]-u 时实际使用的那一步，B 只是先做同余类拆分与换元把多重背包转移变成定长滑窗，再套同一套队列不变量。"},{"id":"pre:luogu/P2627->roj/1603","source":"luogu/P2627","target":"roj/1603","type":"pre","directed":true,"reason":"B 的判定 DP 里『队首出窗后即为窗口最小值的下标』就是 A 教的『窗口极值用单调队列维护断点』那一步（A 取 max、B 取 min），B 只在外面再叠一层二分答案"},{"id":"pre:luogu/P1725->roj/1603","source":"luogu/P1725","target":"roj/1603","type":"pre","directed":true,"reason":"B 的可行性判定 DP 直接复用 A 教的单调队列滑动窗口最值转移，只是把窗口内 dp 最大值换成最小值，再在外层叠加二分答案"},{"id":"pre:roj/1542->roj/1604","source":"roj/1542","target":"roj/1604","type":"pre","directed":true,"reason":"B 把 A 教的单调队列一维滑窗（队尾淘汰/队首过期/队首即最值）封装成 window_best，先横向压行窗口表、再纵向滑列，从而消掉 n^2 因子。"},{"id":"pre:luogu/P2251->roj/1604","source":"luogu/P2251","target":"roj/1604","type":"pre","directed":true,"reason":"B 先做的一维基础正是 A 教的单调队列窗口最小值，用同一套存下标、弹队尾、队首即窗口最值的队列逐行压出 row_min，再纵向滑窗叠加成二维窗口最值。"},{"id":"pre:luogu/P1886->roj/1604","source":"luogu/P1886","target":"roj/1604","type":"pre","directed":true,"reason":"A 教的定长滑动窗口单调队列（存下标、队头弹过期、队尾弹更差候选）正是 B 中 window_best 的实现内容，B 只是把这一个一维步骤先横向逐行、再纵向逐列各用一次，复合出 n×n 方格的最值，再叠加求最小差值。"},{"id":"pre:roj/1604->roj/1605","source":"roj/1604","target":"roj/1605","type":"pre","directed":true,"reason":"A 教的「单调队列存下标、队首即平移窗口最值」正是 B 的 transfer_buy/transfer_sell 对买入窗口 [j-AS,j-1]、卖出窗口 [j+1,j+BS] 所用的那一步，只是被套到 f[p][j']+price*j' 上；B 的额外流程（W 天前驱快照、按 j' 拆项、抢跑 j 方向）属新增环节，A 未教，故记为模板级复用。"},{"id":"pre:luogu/P2627->roj/1605","source":"luogu/P2627","target":"roj/1605","type":"pre","directed":true,"reason":"B 沿用 A 的拆项套路：把不参与比较的 price·j 提出后，窗口内被最大化的量只与决策下标有关，再用单调队列维护随 j 单向平移的窗口最大值，从单个断点窗口扩到买卖两个窗口。"},{"id":"pre:luogu/P1725->roj/1605","source":"luogu/P1725","target":"roj/1605","type":"pre","directed":true,"reason":"B 正解的买入/卖出转移就是把 A 教的「用单调队列保存前驱下标、取合法前驱窗口内 dp 最大值」搬到 f[p][j']+price*j' 上，再叠加 W 天前驱快照与拆项这两个新步骤。"},{"id":"pre:luogu/P1725->roj/1602","source":"luogu/P1725","target":"roj/1602","type":"pre","directed":true,"reason":"A 教的「转移只依赖区间内 dp 最值 → 滑动窗口 → 单调队列保存前驱下标、队首最值、过期出队并弹掉队尾劣者」正是 B 计算 dp[i] = a_i + min(dp[i-m..i-1]) 时实际使用的那一步；B 只是先做约束改写（相邻点火点距离 ≤ m 加尾部豁免）把问题化成滑动窗口最小值，再套用同一队列不变量。"},{"id":"pre:luogu/P2627->roj/1602","source":"luogu/P2627","target":"roj/1602","type":"pre","directed":true,"reason":"B 正解（main.py 单调队列）把 A 教的「窗口随 i 单调右移、用单调队列维护断点取最值」直接套在自己的 dp[i]=a_i+min(dp[i-m..i-1]) 上，只是把 A 的最大值窗口 [i-k,i]（含 dp[j-1]-S[j]）换成最小值窗口 [i-m,i-1]（纯 dp[j]），并额外叠加相邻选中点建模、尾部豁免与哨兵。"},{"id":"pre:roj/1593->roj/1596","source":"roj/1593","target":"roj/1596","type":"pre","directed":true,"reason":"B 沿用 A 教的「按推进方向分段、把影响未来的局部信息压成掩码、只保留最近窗口做滚动 DP」这一状态设计：A 用行掩码只记相邻两行，B 换成 5 位窗口状态只记最近 5 个围栏，再叠加按窗口起点结算得分与枚举初窗口破环。"},{"id":"pre:luogu/P3353->roj/1598","source":"luogu/P3353","target":"roj/1598","type":"pre","directed":true,"reason":"B 的第一步直接复用 A 教的前缀和作差，把限长子段和写成 S_j - S_{i-1}（main.py 里 pres[i]-pres[dq[0]]），再叠加 A 未教的单调队列求窗口最小前缀。"},{"id":"pre:luogu/P8218->roj/1598","source":"luogu/P8218","target":"roj/1598","type":"pre","directed":true,"reason":"B 把 A 教的「前缀和作差得区间和」当作第一步复用（main.py 里 pres[i]-pres[dq[0]]），只是把固定询问区间换成限长子段，再叠加单调队列求窗口最小前缀"},{"id":"pre:luogu/P1725->roj/1599","source":"luogu/P1725","target":"roj/1599","type":"pre","directed":true,"reason":"A 教的是「转移需要区间内 dp 最大值 → 滑动窗口 → 单调队列队首即答案」，B 把这一步原样搬到窗口 [i-K,i-1] 上，只是先把候选值由 dp[j] 拆项成 h(j)=dp[j-1]-S_j，队列的过期出队与劣质淘汰机制完全一致。"},{"id":"pre:roj/3576->roj/1591","source":"roj/3576","target":"roj/1591","type":"pre","directed":true,"reason":"B 直接沿用 A 教的按位三段闭式计数（同一 higher/cur/lower 分解与到顶组补量），只把统计目标从数码 2 推广到 0~9 并加 0 的前导零修正，A 因此是 B 的前置台阶。"},{"id":"pre:luogu/P2858->roj/1580","source":"luogu/P2858","target":"roj/1580","type":"pre","directed":true,"reason":"A 教的是把连续区间当作状态、按区间长度递增递推的区间 DP 骨架，B 的 score[l][r] 与 length 循环直接沿用这一步，只是把 A 的两端收缩转移换成枚举根 k 拼左右子区间，并额外记录 root 还原前序。"},{"id":"pre:luogu/P1775->roj/1580","source":"luogu/P1775","target":"roj/1580","type":"pre","directed":true,"reason":"P1775 用「区间状态 + 枚举断点劈成左右两段 + 按长度递推」的区间 DP 合并石子，1580 直接搬用同一劈区间模板，只把断点换成根、区间和换成因子相乘"},{"id":"pre:luogu/P1775->roj/1574","source":"luogu/P1775","target":"roj/1574","type":"pre","directed":true,"reason":"B 逐行做的区间 DP 直接复用 A 教的两步：把 dp[l][r] 设计成「连续子区间子问题」的状态、再按区间长度（段长）递增递推保证子段先算出（B 代码 for length in range(2, m+1)），只是把 A 的最后一次合并断点转移换成取行首/行尾两候选，并叠加行间独立、权重 2^pick 与 Python 大整数。"},{"id":"pre:luogu/P2858->roj/1574","source":"luogu/P2858","target":"roj/1574","type":"pre","directed":true,"reason":"B 逐行套用 A 的「剩余区间就是完整状态、天数/轮次只由段长决定」这一步，并沿用左右端两候选取 max 的转移，再叠加行间独立与高精度"},{"id":"pre:luogu/P2858->roj/1569","source":"luogu/P2858","target":"roj/1569","type":"pre","directed":true,"reason":"B 的链上 f[i][j] 直接复用 A 教的「连续区间即完整状态、按区间长度从小到大递推」这一步状态设计与递推顺序，只是把 A 的两端收缩转移换成枚举断点 k，并额外叠加复制成链枚举长度 n 窗口来破环。"},{"id":"pre:luogu/P1775->roj/1569","source":"luogu/P1775","target":"roj/1569","type":"pre","directed":true,"reason":"B 的环形石子合并整段照搬 A 教的“最后一次合并断点 k”区间 DP 转移，只是在外层再叠加复制成链并枚举长度 n 的窗口来破环"},{"id":"pre:luogu/P1775->roj/1571","source":"luogu/P1775","target":"roj/1571","type":"pre","directed":true,"reason":"B 把 A 教的“区间状态 + 枚举最后一次断点把区间劈成两段子问题再取 min”整套区间 DP 骨架搬到凸多边形三角剖分上，只把最后的合并代价 sum(l,r) 换成三角形乘积 w_i*w_k*w_j，难度的增量在高精度与剖分建模而非新范式。"},{"id":"pre:luogu/P2234->roj/1567","source":"luogu/P2234","target":"roj/1567","type":"pre","directed":true,"reason":"B 的 F 命令直接复用 A 教的「在树状数组上用二进制倍增按名次 kth」，把第 k 大转成第 staff-k+1 小后调用同一 kth（main.py 的 kth 与 A 的 kth 逐行同构），只在其上叠加全局偏移 bias 与扣薪区间扫描删除作为新流程。"},{"id":"pre:luogu/P2590->roj/1564","source":"luogu/P2590","target":"roj/1564","type":"pre","directed":true,"reason":"B 的 path_spans 在代码里直接复用 A 教的「树链剖分把路径拆成连续 dfn 区间」这一步（同款选链顶较深一侧摘段再跳父亲），再叠加 A 未教的按信仰切分状态、成员 dfn 位置表二分转秩区间，以及树状数组求和/线段树求最大值，属于在树剖模板上叠加多流程的综合题。"},{"id":"pre:roj/1562->roj/1564","source":"roj/1562","target":"roj/1564","type":"pre","directed":true,"reason":"B 的 path_spans 直接复用 A 教的重链头部跳跃拆路径：每次摘下一段连续的 [pos(head), pos(x)] 区间、再跳到链顶父亲，A 用它逐段置 1，B 用它逐段二分求秩后统计，B 只是把单条根到 x 路径推广为 x→y 并叠加按信仰分堆。"},{"id":"pre:luogu/P9244->roj/1546","source":"luogu/P9244","target":"roj/1546","type":"pre","directed":true,"reason":"B 直接沿用 A 教过的「固定右端点、合法左端点构成随右端点单调推进的前缀并一趟增量计数」这一骨架，只是把 A 的单计数器前缀计数换成按色调的 total 减 pend 记账，并叠加最靠右便宜店阈值 g(j) 与清账流程。"},{"id":"pre:leetcodecn/linked-list-cycle->roj/1463","source":"leetcodecn/linked-list-cycle","target":"roj/1463","type":"pre","directed":true,"reason":"A 教的「哈希集合记录已访问节点、再遇到同一节点即判重」正是 B 递推模拟中每一步查走过的值是否出现过的那一步（代码里 seen 集合的 in/add），B 只是把它从链表游走搬到函数迭代出的数列上，再叠加 ρ 形建模与 2×10^6 封顶。"},{"id":"pre:roj/1673->roj/1463","source":"roj/1673","target":"roj/1463","type":"pre","directed":true,"reason":"A 教的\"用一张表记下见过的值、遍历时先查表判重\"被 B 原样搬成递推模拟每一步的哈希集合判重，B 只是把这个判重步骤叠在函数迭代建模与 2×10^6 封顶之上，从入门模板升到普及-。"},{"id":"pre:noi_openjudge/ch0107-09->roj/1402","source":"noi_openjudge/ch0107-09","target":"roj/1402","type":"pre","directed":true,"reason":"A 教的是字母序号平移并在字母表末尾回绕，B 的减法解密正是把这一步按位复用（mod 26 回绕），再叠加密钥循环取位与大小写跟随密文。"},{"id":"pre:noi_openjudge/ch0107-10->roj/1402","source":"noi_openjudge/ch0107-10","target":"roj/1402","type":"pre","directed":true,"reason":"B 的 Vigenère 解法正是把 A 教的「字母转下标后减位移再 % 26 回绕转回字符」这一凯撒移位步骤逐位套用，只把固定位移 5 换成密钥位 k_{i mod n} 并补上密钥循环与大小写跟随"},{"id":"pre:luogu/P1914->roj/1402","source":"luogu/P1914","target":"roj/1402","type":"pre","directed":true,"reason":"B 把 A 教的字母折算 0..25、(x±n)%26 后回写字符这一步逐位套用，只是把固定位移 n 换成循环密钥位 k_{i mod |k|} 的偏移并补上大小写基准"},{"id":"pre:roj/1351->roj/1395","source":"roj/1351","target":"roj/1395","type":"pre","directed":true,"reason":"A 教的 Kahn「把入度为 0（已无未处理前驱）的点入队作起点」正是 B 的拓扑消元起点「把 |cand(i)|=1 的数字放进队列」；B 的 main.py 复用同一队列删点骨架，只把「入度归零」判据换成「候选由位掩码删到只剩一个」，再叠加二分图唯一完美匹配判定与位运算实现。"},{"id":"pre:roj/20022->roj/1373","source":"roj/20022","target":"roj/1373","type":"pre","directed":true,"reason":"B 在枚举最远鱼塘、把每塘递减收益摊平成表之后，直接复用 A 教的「排序取前 m 大」这一贪心步骤，把它改成对各塘合并表降序取前 left 项求和；A 特有的基准转换 d_i=a_i-b_i 未被使用，故属模板级复用。"},{"id":"pre:roj/1235->roj/1373","source":"roj/1235","target":"roj/1373","type":"pre","directed":true,"reason":"B 的 main.py 中 picks.sort(reverse=True); sum(picks[:left]) 直接套用 A 教的「降序排序后取前 k 大」，只是把单张数组换成一排鱼塘合并出的分钟收益表；B 自身的主要难度在枚举最远鱼塘 L、把递减收益摊平成表这两个 A 未教的新步骤，因此该复用属模板级而非决定本题难度。"},{"id":"pre:roj/1255->roj/1360","source":"roj/1255","target":"roj/1360","type":"pre","directed":true,"reason":"A 教的「BFS 首次入队即最短、入队时同时充当访问标记」正是 B 解法的那一步：B 把楼层当无权有向图顶点后，用 dist[A]=0 入队、dist 置 -1 兼作未访问标记、首次到达层号即最少按键次数，只是在同一 BFS 模板上换成动态出边与不可达输出 -1。"},{"id":"pre:roj/8016->roj/1359","source":"roj/8016","target":"roj/1359","type":"pre","directed":true,"reason":"B 的 main.py 直接复用 A 教的四连通洪水填充染色骨架：由种子格出发把可达同类整块标记（B 同为入队前先置 outside 为真），只是把种子从扫描到的单个字母格换成网格最外圈全部 0 格，并在染色后统计未标记的 0 格作为面积"},{"id":"pre:roj/1329->roj/1359","source":"roj/1329","target":"roj/1359","type":"pre","directed":true,"reason":"B 的多源洪水填充直接复用 A 教的那一步——四连通网格 flood fill「入队即标记、每格至多入队一次」的状态设计（B 的代码同样是 outside 置真后才 append），只是把种子从单个细胞格换成网格最外圈的全部 0 格，并把染色对象由细胞翻转为外部格后再数剩余 0。"},{"id":"pre:roj/1331->roj/1356","source":"roj/1331","target":"roj/1356","type":"pre","directed":true,"reason":"B 先用调度场算法把中缀转成后缀，再照搬 A 教的后缀求值「先弹右、后弹左」弹栈步骤；A 的入门题就是这条求值线的起点，B 只在其上叠加优先级与括号流程。"},{"id":"pre:leetcodecn/valid-parentheses->roj/1356","source":"leetcodecn/valid-parentheses","target":"roj/1356","type":"pre","directed":true,"reason":"A 教的两类括号分别入栈/消解至配对左括号这一步，被 B 的调度场括号屏障规则直接复用，B 只是再叠加优先级弹栈与后缀求值"},{"id":"pre:roj/1331->roj/1358","source":"roj/1331","target":"roj/1358","type":"pre","directed":true,"reason":"B 把 A 教的后缀表达式栈式归约（先弹右再弹左）当作流水线最后一段直接复用，代码里 eval_postfix 的 rhs,lhs=stack.pop(),stack.pop() 就是 A 的 eval 那一步，只是前面多叠了状态机校验和调度场转后缀；B 总结也自认「[[roj/1331]] 正是这条链路的后半段」。"},{"id":"pre:roj/1354->roj/1358","source":"roj/1354","target":"roj/1358","type":"pre","directed":true,"reason":"A 教的「中途弹空（右括号多余）与扫完栈非空（左括号多余）两类失败点即配对非法」正是 B 的 tokenize 合法性子过程用的那一步（代码 depth==0 拒右括号、结束后 depth 非 0 拒），B 只把它简化成单类型计数器再叠加记号类别状态机、调度场转后缀与操作数栈求值。"},{"id":"pre:luogu/P1739->roj/1358","source":"luogu/P1739","target":"roj/1358","type":"pre","directed":true,"reason":"B 的 tokenize 合法性子过程直接沿用 A 教的「单计数器记录未匹配左括号、过程中不得为负、末尾归零」这一配对判定（代码里的 depth：expect_operand 位置遇 ) 而 depth==0 判 NO，扫描结束 depth 非 0 判 NO），只在此之上叠加记号类别状态机、调度场转后缀与操作数栈求值。"},{"id":"pre:roj/1351->roj/1352","source":"roj/1351","target":"roj/1352","type":"pre","directed":true,"reason":"B 直接复用 A 教的 Kahn 入度归零入队循环（出队遍历后继、入度减 1、归零入队），只是把出队顺序换成 max(pay[u])+1 的定薪递推，并用出队计数补上判环。"},{"id":"pre:luogu/P1536->roj/1348","source":"luogu/P1536","target":"roj/1348","type":"pre","directed":true,"reason":"B 的 Kruskal 主循环在 B 的代码里逐行指认了 A 教的这一步：find(u)、find(v) 后「两端代表元不同才合并」（father[ru]=rv，main.py 第 61-63 行），A 只教了并查集判连通与合并这一层，B 在它外面叠加按权排序、贪心选满 n-1 条边和复现 std 快排。"},{"id":"pre:luogu/P1551->roj/1348","source":"luogu/P1551","target":"roj/1348","type":"pre","directed":true,"reason":"B 的 Kruskal 把 A 教的「find 取代表元并比较是否同块」直接用作选边判据：代码里 `ru, rv = find(father,u), find(father,v)` 后 `if ru != rv: father[ru] = rv` 才选边，只是在此外层叠加按权排序、割性质贪心与复现 std 不稳定快排的输出处理。"},{"id":"pre:luogu/P3367->roj/1348","source":"luogu/P3367","target":"roj/1348","type":"pre","directed":true,"reason":"B 的 Kruskal 直接用 A 教的 find 找代表元判同集合这一步来决定每条边是否选中（代码 ru!=rv 才 father[ru]=rv），并沿用 A 的路径压缩，只是在外层叠加按权排序、割性质贪心与复现 std 不稳定快排的输出要求。"},{"id":"pre:shumeng/CSP201809B->roj/1333","source":"shumeng/CSP201809B","target":"roj/1333","type":"pre","directed":true,"reason":"B 的多路归并直接复用 A 教的「比较当前两段后推进结束更早一方」这一具体双指针推进规则：A 是 end 更早的区间不可能再相交故前移其指针（main.cpp:40-44），B 是两族队首取小者、被消费的族指针前移一格（main.py:20-21），只把「结束时刻比较」换成「候选值比较」、并在相等时叠加双消费去重与堆模拟对照，属 A→B 的台阶。"},{"id":"pre:shumeng/CSP202006B->roj/1333","source":"shumeng/CSP202006B","target":"roj/1333","type":"pre","directed":true,"reason":"A 教的「两条有序序列各自用只进不退的指针，比较当前队首再决定谁前移、相等则双方同时处理」正是 B 的 2x+1 与 3x+1 两族队首归并取小、相等双消费去重所用的同一步骤；B 只是把它从两个已给的有序稀疏表搬到两条自产流水线上，并叠加了 A 未教的「父小于子 ⇒ 已生成前缀可回喂、生成到第 n 项」这一自产自销机制，故属真实复用且台阶合理。"},{"id":"pre:roj/1255->roj/1330","source":"roj/1255","target":"roj/1330","type":"pre","directed":true,"reason":"B 的正解把 A 教的「首次入队即最短步数、并用同一张表兼作访问标记」这套网格无权图 BFS 引擎与状态设计原样用在 100×100 棋盘上（代码里 dist 字典既存距离又判重），只是把源点从起点搬到终点 (1,1) 做一次反向预处理并叠加 12 种走法。"},{"id":"pre:luogu/P1116->roj/1328","source":"luogu/P1116","target":"roj/1328","type":"pre","directed":true,"reason":"B 在证明「最少交换次数=逆序对数」时直接复用 A 教的那一步——交换相邻逆序对恰使逆序对数减 1（B 下界/上界都靠它），把问题化为数逆序对后再叠加 A 未教的归并排序跨段计数，得到 O(n log n)。"},{"id":"pre:roj/1190->roj/1314","source":"roj/1190","target":"roj/1314","type":"pre","directed":true,"reason":"一维线性递引入重复子问题与计数 DP 的基本形态"},{"id":"pre:luogu/P1116->roj/1310","source":"luogu/P1116","target":"roj/1310","type":"pre","directed":true,"reason":"B 正解的第一步转化直接沿用 A 教的「相邻交换最少次数=初始逆序对数量」，B 代码里 inversions 即最少旋转次数；区别只在 A 用两层循环 O(n²) 统计，B 把统计换成离散化+权值树状数组 O(n log n)。"},{"id":"pre:luogu/P3009->roj/1305","source":"luogu/P3009","target":"roj/1305","type":"pre","directed":true,"reason":"B 的 f 数组直接复用 A 教的\"单飞或接上\"递推，只是把它正反各做一遍再取前缀最大拼接左右两段"},{"id":"pre:luogu/P1115->roj/1305","source":"luogu/P1115","target":"roj/1305","type":"pre","directed":true,"reason":"B 正解第三步的 f_i 逐字复用了 A 教的『以 i 结尾的最大子段和』转移（单飞或接上），只是再叠加反向 h 与枚举分界 k 来拼两段"},{"id":"pre:roj/1146->luogu/P9232","source":"roj/1146","target":"luogu/P9232","type":"pre","directed":true,"reason":"B 的正解把 A 教的「比较两端字符、相等就向内收缩」直接用作递推分支：s[l]==s[r] 时对齐两端后转到 f(l+1,r-1)，只是再叠加第一处不同必在区间内的观察与按右端点滚动的区间 DP。"},{"id":"pre:HDU/4310->roj/10011","source":"HDU/4310","target":"roj/10011","type":"pre","directed":true,"reason":"B 直接复用 A 的相邻交换比值比较：先按 DPS/HP 降序，再把同一比值规则升为块交换引理处理树上约束"},{"id":"pre:luogu/P2676->roj/10006","source":"luogu/P2676","target":"roj/10006","type":"pre","directed":true,"reason":"B 复用 A 教的“先降序排序、再从最大元素开始取”的贪心：涨薪名额给工资最高的人最划算，故工资降序后取前 x 大之和，只是再叠加“开除最小的 n-x-y 人”与快速幂"},{"id":"pre:roj/1326->roj/10006","source":"roj/1326","target":"roj/10006","type":"pre","directed":true,"reason":"B 的 main.cpp 里 ll power(ll base,ll exp)（exp&1 时 result 乘底数、底数自身平方取模、exp>>=1）正是 A 教的「二进制分解快速幂：底数逐位平方升级、位为 1 时把底数乘进答案」这一步，用来算 $3^m$、$2^m$（m 可达 1e9），B 只是在此外层叠加降序排序与「最小 k=n-x-y 人连续两年拿 C 被开除」的贪心结论与 m=1 特判。"},{"id":"pre:roj/1616->roj/10006","source":"roj/1616","target":"roj/10006","type":"pre","directed":true,"reason":"B 的 main.cpp 中 ll power(ll base,ll exp)（result=result*base%MOD、base=base*base%MOD、exp>>=1）与 power(3,m)、power(2,m) 正是复用 A 教的「底数逐位平方、每一步取模」这一模快速幂步骤，用来把 m 年倍数 3^m、2^m 算成模值后代入 3^m S_x+2^m(S_{x+y}-S_x)；B 的 m 可达 1e9 使该步与 A 同样必要（main.py 用 pow(3,m,MOD)），另在其外叠加降序排序、前 k 大求和与剔除 k=n-x-y 人的贪心结论。"},{"id":"pre:roj/1182->roj/10008","source":"roj/1182","target":"roj/10008","type":"pre","directed":true,"reason":"B 直接复用 A 教的「先按关键特征分桶、再在桶内排序、最后按下标取数」这一分桶排序步骤，只是把 2 个性别桶换成 26 个结尾字母桶，并在此外层叠加了同分先读入的二级比较与 O(1) 询问判定。"},{"id":"pre:roj/1236->roj/10001","source":"roj/1236","target":"roj/10001","type":"pre","directed":true,"reason":"B 直接复用 A 的 reach 断口判定，把单遍区间合并中越过 reach 即断开的观察，扩展为逐断档补魔法并输出施法序列"},{"id":"pre:luogu/P1901->roj/10003","source":"luogu/P1901","target":"roj/10003","type":"pre","directed":true,"reason":"B 的 max 栈（弹出条件 top<=x、栈内值严格递减、存下标）就是 A 教的那一步：从左到右扫描并维护高度严格递减的单调栈、栈顶即往左最近的更大站；B 把同一个栈从「取最近更高站」转用为「左端点左移时的极值变化点」，再叠加 min 栈归并、按窗口长度差分入账与值域 100 的段数界这些 A 未涉及的新流程。"},{"id":"pre:luogu/P2947->roj/10003","source":"luogu/P2947","target":"roj/10003","type":"pre","directed":true,"reason":"B 的 max/min 两栈直接复用 A 教的「栈顶不大于当前值就弹出、只保留真正能贡献的候选下标」这一判定（代码里同为 `max_stack[-1][0] <= x`），把它从「求右边第一个更大元素」改成「收集固定右端点下极值变化点」，再叠加双栈按下标归并分段与按窗口长度差分入账。"},{"id":"pre:luogu/P2866->roj/10003","source":"luogu/P2866","target":"roj/10003","type":"pre","directed":true,"reason":"A 教的「弹出所有 `<=` 当前值的栈顶、保留严格递减的高度栈」正是 B 的 max 栈实际使用的同一步（弹出条件 top<=x 收集左移时的记录值位置），B 在其上再叠加 min 栈、按下标归并分段与差分入账。"},{"id":"pre:luogu/P8686->luogu/P9715","source":"luogu/P8686","target":"luogu/P9715","type":"pre","directed":true,"reason":"B 的 sweep_last / sweep_first 直接复用 A 教的后继并查集（代码 nxtR[x]=x+1 后 find 跳下一个），把「已占用就跳到下一个可用位置」从单序列找最小值改造成行列上求最后一次 t=1 与第一次 t=0，再叠加前缀累加计数。"},{"id":"pre:roj/1382->luogu/P9724","source":"roj/1382","target":"luogu/P9724","type":"pre","directed":true,"reason":"B 正解的第一步就是在每步伤害 d-dist(k,v) 构成的非负权图上复用 A 教的堆优化 Dijkstra（main.cpp 同样是 priority_queue + vis 惰性删除的定型模板）求出 dist_shou，再在其外叠加「以第一次传送分两段、等差数列 O(1) 结算」的额外流程。"},{"id":"pre:luogu/P4779->luogu/P9724","source":"luogu/P4779","target":"luogu/P9724","type":"pre","directed":true,"reason":"B 的正解第一段把 A 教的堆优化 Dijkstra 松弛出边原样复用，只是把边权换成 d-dist_k[v] 并在松弛时按 dist_k[v] 分流出首次传送判定"},{"id":"pre:roj/3203->luogu/P9726","source":"roj/3203","target":"luogu/P9726","type":"pre","directed":true,"reason":"B 第三步把「可同时当选断点集合」翻译成二分图独立集、其补集是顶点覆盖并由此得 2n−ν、第四步再用 König 构造最小顶点覆盖证可达，与 A 教的用 König 定理把点覆盖问题转成最大匹配这一步完全同源；B 额外叠加的是区间端点的二选一建模、交错对连边和压位匈牙利实现。"},{"id":"pre:luogu/P2676->luogu/P9728","source":"luogu/P2676","target":"luogu/P9728","type":"pre","directed":true,"reason":"A 教的「贡献独立 + 个数约束 ⇒ 排序优先取最大的若干元素」正是 B 第三步把线性目标配上「恰好选 n-a 个位置」约束后排序取前 n-a 大收益 c_p 的那一步，B 只是在其上叠加贡献法换序求 c_p 的新流程。"},{"id":"pre:roj/20022->luogu/P9728","source":"roj/20022","target":"luogu/P9728","type":"pre","directed":true,"reason":"A 教的是「每个元素两种选项各带收益时，先把选项差量化成单个边际收益，选择问题变成挑最大的若干个正数」；B 正是沿用这一步——先用贡献法把每道不辣菜在位置 p 的增益 c_p 算成独立的边际收益，再在「恰好选 n-a 个位置」的个数约束下取最大的若干项，B 另叠加的贡献法换序求 c_p 是 A 未教的新流程。"},{"id":"pre:roj/3561->luogu/P9728","source":"roj/3561","target":"luogu/P9728","type":"pre","directed":true,"reason":"A 教的「先按位置逐条统计独立贡献、再排序贪心选前 K 大」被 B 直接套用，只是把缝的收益换成每个位置放不辣菜的收益 c_p、把前 K 换成前 n-a"},{"id":"pre:luogu/P1364->luogu/P9245","source":"luogu/P1364","target":"luogu/P9245","type":"pre","directed":true,"reason":"B 的暴力解（brute.cpp 的 path_len）对每对相邻景点各做一次单源 BFS，直接复用 A 教的『树上两点路径唯一、BFS 第一次到达的层数就是边数距离』这一步来量出 d(u,v)；B 的正解在这条路径唯一性上把同一量改写为 dist[u]+dist[v]-2dist[lca] 并用倍增把反复 BFS 压成 O(log N)，额外叠加的 LCA/倍增与线路增量观察是 A 未教的新流程，故为模板级前置。"},{"id":"pre:luogu/P5836->luogu/P9246","source":"luogu/P5836","target":"luogu/P9246","type":"pre","directed":true,"reason":"B 正解第三步给每条路径打 diff[a]++、diff[b]++、diff[w]-=2 时，用的正是 A 教的「路径=两条根路径相加，再把公共前缀（根到 LCA）被多算的两份减掉」这一步，只是把 A 的前缀和计数换成点差分值；B 另外叠加了「删边后不连通等价于该边被路径覆盖」的转化与自底向上子树汇总这两步新流程。"},{"id":"pre:luogu/P3379->luogu/P9246","source":"luogu/P3379","target":"luogu/P9246","type":"pre","directed":true,"reason":"P9246 正解的第三步对每条路径求 w=LCA(a_i,b_i) 后打 diff，其 LCA 完全按 P3379 教的 up[j][u] 倍增表、先对齐深度再同步跳实现，A 的这一步被直接复用为差分的预处理；B 新增的是树上点差分与自底向上求和这一额外流程，构成 普及 -> 普及+/提高- 的台阶。"},{"id":"pre:HDU/1269->luogu/P2169","source":"HDU/1269","target":"luogu/P2169","type":"pre","directed":true,"reason":"A 只教到用 Tarjan 求出 SCC 并数 scc_cnt，B 直接复用 A 的 Tarjan 求 SCC 这一步：用同一套 dfn/low/栈得到 belong[u] 分量编号，把每个强连通分量缩成一个点后再叠加堆优化 Dijkstra 求 belong[1] 到 belong[n] 的最短路。"},{"id":"pre:luogu/P3372->roj/1561","source":"luogu/P3372","target":"roj/1561","type":"pre","directed":true,"reason":"B 在 DFS 序与 depth 拆项把子树加化为区间加后，最后一步沿用 A 教的差分区间加（d[l] 与 d[r+1] 两端修改、点查即前缀和）维护 q1/q2 两棵树状数组，单点加、子树加、询问都复用该模板。"},{"id":"pre:luogu/P3374->roj/1561","source":"luogu/P3374","target":"roj/1561","type":"pre","directed":true,"reason":"B 的最后一层——单点加/子树加/询问全部化为差分树状数组的区间加与点查，A 正是点查即前缀和的 Fenwick 前缀和维护（B 的 bit_add/bit_sum 与 A 的 add/prefix 同构，且 B 把单点加教成子树内每个 y 统一 +a 的区间加），B 在此模板上叠加 DFS 序、贡献视角与 depth 拆两树，故 A 是纯模板、B 在 A 之上叠加额外流程。"},{"id":"pre:luogu/P3368->roj/1561","source":"luogu/P3368","target":"roj/1561","type":"pre","directed":true,"reason":"B 在 DFS 序与贡献视角、depth 拆项之后，把全部操作落到 A 教的差分树状数组区间加、点查上，单点加/子树加/询问都复用该模板"},{"id":"pre:luogu/P1253->roj/1562","source":"luogu/P1253","target":"roj/1562","type":"pre","directed":true,"reason":"B 的线段树 cover 复用了 A 教的「整段覆盖只改摘要和懒标记、进入子树前下传」这一步——整段赋值时直接改 tr[i] 与 tg[i]、下探前把 tg 下推给儿子，只是摘要由区间最大值换成区间和、并去掉了 A 的加法标记；B 的主要新增流程是 A 未教的树链剖分与先序编号。"},{"id":"pre:luogu/P3870->roj/1562","source":"luogu/P3870","target":"roj/1562","type":"pre","directed":true,"reason":"B 的 install 段直接复用 A 教的关键观察「整段取反后数量 = len - sum」：把「段长减区间和」当作区间内未安装（0）的个数，再整段赋 1，B 的 main.py 里 cnt += 段长 - cover(...) 与 A 的结算公式同源，B 只是在其上叠加树链剖分与 DFS 序。"},{"id":"pre:luogu/P2590->roj/1562","source":"luogu/P2590","target":"roj/1562","type":"pre","directed":true,"reason":"B 的 install 直接复用 A 教的树链剖分拆路径方法，沿链头把根到 x 拆成连续区间逐段赋值，只是把 A 的区间查询换成带懒标记的区间赋值"},{"id":"pre:luogu/P3884->roj/1558","source":"luogu/P3884","target":"roj/1558","type":"pre","directed":true,"reason":"B 的费用公式（main.py 里 depth[a]+depth[b]+depth[c]-depth[ab]-depth[bc]-depth[ca]）直接复用 A 教的「先找 LCA 再按深度差拆出树上距离」这一步，只是把 A 的 2/1 加权向上/向下拆解换成标准 dep(u)+dep(v)-2dep(lca)，再把朴素父指针 LCA 升级为重链剖分并叠加「三对 LCA 中最深者即三路公共点」的新判定。"},{"id":"pre:luogu/P3379->roj/1558","source":"luogu/P3379","target":"roj/1558","type":"pre","directed":true,"reason":"B 的解法以「三对 LCA 中最深者即三路公共点」落地，把 A 教的对数级 LCA 当作每问 3 次 LCA 的黑盒调用（实现换成重链剖分，属模板接口级复用）。"},{"id":"pre:luogu/P5903->roj/1554","source":"luogu/P5903","target":"roj/1554","type":"pre","directed":true,"reason":"B 的 dist 每次操作要调用 3 次 LCA，其倍增 LCA 正是直接复用 A 教的「预处理 2^j 级祖先倍增表」这一步（jump[k][v] 与 up[x][j] 同构），A 是纯模板、B 在此之上叠加 DFS 序成环恒等式与有序表增量维护，构成 delta=2 的台阶。"},{"id":"pre:luogu/P8855->roj/1554","source":"luogu/P8855","target":"roj/1554","type":"pre","directed":true,"reason":"B 的 dist 函数直接沿用 A 教的距离公式，AB 的差距在 DFS 序成环与增量维护，而非距离计算"},{"id":"pre:luogu/P3258->roj/1553","source":"luogu/P3258","target":"roj/1553","type":"pre","directed":true,"reason":"B 的树上差分（main.py 中 diff[u]++、diff[v]++、diff[w]-=2 加 reversed(order) 后序累加求父边跨越次数）正是 A 教的“求 lca 后对两端点打差分、自底向上累加统计路径经过次数”这一步，只是把 LCA 的减标记从点差分改成边差分 -=2，再叠加 c_e 的 0/1/≥2 三分类计数。"},{"id":"pre:luogu/P3128->roj/1553","source":"luogu/P3128","target":"roj/1553","type":"pre","directed":true,"reason":"B 的树边跨越计数直接复用 A 教的树上差分：路径加一先化成差分端点并在 LCA 处抵消，再统一后序累加，只是把点差分换成边差分并叠加三分类统计。"},{"id":"pre:luogu/P9246->roj/1553","source":"luogu/P9246","target":"roj/1553","type":"pre","directed":true,"reason":"B 的树上差分逐字复用 A 教的「diff[u]++、diff[v]++、diff[w]-=2」端点打标记与自底向上累加求父边跨越次数这一步，只是把统计对象换成附加边路径，再叠加 c_e 的 0/1/>=2 三分类计数。"},{"id":"pre:luogu/P1816->roj/1549","source":"luogu/P1816","target":"roj/1549","type":"pre","directed":true,"reason":"B 直接复用 A 教的 ST 表两点：相邻半块合并递推与两端 2^k 重叠块查询，只是把按起点分层改成按结尾分层并随追加增量补块，从而支持在线后缀 RMQ。"},{"id":"pre:luogu/P2880->roj/1549","source":"luogu/P2880","target":"roj/1549","type":"pre","directed":true,"reason":"B 询问最后 L 个数时直接复用 A 教的「取 k=⌊log2 L⌋、用两段允许重叠的 2^k 块覆盖区间再取 max」这一步（代码 st[k][size-span] 与 st[k][size-arg] 两段），只是把静态建表换成按区间结尾分层的在线增量补层，并叠加 last_ans 取值依赖。"},{"id":"pre:luogu/P1816->roj/1545","source":"luogu/P1816","target":"roj/1545","type":"pre","directed":true,"reason":"B 的极差正解第一步就是 A 教的『相邻两块 2 的幂区间合并得到下一层』倍增建表，A 建一张 min 表，B 用同一循环建出 max/min 两张表再做极差，是可指认的代码级复用。"},{"id":"pre:luogu/P3374->roj/1539","source":"luogu/P3374","target":"roj/1539","type":"pre","directed":true,"reason":"B 的异或树状数组把 A 教的 Fenwick「单点改沿加 lowbit 上跳、前缀查沿减 lowbit 下拆」原样复用（代码 flip/prefix_xor 与 A 的 add/prefix 同一条 lowbit 链），只把加法换成异或，再叠加 A 未教的异或差分把区间翻转降成两次单点取反。"},{"id":"pre:luogu/P3368->roj/1539","source":"luogu/P3368","target":"roj/1539","type":"pre","directed":true,"reason":"B 直接套用 A 教的差分+树状数组框架：把区间操作化归为两个端点修改、用前缀操作求单点，只是把加法换成异或并把区间加换成区间翻转"},{"id":"pre:OpenJ_Bailian/1300->roj/1528","source":"OpenJ_Bailian/1300","target":"roj/1528","type":"pre","directed":true,"reason":"B 直接复用 A 教的非零度点连通性检查，把无向欧拉路径判定的并查集连通步骤照搬到有向图，再叠加 out-in 度数条件。"},{"id":"pre:luogu/P8604->roj/1521","source":"luogu/P8604","target":"roj/1521","type":"pre","directed":true,"reason":"B 把 A 教的 Tarjan 求点双（A 用它建圆方树数路径割点）直接挪用作块割树三分类统计的基础构件"},{"id":"pre:luogu/P3388->roj/1521","source":"luogu/P3388","target":"roj/1521","type":"pre","directed":true,"reason":"B 的 Tarjan 求点双在 main.py 中用 A 教的 `low[u] >= dfn[p]` 判定割点（cut[p]=True、根数孩子数），把割点当咽喉后再叠加点双双连通分量分解与三分类计数，割点判定正是 B 实际使用的一步。"},{"id":"pre:POJ/1523->roj/1521","source":"POJ/1523","target":"roj/1521","type":"pre","directed":true,"reason":"A 方法二教的 v-BCC 分解与「割点是多个点双的公共交点」正是 B 的第一步：B 的 find_blocks 同时求 cut[] 与各点双 comps，再把相邻点双经公共割点相连的结构用于按块内割点数三分类，在此外层叠加出口计数与方案相乘。"},{"id":"pre:HDU/3836->roj/1518","source":"HDU/3836","target":"roj/1518","type":"pre","directed":true,"reason":"A 教的 Tarjan 求 SCC 后缩点成 DAG 这一步被 B 完整复用（main.py 的 tarjan_scc 与 comp 缩点，正解亦以此为等价变换），B 只是在缩点 DAG 上叠加分量权值合并与最长路 DP，属典型模板级台阶。"},{"id":"pre:HDU/3072->roj/1518","source":"HDU/3072","target":"roj/1518","type":"pre","directed":true,"reason":"B 的 pool[c] 合并正是复用 A 教的「SCC 内部互相可达、整体合并成一点」这一关键观察，只是把 A 的内部费用忽略改成内部收益求和，再叠加缩点 DAG 上的最长路 DP。"},{"id":"pre:luogu/P3275->roj/1518","source":"luogu/P3275","target":"roj/1518","type":"pre","directed":true,"reason":"B 沿用了 A 的“SCC 缩点成 DAG 后求最长路”这一步：A 用分量初值 1 求最小糖果，B 换成以分量点权和 pool 为初值、按分量编号降序做最长路 DP"},{"id":"pre:luogu/P1135->roj/1499","source":"luogu/P1135","target":"roj/1499","type":"pre","directed":true,"reason":"B 直接沿用 A 教的『首次访问即最短』：在 dist[v]<0 的首访分支里用 -1 兼作未访问标记并定下距离，再在此分层基础上叠加最短路计数递推。"},{"id":"pre:luogu/P1126->roj/1495","source":"luogu/P1126","target":"roj/1495","type":"pre","directed":true,"reason":"B 直接复用 A 教的状态设计：只记坐标会漏解，必须把影响未来的附加信息（这里是钥匙掩码）并入状态维度，再做无权网格 BFS 首次到达即最短。"},{"id":"pre:roj/3003->roj/1495","source":"roj/3003","target":"roj/1495","type":"pre","directed":true,"reason":"B 的钥匙态状态设计直接复用 A 教的关键观察：把影响未来的离散集合（A 是已走点集、B 是持有钥匙集）压成整数的二进制位并作为状态的一个维度（B 代码里就是 mask 位与 key_mask 或合并后进 dist[r][c][mask]），A 是纯状压 DP 模板，B 在此之上把该位掩码状态放到网格 BFS 状态图里，再叠加门的钥匙位判定与首次到达即最短路。"},{"id":"pre:roj/3211->roj/1494","source":"roj/3211","target":"roj/1494","type":"pre","directed":true,"reason":"B 的最小环把 A 教的 Floyd 第 k 轮不变量（dist 只含编号小于 k 的中转点）当作受限最短路直接复用，用它把环拆成 i→j 路径加两条到 k 的边"},{"id":"pre:luogu/P1195->roj/1489","source":"luogu/P1195","target":"roj/1489","type":"pre","directed":true,"reason":"B 直接沿用 A 教的按边权升序、并查集合并不同连通块这一步作为合并顺序，只是把合并动作改成用块大小乘积批量累计路径最大边权，并叠加了补边下界 M+1 的新观察。"},{"id":"pre:luogu/P2330->roj/1489","source":"luogu/P2330","target":"roj/1489","type":"pre","directed":true,"reason":"B 的观察二直接复用 A 教的「按边权升序 Kruskal 合并、已处理的更小边保证块内连通」这一步，只把 A 的「最后一条入树边权即最小瓶颈」改写为按块大小乘积 a×b 批量统计所有点对的路径最大边权，再叠加 M+1 与唯一性证明。"},{"id":"pre:luogu/P3366->roj/1489","source":"luogu/P3366","target":"roj/1489","type":"pre","directed":true,"reason":"B 的观察二把 A 教的「边权升序 + 并查集合并两个不同连通块」当作骨架，合并时用块大小乘积批量累计每个点对的路径最大边权，再叠加 A 未教的「补边权取下界 M+1」这一新观察。"},{"id":"pre:luogu/P8306->roj/1484","source":"luogu/P8306","target":"roj/1484","type":"pre","directed":true,"reason":"A 教的是 Trie 插入时沿路径前进、节点代表根到该点的前缀；B 把病毒串按同一路径建成 Trie，并用“节点代表一类安全前缀”这一状态设计压缩无穷前缀，再叠加 A 未教的 fail 链与 Kahn 判环。"},{"id":"pre:roj/3041->roj/1484","source":"roj/3041","target":"roj/1484","type":"pre","directed":true,"reason":"B 的 AC 自动机前半段直接复用 A 教的「从根逐字符下走、缺边新建儿子」的 Trie 建树骨架与根到节点路径即前缀的对应，只是把 A 的共享前缀计数结点升级成安全前缀等价类，再叠加 fail 指针、危险标记传播与安全子图 Kahn 判环。"},{"id":"pre:luogu/P3808->roj/1484","source":"luogu/P3808","target":"roj/1484","type":"pre","directed":true,"reason":"B 直接复用 A 教的三步 AC 自动机模板（Trie 插串、BFS 建 fail、缺失边补全转移），并把 A 的补全转移 `trie[u][c]=trie[fail[u]][c]` 搬成 01 字母表的 `TO[v][b]=TO[fail[v]][b]`，补全后每个节点读 0/1 唯一落点，才得到能判环的完整转移图，额外叠加的是 fail 链传播危险标记与 Kahn 拓扑判环这一新流程。"},{"id":"pre:luogu/P1294->roj/1483","source":"luogu/P1294","target":"roj/1483","type":"pre","directed":true,"reason":"B 直接沿用 A 教的「mask 记录已处理集合 + 一维记录当前结尾」这步：A 用 visited 点集加终点 u，B 换成已用串集合 mask 加结尾串 j，转移同样是把新元素并入 mask。"},{"id":"pre:luogu/P1433->roj/1483","source":"luogu/P1433","target":"roj/1483","type":"pre","directed":true,"reason":"B 的最短母串状压 DP 逐字沿用了 A 教的「mask 记已选集合 + 额外记最后位置」状态设计（B 代码里 best(mask, j) 与 mask|1<<k 累积），只把距离代价换成补尾重合、并叠加字典序 tie-break。"},{"id":"pre:roj/3003->roj/1483","source":"roj/3003","target":"roj/1483","type":"pre","directed":true,"reason":"B 直接沿用 A 教的\"位掩码集合 + 结尾元素\"状态设计，把已走点集换成已用串集合并把终点 j 换成结尾串，只是另加补尾表和字典序 tie-break"},{"id":"pre:roj/3042->roj/1478","source":"roj/3042","target":"roj/1478","type":"pre","directed":true,"reason":"B 用树上前缀异或把路径异或化成端点 d 值后，把 A 教的「01-Trie 逐位贪心走相反位求最大异或对」原样套用到 d 数组上（代码 max_xor 与 A 的 best_xor 同一判定），只额外叠加了求根前缀异或 d 的前置流程。"},{"id":"pre:roj/1472->roj/1478","source":"roj/1472","target":"roj/1478","type":"pre","directed":true,"reason":"B 先新增树上前缀异或把任意路径异或塌缩成 d[u]^d[v]，然后在\"求最大异或对\"这一层原样复用 A 教的 01 字典树反向边贪心下降（B 代码 insert/max_xor 与 A 完全同构，正文自述\"这一步即上一题 [[roj/1472]] 的结论\"\"贪心结论原样复用\"），属基础模板到叠加新流程的台阶。"},{"id":"pre:luogu/P8306->roj/1479","source":"luogu/P8306","target":"roj/1479","type":"pre","directed":true,"reason":"A 教的 Trie 插入与稀疏边表正是 B 建 AC 自动机的必经骨架，B 在其上叠加 fail 指针和 fail 树逆序传播才达到正解难度。"},{"id":"pre:luogu/P3808->roj/1479","source":"luogu/P3808","target":"roj/1479","type":"pre","directed":true,"reason":"B 把 A 教的「沿 fail 链的终止节点就是以当前位置结尾的模式串」这一观察改写成 fail 树上的等价形式：先标记到达状态，再沿 fail 边自底向上合并命中，额外叠加了 BFS 逆序传播这一阶段，而 Trie 插串、BFS 建 fail、扫描文本以及词尾计数累加处理重复串按编号各计一次都沿用 A。"},{"id":"pre:roj/3041->roj/1479","source":"roj/3041","target":"roj/1479","type":"pre","directed":true,"reason":"B 的 build_automaton 直接复用 A 教的 Trie 建树骨架——从根逐字符下走、缺边新建儿子，并把计数挂在结尾结点（cnt[node] += 1，使「重复查询词各算一次」）；A 的原文也点明后续 AC 自动机建立在这套骨架上，B 只是在此之上叠加 A 未教的 fail 指针、BFS 求 fail 与 fail 树逆序传播命中。"},{"id":"pre:luogu/P8306->roj/1476","source":"luogu/P8306","target":"roj/1476","type":"pre","directed":true,"reason":"B 建 01-Trie 时对每条信息沿路径把经过节点累加 through，正是 A 教的「插入时让每个经过节点 cnt 加一」那一步，B 在此基础上额外维护 end 并做查询时沿途 end 累加与终点 through-end 的补算。"},{"id":"pre:roj/3041->roj/1476","source":"roj/3041","target":"roj/1476","type":"pre","directed":true,"reason":"B 的第 1 类统计就是照搬 A 教的「沿查询路径累加结尾计数」，只在其上补了终点 through-end 来处理反向（信息比密码长）的前缀情形"},{"id":"pre:roj/3041->roj/1473","source":"roj/3041","target":"roj/1473","type":"pre","directed":true,"reason":"B 的 01-Trie 直接把 A 教的「从根下走、边不存在就新建儿子」建树骨架搬到位串上（main.cpp 的 trie_insert：nxt==0 就 ++top_node 挂边），再叠加 A 未教的反向位贪心查询与分割点拆分。"},{"id":"pre:roj/3041->roj/1472","source":"roj/3041","target":"roj/1472","type":"pre","directed":true,"reason":"B 的 01 字典树建树（main.py 的 insert）直接复用 A 教的 Trie 插入骨架——逐层从根下走、边存在就复用、缺边就新建儿子，只是把字符换成二进制位；在此之上 B 才叠加 A 未教的高位优先贪心（每层优先走反向边）求最大异或对。"},{"id":"pre:luogu/P8306->roj/1472","source":"luogu/P8306","target":"roj/1472","type":"pre","directed":true,"reason":"B 的正解（main.py 的 insert/best_xor，对应原文第二步）直接复用 A 教的前缀节点路径：A 用路径定位前缀并计数，B 把数按位插入 01 字典树后，用“相反孩子子树是否存在”等价判断该 01 前缀是否存在，再叠加 A 未教的高位优先贪心与最大异或扩展。"},{"id":"pre:roj/1485->roj/1470","source":"roj/1485","target":"roj/1470","type":"pre","directed":true,"reason":"B 的 main.py 逐字复用 A 在 BFS 建 fail 时沿 fail 链下传词尾信息这一步（A 下传 dead 命中标记、B 下传 end 词长），并同样照抄 A 的缺边补全 ch[u][c]=ch[fail[u]][c]，让每字符 O(1) 转移，再在此之上叠加 A 未教的「字符栈 kept + 状态栈 states 命中后整段弹出并回退状态」这一新流程。"},{"id":"pre:leetcodecn/longest-palindromic-substring->roj/1462","source":"leetcodecn/longest-palindromic-substring","target":"roj/1462","type":"pre","directed":true,"reason":"B 的朴素解与最终 while 循环都沿用 A 教的以间隙为中心向两侧扩展比较，只把相等判据换成不等，再叠加 Manacher 镜像继承与右端截断；因此 A 的扩展模板确实作为 B 解法的内层一步被复用"},{"id":"pre:luogu/P2419->roj/1383","source":"luogu/P2419","target":"roj/1383","type":"pre","directed":true,"reason":"B 的传递闭包与 A 的位集 Warshall 是同一段代码（reach[i] |= reach[k]），B 再把闭包第 i 列当 pred 复用 A 的“谁能到我”统计，只额外叠加 SCC 源分量判定"},{"id":"pre:roj/3514->roj/1383","source":"roj/3514","target":"roj/1383","type":"pre","directed":true,"reason":"B 的 transitive_closure 逐字复用 A 教的位集可达集与 Floyd 松弛 g[i] |= g[m] 求出 reach，再叠加取列 pred 与 pred 子集 reach 的判定来数源分量。"},{"id":"pre:luogu/B3611->roj/1383","source":"luogu/B3611","target":"roj/1383","type":"pre","directed":true,"reason":"B 的 bitset 正解第三步直接复用 A 教的 Warshall 位集转移：A 的「i 能到 k 就把 k 的整行可达集并入 i」在 B 代码里就是 `reach = [r | rk if r >> k & 1 else r for r in reach]`，B 在此闭包之上再叠加取转置列 pred、判 pred⊆reach 与 SCC 去重计数的求源分量流程。"},{"id":"pre:luogu/U661986->roj/1299","source":"luogu/U661986","target":"roj/1299","type":"pre","directed":true,"reason":"B 把 A 教的「同一状态只保留最优值、每件物品只选一次」照搬到一维 0/1 背包上：dp[j] 同样只留最优总和并用旧表做一次取/不取转移（代码 old[from]+value 更新 dp[j]），只是把状态从容量换成总和模 K 的余数，并叠加环形回绕 (j - r + k) % k。"},{"id":"pre:roj/1294->roj/1299","source":"roj/1294","target":"roj/1299","type":"pre","directed":true,"reason":"A 教的「一维 dp 表上逐件物品做每件只用一次的 0/1 转移」正是 B 的每一轮循环：B 只把状态维从容量换成模 K 的余数，并用旧表快照代替倒序枚举来维持 0/1 语义，再叠加同余数取最大总和与答案取 dp[0] 的建模。"},{"id":"pre:roj/1281->roj/1286","source":"roj/1281","target":"roj/1286","type":"pre","directed":true,"reason":"最长上升子序列的标准 DP/二分模板；本题把下降方向取负后，复用的就是同一套计算。"},{"id":"common:luogu/P1571--roj/1286","source":"luogu/P1571","target":"roj/1286","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：B 的二分 LIS 里 lower_bound_pos 复用的正是 A 教的「在有序序列里找第一个 >= x 的位置」这一步，只是从判断元素存在变成在 tails 上定位替换点，再叠加取负转 LIS 与正反两方向取最大这两个额外流程。"},{"id":"common:roj/1283--roj/1286","source":"roj/1283","target":"roj/1286","type":"common","directed":false,"reason":"同样对序列正反各做一次 LIS，区别是登山要在峰顶把上升、下降两段拼成一条路线，本题只能取其中一个方向。"},{"id":"pre:luogu/P2678->roj/1278","source":"luogu/P2678","target":"roj/1278","type":"pre","directed":true,"reason":"B 的判定层直接搬用 A 的固定阈值贪心扫描计数：把「最少需删除石头数≤M」换成「最少段数≤k」再二分，随后叠加 A 未教的倒序贪心重建字典序最小方案"},{"id":"pre:luogu/P1182->roj/1278","source":"luogu/P1182","target":"roj/1278","type":"pre","directed":true,"reason":"B 的判定层 people_needed(L) 原样复用 A 的「能装就装」贪心数最少段数，只把段数≤M 换成人数≤k，再叠加 A 未教的倒序贪心重建字典序最小方案，故 A 是 B 的台阶。"},{"id":"pre:OpenJ_Bailian/4135->roj/1278","source":"OpenJ_Bailian/4135","target":"roj/1278","type":"pre","directed":true,"reason":"B 的判定层直接复用 A 的 check(X)：在二分出的上限下从左到右「能装就装」贪心数段，只把段数 ≤M 换成人数 ≤k，再叠加倒序贪心重建字典序最小方案。"},{"id":"pre:luogu/U661988->roj/1272","source":"luogu/U661988","target":"roj/1272","type":"pre","directed":true,"reason":"B 的分组背包直接复用 A 讲的「容量倒序读不到当前物品」这一具体判定：main.cpp 里 cap 从 V 递减，使 dp[cap-W_j] 只含上一组的旧值，从而同组物品不在同一格叠加、天然满足每组最多选一件；A 只用倒序说明 0/1 与完全背包的差异，B 在此基础上叠加「外层枚举组、内层枚举组内物品」的三层流程，难度从入门升到普及-。"},{"id":"pre:roj/1294->roj/1272","source":"roj/1294","target":"roj/1272","type":"pre","directed":true,"reason":"B 的分组背包直接沿用 A 教的「一维 f[v] + 倒序枚举容量避免物品重复叠加」这一步，只在其外再套一层枚举组的循环"},{"id":"pre:luogu/U661986->roj/1272","source":"luogu/U661986","target":"roj/1272","type":"pre","directed":true,"reason":"B 直接复用 A 教的容量倒序枚举，把倒序读旧值的性质从「每件物品只选一次」推广到「每组最多选一件，同组物品不叠加」。"},{"id":"pre:roj/1272->roj/1266","source":"roj/1272","target":"roj/1266","type":"pre","directed":true,"reason":"B 的机器分配就是 A 教的分组背包：每家公司为一组、组内枚举分到几台，再在转移处记 res 回溯方案"},{"id":"pre:luogu/U661986->roj/1266","source":"luogu/U661986","target":"roj/1266","type":"pre","directed":true,"reason":"B 的分组背包沿用 A 教的「相同容量只保留最大价值」这一最优子结构，把一维容量 dp 扩成前 i 组恰好分 j 台的二维 dp，再叠加 res 记决策回溯"},{"id":"pre:luogu/P1757->roj/1266","source":"luogu/P1757","target":"roj/1266","type":"pre","directed":true,"reason":"B 明说自己的转移「即分组背包」，把 A 教的「同组物品互斥、每组最多选一件」直接改成每家分公司为一组、组内只在 0..M 台中选一个 k，用 f[i][j]=max_k f[i-1][j-k]+a[i][k] 套用 A 的组内互斥状态设计，只在其上叠加 res 决策表记录取最优的 k 并倒推输出方案。"},{"id":"pre:leetcodecn/pascals-triangle->roj/1258","source":"leetcodecn/pascals-triangle","target":"roj/1258","type":"pre","directed":true,"reason":"数字金字塔把杨辉三角「由相邻两数相加得下一格」的二维递推骨架换成相邻两数取 max 再累加自身，逐层递推思路直接迁移，只是多了求最大值的取优步骤"},{"id":"pre:roj/1281->roj/1260","source":"roj/1281","target":"roj/1260","type":"pre","directed":true,"reason":"B 的 lis_length 直接复用 A 的 tails+二分找首个 >= v 的定位步骤，只需追加 strict 参数切换 bisect_left/bisect_right，就分别求出第一问的最长不升子序列与第二问的最长严格上升子序列。"},{"id":"common:luogu/P1571--roj/1260","source":"luogu/P1571","target":"roj/1260","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：B 的二分 LIS 直接复用 A 教的「找有序序列中第一个 >= x 的位置」做替换，只是换成在 tails 数组上维护，再叠加 Dilworth 定理完成第二问"},{"id":"pre:roj/1255->roj/1253","source":"roj/1255","target":"roj/1253","type":"pre","directed":true,"reason":"B 的 main.cpp 直接复用 A 教的「无权图 BFS 按层扩展、首次入队即最短步数、dist=-1 兼作访问标记」这一步，只把网格四方向换成 -1/+1/*2 三种操作，并叠加 [0,2K] 搜索上界与 K<=N 时答案 N-K 的收口"},{"id":"pre:roj/1255->roj/1257","source":"roj/1255","target":"roj/1257","type":"pre","directed":true,"reason":"A 教的「BFS 按层扩展、格子第一次入队即最短步数、入队即标记」正是 B 判定马步最少步数的核心一步，B 只把四方向邻接换成 8 个马步落点、单终点答案扩成多组样例，仍靠首次到达终点即返回该层步数。"},{"id":"pre:roj/1255->roj/1252","source":"roj/1255","target":"roj/1252","type":"pre","directed":true,"reason":"M4 反转修正：原登记为 roj/1252->roj/1255，但原 reason 自述的学习顺序与此相反。原 reason 自述「同模型的最简版本…本题在其上增加路径还原」，即 roj/1255（迷宫问题, r0）是基、roj/1252（走迷宫, r1）是加强。箭头确实写反，反转后 Δ=1 合规。"},{"id":"pre:roj/8016->roj/1250","source":"roj/8016","target":"roj/1250","type":"pre","directed":true,"reason":"B 的 main.cpp 完全照搬 A 教的「扫到未访问格 +1 并洪水填充整块」计数连通分量，只在其上叠加墙位标记解码与最大面积统计"},{"id":"pre:roj/1329->roj/1250","source":"roj/1329","target":"roj/1250","type":"pre","directed":true,"reason":"B 直接复制 A 教的「扫描种子格 + 洪水填充整块并计数」骨架，只把连通条件换成墙位判定、再额外统计分量大小求最大值"},{"id":"pre:roj/1335->roj/1250","source":"roj/1335","target":"roj/1250","type":"pre","directed":true,"reason":"B 的 main.cpp 外层双循环与 A 教的「扫到未访问格就开新分量并 flood fill 淹没」完全同构，只把 A 的四连通黑格邻接换成位掩码判墙，再叠加统计分量大小与取最大面积。"},{"id":"pre:roj/1255->roj/1248","source":"roj/1255","target":"roj/1248","type":"pre","directed":true,"reason":"B 原样复用 A 教的「BFS 首次入队即最短步数、入队兼作访问标记」这一层号求最短路步骤，只把四方向二维网格换成六方向三维网格，无新增算法步骤，属模板级复用。"},{"id":"pre:roj/1335->roj/1248","source":"roj/1335","target":"roj/1248","type":"pre","directed":true,"reason":"B 的三维网格 BFS 在六方向扩展时沿用 A 明确点出的「入栈/入队即标记」这一状态设计（先判界再判障碍、命中即就地标记 seen），只是把连接块计数换成 BFS 层号即最短分钟数并叠加三维偏移与多组数据解析，属模板级复用。"},{"id":"pre:roj/1329->roj/1248","source":"roj/1329","target":"roj/1248","type":"pre","directed":true,"reason":"B 的三维六方向 BFS 直接沿用 A 明确点名为「可迁移实现习惯」的入队即标记状态设计（B 代码在 push 前写 seen=true），保证每格只展开一次，只在其上叠加层号当最短分钟数、多组数据解析与三维坐标偏移扩展。"},{"id":"pre:noi_openjudge/ch0111-02->roj/1246","source":"noi_openjudge/ch0111-02","target":"roj/1246","type":"pre","directed":true,"reason":"B 直接复用 A 的实数二分求根模板：A 靠端点函数值异号定位唯一零点，B 靠弦长与 L 的大小比较在同一单调区间上缩半求 θ，只是外面多套了一层弧长弦长的几何建模。"},{"id":"pre:luogu/P1182->roj/1247","source":"luogu/P1182","target":"roj/1247","type":"pre","directed":true,"reason":"B 的 need_remove 判定把 A 教的「固定上限 + 线性贪心 + 贪心计数即最优」整步搬过来：A 靠它证明最少段数不超过 M 就可二分，B 用同一条论证（保留点越靠左后面越松，故贪心即最少移除数）在区间 [1,L] 上判 need_remove(x)<=M，再额外叠加末跳 limit 的端点收口条件。"},{"id":"pre:luogu/P1824->roj/1247","source":"luogu/P1824","target":"roj/1247","type":"pre","directed":true,"reason":"B 的 need_remove 判定直接复用 A 教过的「按最小间距从左到右保留最早可行点」这一线性贪心扫描，只是把段间距离换成相邻间距并叠加终点侧收口"},{"id":"pre:OpenJ_Bailian/4135->roj/1247","source":"OpenJ_Bailian/4135","target":"roj/1247","type":"pre","directed":true,"reason":"A 教的是「二分上限 + 从左到右贪心填满当前段、数段数≤M」这套判定模板，B 的 check 直接套用同一线性贪心判定结构，只把段和换成相邻间距、段数换成最少删点数，再叠加末跳 limit 收口这一额外流程。"},{"id":"pre:roj/1326->roj/1234","source":"roj/1326","target":"roj/1234","type":"pre","directed":true,"reason":"B 的 qpow（res=res*a%mod、a=a*a%mod、e>>=1）就是 A 教的二进制分解快速幂逐位平方、位为 1 累乘这一步，只是额外叠加指数取末四位再取模的数论化简"},{"id":"pre:roj/1305->roj/1224","source":"roj/1305","target":"roj/1224","type":"pre","directed":true,"reason":"B 的 best_segment/内层循环直接复用 A 教的“单飞还是接上”转移，只是把它从整段数组搬到枚举行区间后的列和数组上跑"},{"id":"pre:luogu/P8687->roj/1221","source":"luogu/P8687","target":"roj/1221","type":"pre","directed":true,"reason":"A 教的「不围绕选哪些包、而围绕已覆盖集合设计状态，dp[mask]=覆盖该集合的最少数量」被 B 直接继承：B 明确写出把「分组方案」的枚举换成「集合覆盖」的枚举，代码里 f[S]=1+min f[S\\T] 就是同一状态设计，只是在其上叠加冲突图/独立集建模与 t=(t-1)&S 的子集枚举。"},{"id":"pre:luogu/P1605->roj/1220","source":"luogu/P1605","target":"roj/1220","type":"pre","directed":true,"reason":"B 直接复用 A 教的「前进时标记、回溯时撤销」这一 DFS 回溯步骤，把 vis 布尔标记换成 used[word] 的次数计数（每词至多用两次），在此外层再叠加重叠增益预处理、增益降序枚举与 slack 上界剪枝。"},{"id":"pre:roj/1318->roj/1200","source":"roj/1318","target":"roj/1200","type":"pre","directed":true,"reason":"A 用 start 下界保证非递减使每个无序拆分只数一次，B 把同一「下界收紧」去重步骤迁移到因数分解 f(n,lo) 并叠加记忆化与因子枚举"},{"id":"pre:luogu/P1775->roj/1197","source":"luogu/P1775","target":"roj/1197","type":"pre","directed":true,"reason":"B 的区间划分类 DP 直接复用 A 教的「按最后一步把区间拆成 [l,k] 与 [k+1,r] 枚举断点」这一转移骨架（dp[前段]+段代价取 min、段代价用前缀和求），只是把「合并成一堆的代价」换成分好的「段内到中位村的距离和 cost(l,r)」，并额外叠加 A 未教的连续分段等价性与中位村选址两步新观察。"},{"id":"pre:roj/1335->roj/1191","source":"roj/1335","target":"roj/1191","type":"pre","directed":true,"reason":"B 的按天分层多源 BFS 在四邻扩展时直接沿用 A 教的越界+目标格判定后在入队时立即标记的状态设计，只是把单一淹水源换成全体患者作初始层、把层数当天数；A 的连通块计数主难点（外层开新块）B 未复用，故为模板级。"},{"id":"pre:roj/1255->roj/1191","source":"roj/1255","target":"roj/1191","type":"pre","directed":true,"reason":"B 在多源 BFS 中复用 A 教的「入队时即登记访问标记、每格至多入队一次」这一步来标记已染病格，只是把单起点换成多个 @ 起点并叠加上按天分层的定长出队批次。"},{"id":"pre:roj/1329->roj/1191","source":"roj/1329","target":"roj/1191","type":"pre","directed":true,"reason":"B 的多源 BFS 扩散直接沿用 A 教的入队即标记（每格至多入队一次，原地改格子当访问标记），只是在其上叠加按天分层同步这一额外流程"},{"id":"pre:noi_openjudge/ch0110-03->roj/1180","source":"noi_openjudge/ch0110-03","target":"roj/1180","type":"pre","directed":true,"reason":"B 的排序步骤直接复用 A 教的「负号翻转成降序 + 元组承载次级键」这一两级排序键写法，只是把姓名换成报名号，再在其上叠加取第 ⌊m×150/100⌋ 名定线、按分数统计同分者并输出前缀的额外流程。"},{"id":"pre:noi_openjudge/ch0110-01->roj/1180","source":"noi_openjudge/ch0110-01","target":"roj/1180","type":"pre","directed":true,"reason":"B 先按成绩降序排好序，再取第 r 名选手的成绩当分数线（代码 line = p[r].score），正是 A 教的「排序后按名次取元素」这一步的直接复用；B 在其上叠加了双关键字排序与「同分选手一并入面、按分数统计前缀」的新流程。"},{"id":"pre:roj/1179->roj/1180","source":"roj/1179","target":"roj/1180","type":"pre","directed":true,"reason":"A 教的「三项排序键一次排序后取前几名」被 B 迁移为按 (成绩降序, 报名号升序) 排序再输出排序前缀，B 只是在其上叠加分数线 ⌊1.5m⌋ 与同分扩张统计的额外流程。"},{"id":"pre:noi_openjudge/ch0105-34->roj/1173","source":"noi_openjudge/ch0105-34","target":"roj/1173","type":"pre","directed":true,"reason":"B 复用了 A 教的滚动递推 f_i=f_{i-1}*i 这一具体步骤并累加，只是把整数乘法换成高精度数组乘加以支持 n≤50"},{"id":"pre:luogu/P1075->roj/1154","source":"luogu/P1075","target":"roj/1154","type":"pre","directed":true,"reason":"B 的 sigma(n) 直接复用 A 教的因子成对、小因子不超过 sqrt(n) 的观察，把只枚举 sqrt(n) 一侧的试除扩成求全部真因子和"},{"id":"pre:roj/1150->roj/1154","source":"roj/1150","target":"roj/1154","type":"pre","directed":true,"reason":"B 的 sigma(n) 直接沿用 A 教过的因子成对试除求真因子和，只是把它从区间内逐个判完全数改成对 sigma(a) 再做一次复合判定"},{"id":"pre:roj/1323->luogu/P1325","source":"roj/1323","target":"luogu/P1325","type":"pre","directed":true,"reason":"B 的雷达安装正解先按右端点从小到大排序，再沿用 A 教的 last 边界扫描：A 用 begin_i >= last_end 判断活动相容，B 用 l_i > last 判断区间尚未被覆盖并新建雷达，只是外层多叠加了小岛到区间的几何转化与 -1 判断。"},{"id":"pre:roj/1324->luogu/P1325","source":"roj/1324","target":"luogu/P1325","type":"pre","directed":true,"reason":"B 的雷达安装正解把每个小岛转成 x 轴覆盖区间后，逐字沿用 A 教的右端点升序扫描与取右端点选点贪心，只是在前面加了区间转化与 -1 判定。"},{"id":"common:luogu/P1325--luogu/P1803","source":"luogu/P1325","target":"luogu/P1803","type":"common","directed":false,"reason":"同样按右端点排序的区间贪心：一个是区间调度求最多不相交，一个是区间选点求最少刺穿点。"},{"id":"pre:luogu/P3391->luogu/P4036","source":"luogu/P3391","target":"luogu/P4036","type":"pre","directed":true,"reason":"B 复用 A 的隐式 FHQ Treap 按 size 排名 split/merge 步骤维护可插字符的字符串，再叠加子树哈希与二分 LCQ"},{"id":"pre:roj/1109->luogu/P17414","source":"roj/1109","target":"luogu/P17414","type":"pre","directed":true,"reason":"B 复用 A 教的「按倍数循环做异或翻转」原语：确定要选操作 i 后同样遍历 i 的所有倍数并把当前影响异或翻转，只是额外叠加了按下标递增唯一决定所选操作的贪心消元。"},{"id":"pre:roj/3041->luogu/P17416","source":"roj/3041","target":"luogu/P17416","type":"pre","directed":true,"reason":"B 的查询过程复用了 A 教的「从根沿 Trie 逐字符（此处换成二进制位）下走」的遍历骨架，只是把路径累加计数换成按位贪心选相反分支、取前 k-1 大异或和的新流程。"},{"id":"pre:leetcodecn/daily-temperatures->roj/20025","source":"leetcodecn/daily-temperatures","target":"roj/20025","type":"pre","directed":true,"reason":"B 用单调栈维护后缀最小值阶梯拐点并弹出结算，沿用 A 的单调栈弹出结算候选这一骨架，但核心难度在容斥与奇偶 DP"},{"id":"pre:shumeng/CSP201312C->roj/20025","source":"shumeng/CSP201312C","target":"roj/20025","type":"pre","directed":true,"reason":"B 完全复用 A 的『弹出后新栈顶就是左边第一个更矮位置』这一步：用 cur=弹出后栈顶界定后缀最小值阶梯 min=a[i] 的最后一段并复用 f[cur]，再叠加容斥奇偶 DP 与前缀和"},{"id":"pre:roj/3032->roj/20025","source":"roj/3032","target":"roj/20025","type":"pre","directed":true,"reason":"A 教的「弹出后新栈顶就是左边第一个更矮的位置」正是 B 用 cur = 弹出后栈顶来界定后缀最小值阶梯中 min=a[i] 的最后一段、复用 f[cur] 的那一步。"},{"id":"pre:luogu/P3654->roj/20017","source":"luogu/P3654","target":"roj/20017","type":"pre","directed":true,"reason":"B 沿用 A 的网格上枚举起点加直线方向逐格检查这一枚举骨架，扩到 8 方向与最多一次 90 度拐弯的 L 形匹配"},{"id":"pre:roj/3141->roj/20019","source":"roj/3141","target":"roj/20019","type":"pre","directed":true,"reason":"B 先用交换论证把队内顺序消掉，再把「进 1 号还是 2 号窗口」做成背包式 DP，其内层 j 从 s 递减枚举正是 A 教的「容量倒序枚举保证每件至多选一次」，用来让 f[j-a_i] 停留在上一层旧值；B 额外叠加的是 b 降序交换论证与 f[j]=min 最晚吃完时刻的新状态。"},{"id":"pre:luogu/P1507->roj/20019","source":"luogu/P1507","target":"roj/20019","type":"pre","directed":true,"reason":"B 的背包式 DP 直接复用 A 教的『容量倒序枚举保证物品只选一次』：main.cpp 中 j 从 s 递减到 0 并用 f[j-a_i] 更新，正是 A 二维费用背包两维倒序枚举在单维上的同型写法；B 额外叠加的是交换论证消去队内顺序、以及 f[j]=min 最晚吃完时刻的新状态设计。"},{"id":"pre:luogu/P1734->roj/20019","source":"luogu/P1734","target":"roj/20019","type":"pre","directed":true,"reason":"B 把 A 教的「一维 01 背包容量倒序枚举、每件至多选一次」原样用在滚动数组 f 上（for j 从 s 递减），只是外层先叠加 b 降序交换论证与「两窗口累计打饭之和 = 前缀和」的一维状态压缩，主解仍是 0/1 背包式转移。"},{"id":"pre:luogu/P1020->roj/19998","source":"luogu/P1020","target":"roj/19998","type":"pre","directed":true,"reason":"B 的每位置 k 候选解法建立在 A 教的「二分维护尾值求 LIS」这一步之上（v[L] 最小结尾值表+找最大可接长度），再叠加同位置候选分两步评估写回防接龙与双指针优化。"},{"id":"pre:roj/1294->luogu/P1757","source":"roj/1294","target":"luogu/P1757","type":"pre","directed":true,"reason":"B 的分组背包正是把 A 教的倒序枚举容量 01 转移原样搬进组内层，只把外层从物品换成组来叠加组内互斥约束"},{"id":"pre:luogu/U661986->luogu/P1757","source":"luogu/U661986","target":"luogu/P1757","type":"pre","directed":true,"reason":"B 明说自己是“在 01 背包的基础上加一层组的约束”，复用 A 教的容量倒序枚举来保证一件物品不被重复选，只是把外层循环从“物品”换成“组”，用同一轮 dp 快照实现组内互斥。"},{"id":"pre:roj/1294->luogu/B4141","source":"roj/1294","target":"luogu/B4141","type":"pre","directed":true,"reason":"B 的素数分解正解直接复用 A 教的那一步——一维背包倒序枚举容量保证每件物品只用一次，代码 for(j=n;j>=p;j--) dp[j]=max(dp[j],dp[j-p]+1) 就是 A 的 f[v]=max(f[v],f[v-w]+c)；B 只是把物品换成筛出的素数（重量 p、价值 +1）并额外叠加埃氏筛与 dp 初值 -1 的恰好装满处理，A 入门、B 普及-，属台阶式前置。"},{"id":"pre:luogu/U661986->luogu/B4141","source":"luogu/U661986","target":"luogu/B4141","type":"pre","directed":true,"reason":"B 直接把 A 教的『容量倒序枚举保证物品只选一次』当成 0/1 背包骨架，套在素数集合上（重量 p、价值 +1、目标和恰好为 n），再做埃氏筛与恰好装满的不可达初始化"},{"id":"pre:roj/3141->luogu/P1064","source":"roj/3141","target":"luogu/P1064","type":"pre","directed":true,"reason":"B 的正解 main.cpp 里 `for (int j = n; j >= 0; j--)` 套 `dp[j]=max(dp[j],dp[j-cost]+val)` 直接复用了 A 教的「一维滚动、容量维倒序枚举以保证每个对象至多被选一次」这一步（A 用 f[j]+=f[j-A_i] 计数、B 用 max 取价值，倒序理由完全一致），B 只是在此之上叠加了附件依赖的 2^k 组合枚举与组内互斥。"},{"id":"pre:luogu/P1910->luogu/P1064","source":"luogu/P1910","target":"luogu/P1064","type":"pre","directed":true,"reason":"B 的正解把每个主件及其附件枚举出的组合当成一个 0/1 物品，仍用 A 教过的「每个对象最多选一次、容量维倒序枚举」这一步（main.cpp 的 for (j=n;j>=0;j--) 套 dp[j]=max(dp[j],dp[j-cost]+val)）；A 未教的台阶是附件依赖带来的 2^k 组合枚举与组内互斥，属 A 之上的叠加而非同难度孪生。"},{"id":"pre:luogu/P1734->luogu/P1064","source":"luogu/P1734","target":"luogu/P1064","type":"pre","directed":true,"reason":"B 把 A 教的「一维 dp 内层容量倒序枚举、每件物品至多选一次」原样用于其 main.cpp（for j=n..0 套 dp[j]=max(dp[j],dp[j-cost]+val)），只是把单个物品换成主件+附件枚举出的组合，多出依赖捆绑这一台阶。"},{"id":"pre:roj/1294->luogu/P1466","source":"roj/1294","target":"luogu/P1466","type":"pre","directed":true,"reason":"B 原样复用 A 教的『倒序枚举容量保证每件物品只用一次』这一步（B 代码 for j=target..i 倒序），只把取 max 价值换成方案数累加，再叠加 A 未教的奇数判 0 与除 2 去重。"},{"id":"pre:luogu/U661986->luogu/P1466","source":"luogu/U661986","target":"luogu/P1466","type":"pre","directed":true,"reason":"B 的 main.cpp 原样复用 A 教的『容量倒序枚举保证每件物品只选一次』这一 0/1 背包骨架，只把转移从取 max 价值换成方案数累加 dp[j]+=dp[j-i]，再叠加总和为奇数判 0 与除以 2 去重两个 A 未教的新步骤。"},{"id":"pre:roj/1294->luogu/P1802","source":"roj/1294","target":"luogu/P1802","type":"pre","directed":true,"reason":"B 在 main.cpp 中每个好友都先加 lose_i 再倒序枚举药水量 j 做转移，直接复用 A 教的「依次处理每件物品并倒序枚举容量」的一维 0-1 背包步骤，只是把 A 的 max(f[v],f[v-w]+c) 改成 max(dp[j]+lose_i,dp[j-use_i]+win_i)，倒序枚举保证每个好友最多只打一次。"},{"id":"pre:luogu/U661986->luogu/P1802","source":"luogu/U661986","target":"luogu/P1802","type":"pre","directed":true,"reason":"B 的解法主体就是 A 教的容量倒序枚举，并原样搬用「倒序枚举使 dp[j-use_i] 保留上一轮旧值、从而保证每件物品只选一次」这一理由（原文：这保证了每个好友最多只打一次），只在其上叠加不打也有 lose_i 的保底收益。"},{"id":"pre:luogu/P1734->luogu/P2370","source":"luogu/P1734","target":"luogu/P2370","type":"pre","directed":true,"reason":"B 的 check(L) 把 A 教的「一维 dp 内层容量倒序枚举、每件物品只选一次」原样搬入（main.cpp 第21-22行 for(j=S;j>=w[i];--j) dp[j]=max(dp[j],dp[j-w[i]]+v[i])），只在外面叠加二分答案与文件大小过滤这一额外台阶。"},{"id":"pre:luogu/P2440->luogu/P2370","source":"luogu/P2440","target":"luogu/P2370","type":"pre","directed":true,"reason":"先掌握二分答案 + 计数 check，再把 check 换成 0/1 背包可行性判定。"},{"id":"pre:roj/3141->luogu/P4141","source":"roj/3141","target":"luogu/P4141","type":"pre","directed":true,"reason":"B 的第一步正是 A 教的「f[j] 表示凑出和 j 的方案数、倒序枚举做 0/1 背包计数」这一模板（B 代码 main.cpp 的 f[j]=(f[j]+f[j-w[i]])%10 与 A 逐字同构），在前置模板之上再叠加 g[j]=f[j]-g[j-w_i] 的补集递推作为新增难点。"},{"id":"pre:roj/3142->luogu/U661993","source":"roj/3142","target":"luogu/U661993","type":"pre","directed":true,"reason":"B 的 s=0 分支直接套用 A 教的「完全背包内层正序枚举使 dp[j-v] 取到本轮值、从而允许同一物品重复取用」这一步，只是把计数式 f[j]+=f[j-i] 换成取 max 的转移，并在同一 dp 上再并列一条 A 同时讲过的 01 倒序分支。"},{"id":"pre:roj/3141->luogu/U661993","source":"roj/3141","target":"luogu/U661993","type":"pre","directed":true,"reason":"B 的 01 物品分支直接搬用 A 教的「倒序枚举容量→dp[j-v] 是上一轮旧值→每件只取一次」，只在其上按 s_i 追加正序的完全背包分支，属于同一 dp 上的台阶式叠加。"},{"id":"pre:luogu/P1510->luogu/U661993","source":"luogu/P1510","target":"luogu/U661993","type":"pre","directed":true,"reason":"B 的混合背包把 A 教的 01 背包「倒序枚举容量保证只取一次」整段搬成 s=-1 分支的转移（main.cpp 的 for(j=V;j>=v;j--) 与 A 的 for(j=c;j>=m;j--) 同形），只在其上追加正序的完全背包分支，属于同一 dp 上的叠加扩展；两者难度 普及/提高- → 普及+/提高-，方向正确。"},{"id":"pre:roj/1294->luogu/U661994","source":"roj/1294","target":"luogu/U661994","type":"pre","directed":true,"reason":"B 把 A 教的『倒序枚举容量保证每件物品只选一次』原样扩成二维费用背包的两维倒序枚举，转移式只是在一维 0-1 背包基础上加一个费用维"},{"id":"pre:luogu/U661986->luogu/U661994","source":"luogu/U661986","target":"luogu/U661994","type":"pre","directed":true,"reason":"A 教的『容量倒序枚举保证每件物品只选一次』被 B 直接用作二维费用背包的两维倒序循环（B 的 main.cpp 中 j、k 两层均倒序，并注明“这和 01 背包倒序的道理完全一样”），B 只在此基础上叠加 A 未涉及的第二个费用维；难度由入门升到普及-，台阶性与方向均成立。"},{"id":"pre:roj/3141->luogu/U661995","source":"roj/3141","target":"luogu/U661995","type":"pre","directed":true,"reason":"B 的分组背包把 A 教的「倒序枚举让物品至多选一次、取旧值」当作组内物品层转移规则，再叠加 previous 冻结上一组状态来实现组内互斥"},{"id":"pre:luogu/P1734->luogu/U661995","source":"luogu/P1734","target":"luogu/U661995","type":"pre","directed":true,"reason":"B 的分组背包把 A 教的「容量倒序遍历让物品至多选一次」当作物品层转移规则，再叠加 previous 冻结上一组状态；倒序是 B 每件物品由未选中态转移所必需的步骤，但仅此还不够隔离同组互斥，组内隔离是 B 新增的台阶。"},{"id":"pre:luogu/P1910->luogu/U662039","source":"luogu/P1910","target":"luogu/U662039","type":"pre","directed":true,"reason":"B 的方案枚举建立在同一 0/1 背包转移上（f[i-1][c-v[i]]+w[i] 与 A 的 dp=max(dp,dp[减去消耗]+收益) 同形，逐件选/不选），再叠加 A 未教的二维 DP 表保留、DFS 回溯枚举所有最优方案与字典序排序。"},{"id":"pre:roj/1266->luogu/U662039","source":"roj/1266","target":"luogu/U662039","type":"pre","directed":true,"reason":"A 教的「背包 DP 表保存了每层每容量的最优值，从末状态倒推即可还原出取到最优的方案」正是 B 的第一步：B 把同一 f 表从物品 N 到 1 倒序回溯，只是把 A 用 res 记决策倒推一条路径改成用前驱等式判定分支、DFS 枚举所有最优路径，并额外叠加求 best_val、对所有最优容量分别回溯与字典序排序。"},{"id":"pre:roj/3141->luogu/U662039","source":"roj/3141","target":"luogu/U662039","type":"pre","directed":true,"reason":"A 教的「把第 i 件物品的决策拆成不选 f[i-1][j] 与选 f[i-1][j-A_i] 两支」正是 B 的 DFS 回溯判定所用的同一对前驱（f[i-1][c] 对应不选、f[i-1][c-v[i]]+w[i] 对应选），B 只是把 A 的计数相加换成相等性检查来枚举所有最优路径，并叠加找最优值、容量遍历与字典序排序。"},{"id":"pre:roj/3141->luogu/U662097","source":"roj/3141","target":"luogu/U662097","type":"pre","directed":true,"reason":"B 的 cnt 计数转移直接套用 A 教的「倒序枚举容量保证每个数至多被选一次」这一步（main.cpp:32 与 A main.py:29 同形），再叠加『更大覆盖、相等累加』与初值全 1 的新流程。"},{"id":"pre:luogu/P1734->luogu/U662097","source":"luogu/P1734","target":"luogu/U662097","type":"pre","directed":true,"reason":"B 的主解 main.cpp 直接搬用 A 教的一维 01 背包「容量倒序枚举 dp[c]=max(dp[c],dp[c-v]+w)、每件至多选一次」这一步（B main.cpp:32 for(c=V;c>=v[i];--c) 与 A main.cpp:36 同形），只在其上叠加 cnt 平行数组的『更大覆盖、相等累加』与初值全 1 这一新流程，属台阶式叠加。"},{"id":"pre:roj/3141->luogu/U662107","source":"roj/3141","target":"luogu/U662107","type":"pre","directed":true,"reason":"B 的 cnt 数组就是 A 教的「恰好为 j 的方案数」表：同样 cnt[0]=1、倒序扫描、cnt[c]+=cnt[c-v] 逐项累加方案数，B 只是在外面加了一层 -INF 可达性判断与 dp 最优值合并来选恰好装满的最优方案，属台阶式叠加。"},{"id":"pre:luogu/P2347->luogu/U662107","source":"luogu/P2347","target":"luogu/U662107","type":"pre","directed":true,"reason":"B 的恰好装满解法直接沿用 A 的「用可达性初值区分可达与不可达」这一步，把 A 的 dp[0]=true 换成 dp[0]=0 可达、其余 -INF 不可达，并再加一层前驱可达性判断来做方案计数（main.cpp:35）。"},{"id":"pre:roj/1294->luogu/U661992","source":"roj/1294","target":"luogu/U661992","type":"pre","directed":true,"reason":"B 的多重背包正解（main.cpp 第 21 行 for (int c = V; c >= 0; --c) 内套 k 循环）原样复用 A 教的『倒序枚举容量保证每件物品至多选一次』这一步，只把每件物品换成展平后的 s_i 个 01 副本并在内层叠加件数枚举，B 文中也明说与标准 01 背包的倒序要求完全一样。"},{"id":"pre:luogu/U661988->luogu/U661992","source":"luogu/U661988","target":"luogu/U661992","type":"pre","directed":true,"reason":"A 讲清的「倒序保证每件物品只用一次」被 B 直接当作展平后 01 物品的枚举方向依据，B 只是在同一 dp[c] 转移上内层叠加件数枚举"},{"id":"pre:luogu/U661986->luogu/U661992","source":"luogu/U661986","target":"luogu/U661992","type":"pre","directed":true,"reason":"B 把每种物品的 s_i 件展平成多个 01 物品后，仍沿用 A 教的『容量倒序枚举保证每件物品只选一次』（main.cpp 第 21 行 for (int c = V; c >= 0; --c)），只是在倒序容量循环内层再枚举选取件数 k，属于在 01 背包模板上叠加一层枚举。"},{"id":"pre:roj/1294->luogu/U663298","source":"roj/1294","target":"luogu/U663298","type":"pre","directed":true,"reason":"B 的 dp[c] 状态定义与「倒序枚举容量保证每件只用一次」这一步直接照搬 A 的 0-1 背包一维写法（代码里同为 for(c=V;c>=v;--c)），只在 dp[0]=1 初始化、加法累加与取模上把求最大价值换成计数恰好装满的方案数。"},{"id":"pre:luogu/U661986->luogu/U663298","source":"luogu/U661986","target":"luogu/U663298","type":"pre","directed":true,"reason":"B 直接沿用 A 教的 01 背包一维 dp[c] 状态与容量倒序枚举（main.cpp 里同一层 `for (c = V; c >= v; --c)` 只把 max 换成加法累加），仅在 dp[0]=1 初始化与取模上把「求最大价值」改成「计数方案数」。"},{"id":"pre:roj/1294->luogu/U663710","source":"roj/1294","target":"luogu/U663710","type":"pre","directed":true,"reason":"B 沿用 A 教的一维 dp[c] 容量状态骨架与「枚举方向决定物品复用次数」这一具体判定，把 A 的倒序翻转成正序，使 dp[c-v] 含当前物品贡献从而实现无限件，再把 A 的 max 转移换成 dp[c]+=dp[c-v] 的组合计数（dp[0]=1、取模）并叠加先物品后容量的组合语义去重。"},{"id":"pre:luogu/U661986->luogu/U663710","source":"luogu/U661986","target":"luogu/U663710","type":"pre","directed":true,"reason":"B 直接沿用 A 的一维容量 dp 与「枚举方向决定物品复用次数」这一步：A 用倒序保证物品只用一次，B 反过来用正序枚举容量，让 dp[c-v] 被当前物品反复更新，从而把同一状态骨架改成完全背包的组合计数。"},{"id":"pre:luogu/U661988->luogu/U663710","source":"luogu/U661988","target":"luogu/U663710","type":"pre","directed":true,"reason":"B 的 main.cpp（第 21-22 行 for(c=v;c<=V;++c)）直接复用 A 教的『容量正序枚举使 dp[c-v] 已含当前物品』这一无限件机制，只把取 max 的价值转移换成 dp[c]+=dp[c-v] 的方案数转移"},{"id":"pre:roj/1604->luogu/U663797","source":"roj/1604","target":"luogu/U663797","type":"pre","directed":true,"reason":"A 教的『队列存下标、队首即窗口最值』被 B 直接用来在余数分组的等差窗口内 O(1) 取最优前驱算 dp[r+kv]，只是先叠加了按体积取模分组与 g 备份。"},{"id":"pre:luogu/P2627->luogu/U663797","source":"luogu/P2627","target":"luogu/U663797","type":"pre","directed":true,"reason":"B 的 main.cpp 按体积余数分组后把多重背包转移整理成 kw + 窗口内 (g[r+jv]-jw) 最大值，正是 A 教的『转移化成常数项加窗口内 dp[j-1]-S[j] 最大值、用单调队列维护右移窗口取队首』这一步的直接套用，只是窗口从 1D 断点 j 搬到等差数列组内下标 k，并叠加余数分组、g 备份这两个 A 未教的新步骤。"},{"id":"pre:luogu/P1725->luogu/U663797","source":"luogu/P1725","target":"luogu/U663797","type":"pre","directed":true,"reason":"B 的多重背包单调队列优化直接复用 A 教的滑窗单调队列三步（过期下标出队、弹掉队尾更差元素、用队头求最优），只是先把容量按体积余数分组成等差数列，使同一套队头转移能 O(1) 取每组窗口最优前驱。"},{"id":"pre:luogu/P3374->luogu/P2345","source":"luogu/P3374","target":"luogu/P2345","type":"pre","directed":true,"reason":"B 沿用 A 的树状数组单点修改与前缀和查询模板维护数量、坐标和，再叠加按听力排序消掉 max 与绝对值拆左右"},{"id":"pre:luogu/P2866->luogu/P7167","source":"luogu/P2866","target":"luogu/P7167","type":"pre","directed":true,"reason":"B 沿用 A 的单调栈弹出所有不超过当前者、栈顶即第一个更高位置这一机制求溢出去向 nxt[i]，再叠加倍增链上跳跃"},{"id":"pre:luogu/P1901->luogu/P7167","source":"luogu/P1901","target":"luogu/P7167","type":"pre","directed":true,"reason":"B 用 A 的递减单调栈步骤求出每个圆盘下方第一个更大者的 nxt 链，再叠加倍增跳链扣容量"},{"id":"pre:luogu/P2947->luogu/P7167","source":"luogu/P2947","target":"luogu/P7167","type":"pre","directed":true,"reason":"B 的第一阶段直接复用 A 的「单调栈求下一个更大元素」这一步求出每个圆盘的溢出去向 nxt[i]，再叠加把链上的容量前缀和塞进倍增表加速询问"},{"id":"pre:luogu/P4735->luogu/P4592","source":"luogu/P4735","target":"luogu/P4592","type":"pre","directed":true,"reason":"B 的版本差机制直接复用 A 的可持久化 Trie 版本思路：A 用 root 版本加 time_id 约束候选下标区间，B 把候选点集表示成版本差、sz 计数差大于 0 才走分支，再叠加 A 未教的 DFS 序区间化子树与根链四版本容斥拆路径。"},{"id":"pre:luogu/P3038->luogu/P4592","source":"luogu/P3038","target":"luogu/P4592","type":"pre","directed":true,"reason":"B 的路径查询把 A 教的路径拆成根路径容斥 u+v−lca−fa[lca] 迁移成可持久化 Trie 的四版本差，子树查询同理复用 A 的 dfn 连续区间子树和"},{"id":"pre:leetcodecn/merge-k-sorted-lists->luogu/P2048","source":"leetcodecn/merge-k-sorted-lists","target":"luogu/P2048","type":"pre","directed":true,"reason":"B 复用 A 的堆维护多路候选弹出即取值再补入下一候选的归并模型，把链表 next 换成弹出后分裂出的两个子区间候选"},{"id":"pre:luogu/P1631->luogu/P2048","source":"luogu/P1631","target":"luogu/P2048","type":"pre","directed":true,"reason":"多路归并思想的基础版本，用堆合并有序流取最小 N 个和"},{"id":"common:acwing/1018--luogu/P1002","source":"acwing/1018","target":"luogu/P1002","type":"common","directed":false,"reason":"同样是只能向右/向下走的网格路径 DP，过河卒是带障碍的路径计数版"},{"id":"pre:luogu/P1255->acwing/1015","source":"luogu/P1255","target":"acwing/1015","type":"pre","directed":true,"reason":"B 把 A 教的「按最后一步分类、从前驱方案递推」这一步从一维爬楼梯推广到二维网格（每格只可能从上方或左方走来，取更优前驱加 a[i][j]），并叠加网格边界与多组数据处理。"},{"id":"pre:leetcodecn/partition-equal-subset-sum->shumeng/CSP202603B","source":"leetcodecn/partition-equal-subset-sum","target":"shumeng/CSP202603B","type":"pre","directed":true,"reason":"B 的普通任务部分复用 A 的 0/1 背包倒序更新这一步：dp[u] 取 max(dp[u], dp[u-c]+g) 的背包转移保证每件任务只选一次，再叠加 A 未教的灵活任务按单位收益贪心与两类任务按杯数合并枚举。"},{"id":"pre:luogu/P2307->shumeng/CSP202603E","source":"luogu/P2307","target":"shumeng/CSP202603E","type":"pre","directed":true,"reason":"B 把每次翻修边解释为 A 教的并查集分量合并（小并大），在其上叠加分量邻接集合维护、端点对双向链表与路径段可行性判定"},{"id":"pre:shumeng/CSP202603E->shumeng/CSP202603E2","source":"shumeng/CSP202603E","target":"shumeng/CSP202603E2","type":"pre","directed":true,"reason":"B 的 X=1 在线分支原文自述复用 Easy 版整体框架：复制维修站边界建局部树、并查集收缩已翻修边、小并大维护分量邻接与端点对记录，X=0 离线分支另叠树剖线段树"},{"id":"pre:roj/3019->shumeng/CSP202605D","source":"roj/3019","target":"shumeng/CSP202605D","type":"pre","directed":true,"reason":"B 的区间调度贪心复用了 A 教的关键观察——都按右端点升序并优先取结束最早的区间：A 据此把点放在线段右端点，B 据此逆序预处理 best_end（最早结束位置）后逐次跳转，只是 B 在此之上又叠加了前缀异或判定与倍增回答询问。"},{"id":"pre:luogu/P2866->shumeng/CSP201312C","source":"luogu/P2866","target":"shumeng/CSP201312C","type":"pre","directed":true,"reason":"B 的最大矩形复用 A 教的「单调栈弹出维护、弹出即结算」这一步（弹后新栈顶给出左界、当前 i 给出右界），再叠加同高旧柱继承扩展范围的处理与末尾 0 哨兵清栈。"},{"id":"pre:luogu/P2947->shumeng/CSP201312C","source":"luogu/P2947","target":"shumeng/CSP201312C","type":"pre","directed":true,"reason":"B 的矩形结算复用 A 的单调栈「被挡住者弹出、弹栈确定边界」这一步：高度低于栈顶就弹出并以当前柱 i-1 定右边界、新栈顶定左边界，再叠加 A 未教的固定最低柱建模、相等高度处理与 0 高度哨兵。"},{"id":"pre:luogu/P2224->shumeng/CSP201403E","source":"luogu/P2224","target":"shumeng/CSP201403E","type":"pre","directed":true,"reason":"B 复用 A 的按机器累计负载设状态并在全部状态上取 max(负载) 最小值的模型，只是从两机扩到两 CPU 加 GPU 再加串行代价"},{"id":"pre:roj/1329->shumeng/CSP201409D","source":"roj/1329","target":"shumeng/CSP201409D","type":"pre","directed":true,"reason":"B 的多源 BFS 直接复用 A 教的网格 BFS「入队即标记、每格至多入队一次」这一状态设计（B 代码同样在入队时写 dist 并靠 dist!=INF 阻止重复入队），据此保证每个格点只被首次到达一次，从而把首次到达的层数当作到最近分店的最短距离，只在其上叠加多源起点与按需求量计费。"},{"id":"pre:roj/1255->shumeng/CSP201409D","source":"roj/1255","target":"shumeng/CSP201409D","type":"pre","directed":true,"reason":"B 复用 A 教的「BFS 首次到达即最短步数」这一关键观察，只是把单源起点换成所有分店同时入队，再按需求量线性加权"},{"id":"pre:HDU/1213->shumeng/CSP201412D","source":"HDU/1213","target":"shumeng/CSP201412D","type":"pre","directed":true,"reason":"B 的 Kruskal 循环复用 A 教的「并查集 find 判是否同一集合、不同才合并」这一步来决定每条边取舍，再叠加 A 未教的按费用排序贪心与最小生成树选边计费。"},{"id":"pre:luogu/P1551->shumeng/CSP201412D","source":"luogu/P1551","target":"shumeng/CSP201412D","type":"pre","directed":true,"reason":"B 的 Kruskal 复用 A 的并查集合并与代表元判连通步骤，按费用序决定每条水渠选或不选"},{"id":"pre:luogu/P3367->shumeng/CSP201412D","source":"luogu/P3367","target":"shumeng/CSP201412D","type":"pre","directed":true,"reason":"B 的 Kruskal 用 A 教的并查集基本操作（查是否同属一个集合、不同则合并）维护连通块并判成环，再叠加 A 未教的边排序贪心与最小生成树计费。"},{"id":"pre:roj/1255->shumeng/CSP201503D","source":"roj/1255","target":"shumeng/CSP201503D","type":"pre","directed":true,"reason":"B 的 bfs() 直接复用 A 教的「无权图 BFS 按层扩展、首次入队即最短步数、dist=-1 兼作访问标记」这一步来量出到各点距离，两次 BFS 正是靠它找最远节点 u 再量出树直径，只是在此外层叠加了把电脑当叶子的建树与直径观察；A 特有的前驱回溯未被使用，故为模板级复用。"},{"id":"pre:roj/1485->shumeng/CSP201509E","source":"roj/1485","target":"shumeng/CSP201509E","type":"pre","directed":true,"reason":"B 复用 A 在 BFS 建 fail 时沿 fail 链下传词尾信息这一步：A 下传 dead 命中标记，B 在同一处把 score 沿 fail 链累加，使落点的命中/得分信息完整，再叠加矩阵快速幂。"},{"id":"pre:roj/1329->shumeng/CSP201512C","source":"roj/1329","target":"shumeng/CSP201512C","type":"pre","directed":true,"reason":"B 的填充操作直接复用 A 教的四连通 flood fill 从种子格扩散并维护访问标记这一步，main.cpp 的 flood_fill 同样是 dx/dy 四方向入队，只是把可扩散条件从「不是 '0'」换成「不是线段 - | +」，并叠加画线模拟与坐标倒序输出这些新流程，难度仅升一级（入门→普及-）。"},{"id":"pre:roj/1255->shumeng/CSP201604D","source":"roj/1255","target":"shumeng/CSP201604D","type":"pre","directed":true,"reason":"A 教的「无权网格 BFS 第一次入队即最短步数」正是 B 第二段所依赖的同一结论：B 只在第一阶段叠加时间维动态危险判定，第二阶段网格静态后仍按首次到达即最短的规则从多源点扩展求剩余步数，并额外补上多源点初值时刻与答案拼接。"},{"id":"pre:roj/1635->shumeng/CSP202104E","source":"roj/1635","target":"shumeng/CSP202104E","type":"pre","directed":true,"reason":"B 的换乘会面时刻就是照搬 A 教的非互质同余合并步骤：先看偏移差能否被 gcd 整除，再约去公因子求逆元取模，只是外面再套一层时间依赖 Dijkstra"},{"id":"pre:luogu/P1075->shumeng/CSP202312B","source":"luogu/P1075","target":"shumeng/CSP202312B","type":"pre","directed":true,"reason":"B 复用 A 的「从 2 开始逐个试除到根号分解质因子」这一步，只是扩展到统计每个质因子的指数并按阈值 k 取舍，外加除尽后剩余大质因子的兜底处理"},{"id":"pre:luogu/P1294->shumeng/CSP202312E","source":"luogu/P1294","target":"shumeng/CSP202312E","type":"pre","directed":true,"reason":"B 复用 A 的集合加末位 (mask,u) 状压状态设计做两侧最大长度 DP，再叠加折半与位集合并"},{"id":"pre:shumeng/CSP202406A->shumeng/CSP202406B","source":"shumeng/CSP202406A","target":"shumeng/CSP202406B","type":"pre","directed":true,"reason":"B 的主解完全建立在 A 教的「一维行优先序列 + 线性下标映射、重塑不动数组只改形状」这一步上（重塑 O(1) 改 n,m、查询 x*m+y），再叠加转置时按下标重排与 n,m 交换的新操作。"},{"id":"pre:roj/1126->shumeng/CSP202406B","source":"roj/1126","target":"shumeng/CSP202406B","type":"pre","directed":true,"reason":"B 的转置步骤复用了 A 教的 (i,j)->(j,i) 下标交换：brute 直接照搬二维交换，main 把它改写成 i*m+j -> j*n+i 的一维映射。"},{"id":"pre:luogu/P1135->shumeng/CSP202412D","source":"luogu/P1135","target":"shumeng/CSP202412D","type":"pre","directed":true,"reason":"B 的关键优化建立在 A 教的「BFS 首次到达时距离已经最小」这一步上（第一次覆盖落点即最优、之后不可能更短，故每个落点只处理一次），再叠加 A 未教的并查集维护下一个未删除落点把区间枚举压成近线性。"},{"id":"pre:luogu/P8686->shumeng/CSP202412D","source":"luogu/P8686","target":"shumeng/CSP202412D","type":"pre","directed":true,"reason":"B 的区间落点去重复用 A 的「后继并查集」这一步：find(x) 一路跳到第一个未删除下标、处理后并到下一个，使每个落点只被扫描一次，再叠加 A 未教的 BFS 最短路与首次覆盖即最优的证明。"},{"id":"pre:leetcodecn/daily-temperatures->shumeng/CSP202412E","source":"leetcodecn/daily-temperatures","target":"shumeng/CSP202412E","type":"pre","directed":true,"reason":"B 求两侧最近不小于自己的边界 L、R 就是 A 的单调栈弹出结算设备的两侧版本，再叠加端点贪心、嵌套区间限制与计数排序"},{"id":"pre:roj/1485->shumeng/CSP202506D","source":"roj/1485","target":"shumeng/CSP202506D","type":"pre","directed":true,"reason":"B 的段内计数直接复用 A 的关键状态设计：同样把「已匹配模式前缀的最长后缀」当作自动机状态逐字符转移，并沿用 A 的「落点已命中模式则该路径不计入」得到 no_ccf/no_spark（对应 main.cpp 的 next_state 与 output_mask 过滤）；在此之上叠加 # 切段、全局三状态顺序衔接、safe_order 计数与 Kitamasa 求长段递推项，A 的计数 DP 恰好是 B 每个自由段里的子过程。"},{"id":"pre:luogu/P1047->shumeng/CSP202509B","source":"luogu/P1047","target":"shumeng/CSP202509B","type":"pre","directed":true,"reason":"B 复用 A 的「区间加在 left 处 +1、right+1 处 -1，最后统一前缀和统计覆盖次数」这一步差分合并，只是区间来源换成每个窗口可行的阈值区间，再叠加黑白位置灰度取区间端点"},{"id":"pre:luogu/P2590->shumeng/CSP202509D","source":"luogu/P2590","target":"shumeng/CSP202509D","type":"pre","directed":true,"reason":"B 复用 A 的树链剖分拆 dfn 区间步骤，把路径拆成区间组后扫相邻空隙用线段树求路径补集最小 mex"},{"id":"pre:roj/1562->shumeng/CSP202509D","source":"roj/1562","target":"shumeng/CSP202509D","type":"pre","directed":true,"reason":"A 教的关键观察——用树链剖分让根到 x 的路径拆成 O(log n) 段连续区间——正是 B 解法的第一步：B 在 main.cpp 里用同样的链头跳转把路径拆成剖分区间，再叠加「路径补集=相邻区间空隙」的扫描，用区间最小值线段树回答 mex。"},{"id":"pre:roj/1318->luogu/P2386","source":"roj/1318","target":"luogu/P2386","type":"pre","directed":true,"reason":"B 正解的 DFS 下界剪枝直接复用 A 教的“加数不小于前一个加数”这一去重观察，只是把 A 的输出所有拆分改为计数、并叠加盘数与剩余苹果上界"},{"id":"pre:roj/1058->noi_openjudge/ch0104-20","source":"roj/1058","target":"noi_openjudge/ch0104-20","type":"pre","directed":true,"reason":"B 把 A 教的「先算判别式、按其符号分支输出、两根排序保证题面顺序、把负零规范化为 0」整段搬到自己的 D>0/D=0 两支（代码同样调用排序与 normalize_zero），仅在 D<0 支把 A 的 No answer! 换成共轭复根这一额外流程，属于在判别式分类模板上叠加新分支。"},{"id":"pre:roj/1151->noi_openjudge/ch0105-44","source":"roj/1151","target":"noi_openjudge/ch0105-44","type":"pre","directed":true,"reason":"B 的第 n 小质数解法直接复用 A 教的埃氏筛「从 p 的平方起标记倍数」这一具体步骤，只是把统计素数个数改为按下标取第 n 个质数，额外确定 104729 上界。"},{"id":"pre:noi_openjudge/ch0107-07->noi_openjudge/ch0107-11","source":"noi_openjudge/ch0107-07","target":"noi_openjudge/ch0107-11","type":"pre","directed":true,"reason":"B 复用 A 的按字符映射表逐字符翻译这一步，再叠加双向一一映射冲突校验与恰有 26 个密字的检查"},{"id":"pre:noi_openjudge/ch0107-04->noi_openjudge/ch0107-11","source":"noi_openjudge/ch0107-04","target":"noi_openjudge/ch0107-11","type":"pre","directed":true,"reason":"B 的双射映射沿用 A 教的字典映射表建模与查表判定（扫描时查表判断是否与已有记录冲突），再叠加双向两表一致性与 26 字母全覆盖校验"},{"id":"pre:roj/1673->noi_openjudge/ch0107-11","source":"roj/1673","target":"noi_openjudge/ch0107-11","type":"pre","directed":true,"reason":"B 复用 A 教的「扫描时用字典查表判定已有记录」这一步：A 用它判重复并取首次位置，B 用同一查表判定检验密文→明文与明文→密文两向映射是否与已有记录冲突，再叠加 26 字母全覆盖校验与翻译，属于在 A 基础上叠加额外流程。"},{"id":"pre:noi_openjudge/ch0107-18->noi_openjudge/ch0107-19","source":"noi_openjudge/ch0107-18","target":"noi_openjudge/ch0107-19","type":"pre","directed":true,"reason":"B 复用 A 的子串存在性判断这一步，只是先把原串拼接自身把循环移位转化为普通子串问题"},{"id":"pre:roj/1140->noi_openjudge/ch0107-19","source":"roj/1140","target":"noi_openjudge/ch0107-19","type":"pre","directed":true,"reason":"B 把 A 教的子串判定当作子过程，先构造 longer+longer 覆盖全部循环移位，再对 shorter 做一次子串判定；A 只教单次子串判定，B 额外叠加了移位拼接这一流程。"},{"id":"pre:noi_openjudge/ch0107-19->noi_openjudge/ch0107-30","source":"noi_openjudge/ch0107-19","target":"noi_openjudge/ch0107-30","type":"pre","directed":true,"reason":"B 的环展开直接复用 A 的「原串拼接自身、倍长后环上连续片段变普通子串」这一步（两个环都扩成双倍串），再叠加 A 未教的最长公共后缀 DP、不绕超过一圈的长度截断与滚动数组。"},{"id":"pre:roj/1297->noi_openjudge/ch0107-30","source":"roj/1297","target":"noi_openjudge/ch0107-30","type":"pre","directed":true,"reason":"B 把 A 教的「字符相等时左上角加一」DP 转移和两行滚动数组，从 LCS 前缀状态直接搬到双倍环上的最长公共后缀状态，只把不等时的取 max 改成清零"},{"id":"pre:luogu/P2670->noi_openjudge/ch0108-16","source":"luogu/P2670","target":"noi_openjudge/ch0108-16","type":"pre","directed":true,"reason":"B 的逐格演化复用 A 教的「逐格枚举邻格并做边界判断」这一步（方向由八个缩到四邻），再叠加克制者字典与所有格子同步更新避免连锁覆盖的规则。"},{"id":"pre:noi_openjudge/ch0108-15->noi_openjudge/ch0108-16","source":"noi_openjudge/ch0108-15","target":"noi_openjudge/ch0108-16","type":"pre","directed":true,"reason":"B 复用 A 的「每天先把结果写进新矩阵、全部判断完再整体替换」这一步同步更新纪律，只是把八邻扩散规则换成四邻克制占领规则"},{"id":"pre:roj/1128->noi_openjudge/ch0108-16","source":"roj/1128","target":"noi_openjudge/ch0108-16","type":"pre","directed":true,"reason":"B 直接复用 A 教的读旧数组、写新数组避免脏读的双缓冲做法，把它从单次矩阵变换扩展到 n 天需要同步更新的迭代。"},{"id":"pre:luogu/P5731->noi_openjudge/ch0108-25","source":"luogu/P5731","target":"noi_openjudge/ch0108-25","type":"pre","directed":true,"reason":"B 沿用 A 的顺时针螺旋顺序逐格填矩阵这一骨架（实现换成四边界收缩），把填数字换成填五位二进制密文"},{"id":"pre:noi_openjudge/ch0108-23->noi_openjudge/ch0108-25","source":"noi_openjudge/ch0108-23","target":"noi_openjudge/ch0108-25","type":"pre","directed":true,"reason":"B 的螺旋填入复用 A 的四边界螺旋遍历这一步（走上、右、下、左边再整体收缩四条边界、单行单列跳过重复边），再叠加 A 未教的字符五位二进制编码、不足容量补零与按行拼接输出。"},{"id":"pre:noi_openjudge/ch0110-01->noi_openjudge/ch0110-04","source":"noi_openjudge/ch0110-01","target":"noi_openjudge/ch0110-04","type":"pre","directed":true,"reason":"B 沿用 A 的把学生记成元组后按成绩排序取名次这一步，扩成总分、语文、学号三关键字并取前五名"},{"id":"pre:noi_openjudge/ch0110-03->noi_openjudge/ch0110-04","source":"noi_openjudge/ch0110-03","target":"noi_openjudge/ch0110-04","type":"pre","directed":true,"reason":"B 直接复用 A 的「多级排序键用元组加负号转降序」这一步写法，只是从两级规则扩到三级（总分降序、语文降序、学号升序）"},{"id":"pre:noi_openjudge/ch0110-01->noi_openjudge/ch0110-05","source":"noi_openjudge/ch0110-01","target":"noi_openjudge/ch0110-05","type":"pre","directed":true,"reason":"B 的分数线划定复用 A 教的排序后按名次取元素（名次转下标要减一）：第 int(m*1.5) 名的成绩对应下标 int(m*1.5)-1，再叠加 key 元组排序与同线全保留"},{"id":"pre:noi_openjudge/ch0110-03->noi_openjudge/ch0110-05","source":"noi_openjudge/ch0110-03","target":"noi_openjudge/ch0110-05","type":"pre","directed":true,"reason":"B 的排序直接复用 A 的「(-分数, 次键) 元组键」这一步：负号让分数降序、报名号升序做次键，一条 sort 表达两级规则，再叠加 A 未教的 150% 分数线取第 int(m*1.5) 名成绩与不低于线的筛选。"},{"id":"common:luogu/P1093--noi_openjudge/ch0110-05","source":"luogu/P1093","target":"noi_openjudge/ch0110-05","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：B 直接复用 A 教的多关键字 key 元组写法（降序字段取负、升序字段取原值）按分数降序报名号升序排序，再叠加 150% 分数线划定与同线全保留"},{"id":"pre:noi_openjudge/ch0107-17->noi_openjudge/ch0112-05","source":"noi_openjudge/ch0107-17","target":"noi_openjudge/ch0112-05","type":"pre","directed":true,"reason":"B 沿用 A 的统一转小写实现大小写无关匹配这一步，再叠加 split 分词、起始下标累计与首次位置输出"},{"id":"pre:noi_openjudge/ch0107-16->noi_openjudge/ch0112-05","source":"noi_openjudge/ch0107-16","target":"noi_openjudge/ch0112-05","type":"pre","directed":true,"reason":"B 的不区分大小写单词统计复用 A 教的「先统一转小写规范化再比较」这一步，再叠加 split(\" \") 按完整单词切分与累计段长加分隔空格求起始下标。"},{"id":"pre:noi_openjudge/ch0107-21->noi_openjudge/ch0112-05","source":"noi_openjudge/ch0107-21","target":"noi_openjudge/ch0112-05","type":"pre","directed":true,"reason":"B 复用 A 的切词后只对完整单词做比较这一步骤，逐词统计匹配次数并累计起始下标"},{"id":"pre:noi_openjudge/ch0107-09->noi_openjudge/ch0112-08","source":"noi_openjudge/ch0107-09","target":"noi_openjudge/ch0112-08","type":"pre","directed":true,"reason":"B 把 A 的「字母按字母表循环移位、越过 z/Z 回绕」这一步推广成任意位移模 26 并按大小写选 ASCII 基准，逐字符映射的处理框架完全相同"},{"id":"pre:luogu/P1914->noi_openjudge/ch0112-08","source":"luogu/P1914","target":"noi_openjudge/ch0112-08","type":"pre","directed":true,"reason":"B 沿用 A 的字符编号与 ASCII 基准互转加位移取模 26 的做法，把凯撒平移扩成密钥循环的 Vigenere 解密并处理大小写基准"},{"id":"pre:noi_openjudge/ch0107-10->noi_openjudge/ch0112-08","source":"noi_openjudge/ch0107-10","target":"noi_openjudge/ch0112-08","type":"pre","directed":true,"reason":"B 的逐字母解密复用 A 的「字母转 0-25 下标、减位移、模 26 回绕、转回字符」这一步，只是把固定位移 5 换成按 key[i%len] 变化的位移，再叠加 A 未教的大小写 ASCII 基准选择与循环密钥取模。"},{"id":"pre:noi_openjudge/ch0108-11->noi_openjudge/ch0112-09","source":"noi_openjudge/ch0108-11","target":"noi_openjudge/ch0112-09","type":"pre","directed":true,"reason":"B 的顺时针旋转复用 A 教的「新行=原列自下而上」行列映射（用 zip(*image[::-1]) 实现同一效果），再叠加逆时针、两种翻转与按操作串依次执行的变换序列。"},{"id":"pre:noi_openjudge/ch0108-10->noi_openjudge/ch0112-09","source":"noi_openjudge/ch0108-10","target":"noi_openjudge/ch0112-09","type":"pre","directed":true,"reason":"B 复用 A 的「zip(*matrix) 转置」这一步作为构件，顺时针旋转=逆序行后转置、逆时针=转置后逆序行，再叠加四种变换按给定顺序模拟"},{"id":"pre:roj/1126->noi_openjudge/ch0112-09","source":"roj/1126","target":"noi_openjudge/ch0112-09","type":"pre","directed":true,"reason":"B 的顺时针旋转把 A 教的转置当作构件（先逆序行，再转置），在转置之上叠加旋转/翻转操作序列模拟，方向由入门基础题指向普及-综合题"},{"id":"pre:roj/1151->noi_openjudge/ch0112-10","source":"roj/1151","target":"noi_openjudge/ch0112-10","type":"pre","directed":true,"reason":"B 的 Python 解法直接复用 A 教的埃氏筛（平方起标记合数）先打出素数表，再叠加枚举差为 2 的素数对"},{"id":"pre:noi_openjudge/ch0107-02->noi_openjudge/ch0113-04","source":"noi_openjudge/ch0107-02","target":"noi_openjudge/ch0113-04","type":"pre","directed":true,"reason":"B 沿用 A 的 Counter 频次统计这一步决定直方图高度，再叠加逐层按频次输出星号的打印流程"},{"id":"pre:noi_openjudge/ch0107-18->noi_openjudge/ch0113-06","source":"noi_openjudge/ch0107-18","target":"noi_openjudge/ch0113-06","type":"pre","directed":true,"reason":"B 复用 A 的「子串包含判定」这一步来检查各乘积是否为原串的循环移位，只是判定对象换成 s+s 中的出现并叠加 zfill 补前导零与逐倍数枚举"},{"id":"pre:leetcodecn/daily-temperatures->leetcodecn/largest-rectangle-in-histogram","source":"leetcodecn/daily-temperatures","target":"leetcodecn/largest-rectangle-in-histogram","type":"pre","directed":true,"reason":"B 的柱状图最大矩形复用 A 教的「单调栈存下标、弹出时用下标差结算未解决候选」这一步（弹出后新栈顶即左界、当前 i 即右界），再叠加两侧第一个更矮的宽度语义与末尾补 0 哨兵。"},{"id":"pre:leetcodecn/letter-combinations-of-a-phone-number->leetcodecn/word-search","source":"leetcodecn/letter-combinations-of-a-phone-number","target":"leetcodecn/word-search","type":"pre","directed":true,"reason":"B 沿用 A 的选择-递归-恢复三步对称的回溯纪律（标记→递归→恢复成对），只把候选从按键字母换成网格四方向"},{"id":"pre:leetcodecn/find-all-anagrams-in-a-string->leetcodecn/minimum-window-substring","source":"leetcodecn/find-all-anagrams-in-a-string","target":"leetcodecn/minimum-window-substring","type":"pre","directed":true,"reason":"B 复用 A 教的「滑动窗口增量维护字符计数」这一步（右扩左缩都只改进出字符的计数），只是把固定长异位词匹配扩成变长最小覆盖并用 need/have 计数替代整数组比较。"},{"id":"pre:leetcodecn/container-with-most-water->leetcodecn/trapping-rain-water","source":"leetcodecn/container-with-most-water","target":"leetcodecn/trapping-rain-water","type":"pre","directed":true,"reason":"B 的双指针解法复用 A 的移动较矮一侧这一步骤，较矮侧水量定死后移动该侧指针向内收缩"},{"id":"pre:luogu/P3038->luogu/P1600","source":"luogu/P3038","target":"luogu/P1600","type":"pre","directed":true,"reason":"B 把每条路径压成端点挂 +1/-1 的树上差分事件，正是 A 教的 c[u]+=1、c[v]+=1、c[lca]-=2 差分套路在特征值桶上的迁移，再叠加深度等式与桶汇总"},{"id":"pre:roj/1553->luogu/P1600","source":"roj/1553","target":"luogu/P1600","type":"pre","directed":true,"reason":"B 的 main.cpp 把每条玩家路径拆成两段后挂 event_up/event_down 的 +1/-1 事件，正是 A 教的「用 LCA 做端点加、LCA 处抵消的树上差分，再后序累加」这一步，只是把数值累加换成两个桶的前后读数之差，并叠加 A 未教的「到达时刻写成深度等式→特征值入桶」维度，故为 提高→省选/NOI- 的台阶式前置。"},{"id":"pre:luogu/P3374->luogu/P1438","source":"luogu/P3374","target":"luogu/P1438","type":"pre","directed":true,"reason":"B 的解法一把常数系数与下标系数的差分各交给一个 Fenwick，直接复用 A 的「lowbit 单点修改+前缀和查询」这一步树状数组机制，再叠加等差数列拆系数"},{"id":"pre:luogu/P3368->luogu/P1438","source":"luogu/P3368","target":"luogu/P1438","type":"pre","directed":true,"reason":"B 的解法一把等差数列拆成常数系数与下标系数后，两次复用 A 教的差分两端点修改 + Fenwick 前缀和恢复单点值，再叠加拆系数观察与双 Fenwick 组合"},{"id":"pre:luogu/P3372->luogu/P1438","source":"luogu/P3372","target":"luogu/P1438","type":"pre","directed":true,"reason":"B 的解法二把 A 的区间加懒标记线段树推广为等差数列懒标记（整段命中 O(1) 结算、进子树前 push 下传），这是复用点；解法一（主解）的双 Fenwick 差分（拆常数系数与下标系数、四次端点修改）是 A 未教的等价写法。"},{"id":"pre:luogu/P1253->luogu/P1471","source":"luogu/P1253","target":"luogu/P1471","type":"pre","directed":true,"reason":"B 复用 A 的线段树懒标记整段更新、进子树前下传这一步，把摘要从最大值换成和与平方和并推出平方和的封闭更新式"},{"id":"pre:luogu/P3372->luogu/P1471","source":"luogu/P3372","target":"luogu/P1471","type":"pre","directed":true,"reason":"区间加懒标记骨架，本题多维护一个平方和字段"},{"id":"pre:luogu/P3372->luogu/P1558","source":"luogu/P3372","target":"luogu/P1558","type":"pre","directed":true,"reason":"复用 A 的线段树懒标记下传骨架（整段命中改节点摘要并打标记、进子树前 push 下传），B 把 A 的加法标记语义换成赋值标记（整段变单色掩码 1<<(c-1)），再叠加位掩码集合合并与 popcount 统计颜色数。"},{"id":"pre:luogu/P3374->luogu/P1637","source":"luogu/P3374","target":"luogu/P1637","type":"pre","directed":true,"reason":"B 的两次扫描计数复用 A 的树状数组单点修改与前缀查询这一步（left[i]=前缀和(rk[i]-1) 数左侧更小个数、插入排名更新树），再叠加 A 未教的固定中间点拆分、贡献法乘法合并与严格排除相等值的排名处理。"},{"id":"pre:luogu/P1908->luogu/P1637","source":"luogu/P1908","target":"luogu/P1637","type":"pre","directed":true,"reason":"逆序对一维计数扩展为二维偏序的三元组计数"},{"id":"pre:luogu/P3379->luogu/P3128","source":"luogu/P3379","target":"luogu/P3128","type":"pre","directed":true,"reason":"B 每条路径做四点差分前都要复用 A 教的倍增 LCA 求出 g=lca(u,v)，这是 A 教的具体查询步骤；B 在此之上叠加树上点差分四端点标记与子树汇总的新流程。"},{"id":"pre:roj/1547->luogu/P3372","source":"roj/1547","target":"luogu/P3372","type":"pre","directed":true,"reason":"B 的双树状数组解法把 A 教的「两次前缀和相减求区间和」原样保留（prefix(r)-prefix(l-1)），并在 A 的 Fenwick lowbit 拆段前缀和之上叠加差分数组与第二棵 i*d[i] 树，把单点加升级为区间加。"},{"id":"pre:roj/1535->luogu/P3372","source":"roj/1535","target":"luogu/P3372","type":"pre","directed":true,"reason":"B 的双树状数组解法沿用 A 教的 Fenwick lowbit 链单点加与「两次前缀和相减」框架，区间和仍用 prefix(r)-prefix(l-1) 求；只是把单点加扩成差分边界的两次 add，并加第二棵 Fenwick 维护 i*d[i] 还原区间和。"},{"id":"pre:luogu/P7076->luogu/P5522","source":"luogu/P7076","target":"luogu/P5522","type":"pre","directed":true,"reason":"B 复用 A 的自由位计数步骤：约束压成掩码后统计自由位，答案为 2 的自由位数次幂"},{"id":"pre:luogu/P1558->luogu/P5522","source":"luogu/P1558","target":"luogu/P5522","type":"pre","directed":true,"reason":"位掩码状态的线段树维护，本题改为单点改+按位或合并"},{"id":"pre:roj/1255->luogu/P5908","source":"roj/1255","target":"luogu/P5908","type":"pre","directed":true,"reason":"B 的 main.cpp 沿用了 A 教的分层 BFS 状态设计：用 dist 数组（-1 兼作访问标记）按层 +1，靠「第一次入队即最短步数」把 BFS 层数当成到 1 号点的距离，只额外叠加限深（dist[u]==d 停止扩展）与根节点不计数的收口，并用一次遍历取代每只企鹅各自 DFS 的 O(n^2) 暴力。"},{"id":"pre:luogu/P3378->luogu/P1168","source":"luogu/P3378","target":"luogu/P1168","type":"pre","directed":true,"reason":"B 的双堆写法复用 A 教的 heapq 三个基本接口：heappush/heappop 跨堆搬运元素、堆顶 -lower[0] 直接取中位数，接口之外 B 自己新增上下两半中点不变式与负数最大堆写法。"},{"id":"pre:luogu/P1168->luogu/P1801","source":"luogu/P1168","target":"luogu/P1801","type":"pre","directed":true,"reason":"B 的双堆直接复用 A 的「lower 负数最大堆+upper 小根堆按分界线维护两半」这一步，只是平衡目标从中位数换成第 i 小，并按 GET 次数把右堆顶移入左堆推进排名"},{"id":"common:luogu/P1878--luogu/P3378","source":"luogu/P1878","target":"luogu/P3378","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：B 的核心数据结构就是 A 教的 heapq 小根堆三接口操作（压入候选对、取堆顶最小差值对、弹出），再叠加双向链表维护相邻关系与弹出时检查存活且仍相邻的懒删除。"},{"id":"pre:luogu/P1908->luogu/P2161","source":"luogu/P1908","target":"luogu/P2161","type":"pre","directed":true,"reason":"B 复用 A 教的「值域桶放在 Fenwick 上维护出现情况、单点加配合前缀和」这一步（扩展为 0/1 存在桶），再叠加 kth 二进制提升按秩查找模拟有序集合与区间不交的前驱后继删除流程。"},{"id":"pre:luogu/P3374->luogu/P2161","source":"luogu/P3374","target":"luogu/P2161","type":"pre","directed":true,"reason":"B 的 kth 二进制提升正是建立在 A 教的「tree[x] 保存 lowbit 决定的段和、沿 lowbit 增删与查询」这一步之上（整段跳过即段和比较），再叠加 0/1 存在桶、前驱后继与区间不交删除流程。"},{"id":"pre:luogu/P2367->luogu/P3368","source":"luogu/P2367","target":"luogu/P3368","type":"pre","directed":true,"reason":"B 的区间加单点查把 A 教的差分两端点修改逐字搬进树状数组（d[l]+=x、d[r+1]-=x，位置值是差分前缀和），再叠加 Fenwick 维护与越界 r+1 判断"},{"id":"pre:roj/1547->luogu/P3368","source":"roj/1547","target":"luogu/P3368","type":"pre","directed":true,"reason":"B 的 add/prefix 与 A 教的 Fenwick 单点修改上跳、前缀下跳模板逐字相同（代码里都是 index += index & -index 与 index -= index & -index），B 只是把维护对象换成差分数组，再叠加 A 未教的「区间加=两端点差分修改」这一步。"},{"id":"pre:roj/1535->luogu/P3368","source":"roj/1535","target":"luogu/P3368","type":"pre","directed":true,"reason":"B 复用 A 教的沿 lowbit 向上 add、向下 prefix 的 Fenwick 模板，只是把维护对象从原数组换成差分数组，从而把单点加/区间查对偶成区间加/单点查。"},{"id":"pre:luogu/P8218->luogu/P3374","source":"luogu/P8218","target":"luogu/P3374","type":"pre","directed":true,"reason":"B 把 A 的前缀差求区间和一步原样保留在 Fenwick 查询里，再叠加 lowbit 单点修改支持动态更新"},{"id":"pre:luogu/P4568->luogu/P1078","source":"luogu/P4568","target":"luogu/P1078","type":"pre","directed":true,"reason":"B 沿用 A 的状态扩展建模步骤，把额外约束并入状态后再跑 Dijkstra，只是把层号换成文化掩码"},{"id":"pre:roj/1255->luogu/P2960","source":"roj/1255","target":"luogu/P2960","type":"pre","directed":true,"reason":"B 的 main.py 直接复用 A 教的那一步「BFS 按层扩展、格子第一次入队即得最短步数」：把 distance 字典当分层距离表，靠 BFS 距离就是首次被占领时间，答案取最大距离，只是把四方向换成八邻域、起点行做 height-start_y 翻转，并额外叠加按最长距离取代到终点距离的收口。"},{"id":"pre:roj/3082->luogu/P5507","source":"roj/3082","target":"luogu/P5507","type":"pre","directed":true,"reason":"B 的 backward() 复用了 A 教的“反向侧沿逆规则扩展”这一步，只是机关题的逆操作要先由旧状态推出联动旋钮，再叠加状态压缩与路径恢复。"},{"id":"pre:luogu/P2580->luogu/P1481","source":"luogu/P2580","target":"luogu/P1481","type":"pre","directed":true,"reason":"B 的集合解法直接复用 A 的『完整字符串入集合 O(1) 判存在』这一步：对每个单词枚举其非空前缀并逐一查集合，统计最长词链；A 未教的 Trie 视角（沿树下行累计终结标记）作为等价实现叠加。"},{"id":"pre:luogu/P4391->luogu/P2375","source":"luogu/P4391","target":"luogu/P2375","type":"pre","directed":true,"reason":"B 复用 A 的前缀函数求最长 border 步骤，并沿 pi 链回退扩展为枚举全部 border"},{"id":"pre:luogu/P3375->luogu/P2375","source":"luogu/P3375","target":"luogu/P2375","type":"pre","directed":true,"reason":"B 的滑动 j 失配时沿 pi 链回退正是 A 教的 j=prefix[j-1] border 链回退规则（B 自己也注明与构造 pi 的失配处理完全相同），只在其上叠加 border_count 查表与不重叠约束"},{"id":"pre:luogu/P3435->luogu/P2375","source":"luogu/P3435","target":"luogu/P2375","type":"pre","directed":true,"reason":"失配链递推的深度进阶：从求最短 border 到统计链上长度不超过一半的节点数"},{"id":"pre:luogu/P3370->luogu/P2580","source":"luogu/P3370","target":"luogu/P2580","type":"pre","directed":true,"reason":"B 沿用 A 的完整字符串集合做 O(1) 判重存在查询，扩成 valid/called 两个集合的状态记录，再叠加点名三态输出"},{"id":"pre:luogu/P5250->luogu/P3369","source":"luogu/P5250","target":"luogu/P3369","type":"pre","directed":true,"reason":"B 的解法一（离线坐标压缩+Fenwick）直接复用 A 的这一步：小于 x 的数量定位第 count 小即前驱、kth 二进制提升取第 k 小，再叠加 A 未教的可重集合计数语义（排名、删除一个 x）；B 的 FHQ-Treap 解法不涉及此步。"},{"id":"pre:luogu/P2234->luogu/P3369","source":"luogu/P2234","target":"luogu/P3369","type":"pre","directed":true,"reason":"B 沿用 A 的坐标压缩加 Fenwick 前缀计数、kth 二进制提升查前驱后继这一步实现平衡树六操作"},{"id":"pre:luogu/P3375->luogu/P3435","source":"luogu/P3375","target":"luogu/P3435","type":"pre","directed":true,"reason":"B 求最短非空 border 直接复用 A 的失配 border 链这一具体步骤：A 退到次长 border 的链式跳转在 B 中被推广为沿 pi 反复跳枚举全部 border，再叠加 mini[i] 继承式递推避免 O(n^2) 跳链。"},{"id":"pre:luogu/P4391->luogu/P3435","source":"luogu/P4391","target":"luogu/P3435","type":"pre","directed":true,"reason":"单点周期结论的推广：从对整个串求一次周期，到对每个前缀沿失配链求最短 border"},{"id":"pre:luogu/P9236->luogu/P4551","source":"luogu/P9236","target":"luogu/P4551","type":"pre","directed":true,"reason":"B 的第一把钥匙直接复用 A 教的前缀异或化区间为两前缀相异或这一步，只是把数组区间搬到树根路径上，再叠加 01-Trie 求最大异或对"},{"id":"pre:luogu/P6824->luogu/P4735","source":"luogu/P6824","target":"luogu/P4735","type":"pre","directed":true,"reason":"B 复用 A 的 01-Trie 按位比较下行这一步（A 按 k 的位决策下行打子树标记，B 换成优先相反位的贪心查询），再叠加 A 未教的可持久化 root[i] 版本序列与『区间转版本』（后缀异或化为 s[N]^s[p-1]，查询落在版本 [l-1, r-1]）。A 的区间贡献覆盖与懒标记在 B 中并未使用。"},{"id":"pre:leetcodecn/merge-k-sorted-lists->luogu/P5283","source":"leetcodecn/merge-k-sorted-lists","target":"luogu/P5283","type":"pre","directed":true,"reason":"B 的多路归并调度直接复用 A 的「堆中弹出当前最优、随即把同一路的下一个候选压回堆」这一步，只是从 k 条链表头换成 n 口井的 (val,i,rank) 懒加载，并叠加 Trie 上第 rank 大查询"},{"id":"pre:luogu/P2048->luogu/P5283","source":"luogu/P2048","target":"luogu/P5283","type":"pre","directed":true,"reason":"堆多路归并取前 k 大的经典模型，异或版本在此基础上换用可持久化 Trie"},{"id":"pre:luogu/P4735->luogu/P5283","source":"luogu/P4735","target":"luogu/P5283","type":"pre","directed":true,"reason":"可持久化 01-Trie + 前缀异或的基础模型，先掌握区间转版本，再学第 rank 大与堆归并"},{"id":"pre:luogu/P4551->luogu/P6824","source":"luogu/P4551","target":"luogu/P6824","type":"pre","directed":true,"reason":"B 复用 A 的 01-Trie 机制（高位建树、沿位下行比较）：A 的查询每层优先走相反位贪心，B 换成按 k 的位决策下行并给整棵子树打懒标记，再叠加 DFS 下传求覆盖最大值。B 与『树上路径转根前缀异或』无关，原 reason 该半句失实。"},{"id":"pre:leetcodecn/single-number->luogu/P1469","source":"leetcodecn/single-number","target":"luogu/P1469","type":"pre","directed":true,"reason":"B 直接复用 A 的「全部异或、成对抵消」这一步异或技巧求落单筷子长度，只是把其余出现两次推广到出现偶数次并叠加 8MB 内存限制下的 fread 流式读入"},{"id":"pre:luogu/P1835->luogu/P3601","source":"luogu/P1835","target":"luogu/P3601","type":"pre","directed":true,"reason":"B 的分段质因数分解复用 A 教的「先筛 sqrt 内基础质数、再从区间内第一个 p 的倍数（max(p*p,ceil(L/p)*p)）开始处理」这一步，只是把标记合数换成更新 phi 并除尽 remaining，再叠加 A 未教的欧拉函数转化与剩余大质因子判定。"},{"id":"pre:roj/1352->luogu/P1347","source":"roj/1352","target":"luogu/P1347","type":"pre","directed":true,"reason":"A 教的 Kahn 拓扑排序加「出队点数少于 n 即有环」的环检测，被 B 的 topological_status（len(order)<n 返回 inconsistent）原样复用；B 只是在其上叠加每加一条边重跑与队列分支数判唯一。"},{"id":"pre:roj/1352->luogu/P4017","source":"roj/1352","target":"luogu/P4017","type":"pre","directed":true,"reason":"B 直接复用 A 教的「入度归零即所有前驱已定型、此时沿边由前驱递推」这一步，只把取 max 下界换成 ways[v]+=ways[u] 计数并补上生产者/消费者与取模。"},{"id":"pre:luogu/P1359->luogu/P4017","source":"luogu/P1359","target":"luogu/P4017","type":"pre","directed":true,"reason":"B 的转移复用了 A 教的「DAG 上按拓扑序沿有向边由前驱状态更新后继状态」这一步：A 靠编号天然拓扑序做 dp[j]=min(dp[j],dp[i]+cost)，B 显式 Kahn 拓扑排序后做 ways[v]+=ways[u]，仅把取 min 换成计数累加并补上生产者/消费者判定与取模。"},{"id":"pre:roj/1547->luogu/P1908","source":"roj/1547","target":"luogu/P1908","type":"pre","directed":true,"reason":"B 的 main.py/main-fenwick.cpp 直接复用 A 教的 lowbit 链模板（查询 i-=lowbit 拆前缀、修改 i+=lowbit 上跳），只把 tree 的语义从区间和换成效值域桶的出现次数，再叠加 A 未教的离散化与二维偏序扫描。"},{"id":"pre:roj/1535->luogu/P1908","source":"roj/1535","target":"luogu/P1908","type":"pre","directed":true,"reason":"B 的 main.py 直接复用 A 教的 i-=lowbit 前缀和与 i+=lowbit 单点加，把 A 的区间和树状数组换成语义为计数的值域桶，再叠加 A 未教的离散化与二维偏序统计。"},{"id":"common:luogu/P1908--luogu/P3372","source":"luogu/P1908","target":"luogu/P3372","type":"common","directed":false,"reason":"同型/平行题（M4 自 pre 移入）：原 reason 写「权值线段树单点加+区间查计数用法，骨架同 P3372」——「同」是 common（同型/平行）语汇，不是 pre。且 P1908 逆序对通常先于线段树教学。移出 pre，应登记为 common。"},{"id":"pre:roj/3015->luogu/P1966","source":"roj/3015","target":"luogu/P1966","type":"pre","directed":true,"reason":"A 用相邻交换恰消一个逆序对证明最少交换次数=逆序对数，B 直接复用这一结论把配对后的 target 排列交给树状数组求逆序对数"},{"id":"pre:luogu/P3369->luogu/P7562","source":"luogu/P3369","target":"luogu/P7562","type":"pre","directed":true,"reason":"B 维护空闲段用的随机 Treap 正是 A 的按值分裂/合并那一步：main.py 里 split(root,key)/merge 按左端点切分空闲段，前驱、插入、删除与 A 的有序多重集接口一一对应，再叠加区间调度贪心与 next_end 倍增计数。"},{"id":"pre:luogu/P3366->luogu/P1621","source":"luogu/P3366","target":"luogu/P1621","type":"pre","directed":true,"reason":"B 沿用 A 的并查集判两端不同集合才合并并计数的步骤，用筛法枚举质因数把区间倍数两两 union 出最终集合数"},{"id":"pre:HDU/1213->luogu/P2814","source":"HDU/1213","target":"luogu/P2814","type":"pre","directed":true,"reason":"B 的查询沿父亲指针走到 parent[name]==name 的根并做路径压缩，复用 A 教的「parent 指针维护集合、find(x)==x 即集合根」这一步，再叠加 #name/+child/?name 输入语义与 setdefault 防重置细节。"},{"id":"pre:luogu/P1551->luogu/P2814","source":"luogu/P1551","target":"luogu/P2814","type":"pre","directed":true,"reason":"B 的找最早祖先复用 A 的「find 沿父指针取代表元 + 路径压缩」这一步（查询后把途中人物直接指向根），只是把整数下标换成字典按名字存父指针，再叠加 A 未教的 #/+/? 输入解析与 setdefault 防覆盖。"},{"id":"pre:luogu/P3367->luogu/P2814","source":"luogu/P3367","target":"luogu/P2814","type":"pre","directed":true,"reason":"B 的查询直接复用 A 教的「find 沿父指针找代表元 + 路径压缩把途中点挂到根上」这一步，只是集合用名字字典实现、根定义为最早祖先，再叠加 #name/+child/?name 输入语义与 setdefault 细节。"},{"id":"pre:luogu/P3368->luogu/P5250","source":"luogu/P3368","target":"luogu/P5250","type":"pre","directed":true,"reason":"B 复用 A 的 Fenwick 单点加与前缀和查询模板维护长度存在性，再叠加离散化与树状数组上二分求第 k 小的前驱后继"},{"id":"pre:luogu/P3374->luogu/P5250","source":"luogu/P3374","target":"luogu/P5250","type":"pre","directed":true,"reason":"B 复用 A 的树状数组单点修改与前缀和查询模板维护库存存在性，再叠加离散化与树上二分求第 k 小的前驱后继"},{"id":"pre:luogu/P2947->luogu/P1950","source":"luogu/P2947","target":"luogu/P1950","type":"pre","directed":true,"reason":"B 的递增栈沿用 A 教的单调栈弹出被挡住候选、保持栈单调的结算纪律，再叠加 (height,width_count) 合并等高宽度与子数组最小值贡献记账"},{"id":"pre:shumeng/CSP201312C->luogu/P1950","source":"shumeng/CSP201312C","target":"luogu/P1950","type":"pre","directed":true,"reason":"B 的矩形计数复用 A 教的「每根柱子作为最低高度的左右控制区间」这一步（直方图子数组最小值正由单调栈按控制区间统计），只是从求最大矩形面积变成把每个控制区间贡献求和，并叠加 A 未教的逐行高度转换与 (height,width_count) 合并宽度技巧。"},{"id":"pre:roj/3032->luogu/P1950","source":"roj/3032","target":"luogu/P1950","type":"pre","directed":true,"reason":"B 把 A 教的单调递增栈弹出时定出左右边界这一步搬到逐行直方图上，只是把结算最大面积换成用 width_count 合并等高宽度来累加子数组最小值之和"},{"id":"pre:luogu/P1714->luogu/P2216","source":"luogu/P1714","target":"luogu/P2216","type":"pre","directed":true,"reason":"B 把 A 教的单调队列窗口最值原语在横、纵两个方向各套一遍求正方形内最值差，A 的前缀和换法只是它的一个应用"},{"id":"pre:luogu/P1440->luogu/P2216","source":"luogu/P1440","target":"luogu/P2216","type":"pre","directed":true,"reason":"B 的两遍扫描都直接复用 A 的一维滑动窗口单调队列取最值这一步（弹过期后队头即窗口最值），先逐行再逐列把二维正方形窗口拆成两次一维，再叠加 A 未教的行列两级拆解与 max-min 差值目标。"},{"id":"pre:luogu/P2032->luogu/P2216","source":"luogu/P2032","target":"luogu/P2216","type":"pre","directed":true,"reason":"B 把 A 教的「一维单调队列滑窗最值」这一步在横竖两个方向各执行一次合成正方形整体最值（拆成两次一维单调队列），再叠加 A 未教的二维拆解次序与最大减最小的差值目标。"},{"id":"pre:luogu/P1147->luogu/P3143","source":"luogu/P1147","target":"luogu/P3143","type":"pre","directed":true,"reason":"B 求 right[i] 时复用 A 的「左右指针只向右移动、按条件推进窗口得到线性扫描」这一步双指针扫法，再叠加排序与枚举第一柜+后缀最优第二柜的组合"},{"id":"pre:luogu/P3467->luogu/P4147","source":"luogu/P3467","target":"luogu/P4147","type":"pre","directed":true,"reason":"B 的直方图最大矩形复用 A 教的「递增高度栈、遇更矮就弹出已结束高度层」这一步（只是弹出时改以当前列为右边界结算面积），再叠加逐行 heights 清零建直方图与高度 0 哨兵清栈。"},{"id":"pre:shumeng/CSP201312C->luogu/P4147","source":"shumeng/CSP201312C","target":"luogu/P4147","type":"pre","directed":true,"reason":"B 把 A 教的直方图最大矩形单调栈原语（弹栈新栈顶为左界、当前为右界、0 哨兵清空栈）逐行套用，额外只做 heights 二维转一维"},{"id":"pre:roj/3032->luogu/P4147","source":"roj/3032","target":"luogu/P4147","type":"pre","directed":true,"reason":"B 把 A 教的单调栈弹出结算（新栈顶为左界、当前列为右界、末尾补 0 哨兵）作为内层原语，逐行维护 heights 压成直方图后再套用一次，额外流程只是二维转一维。"},{"id":"common:luogu/P1571--luogu/P5076","source":"luogu/P1571","target":"luogu/P5076","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：B 的全部查询建立在 A 教的有序序列 lower_bound 二分定位（第一个≥x 的位置）语义上，bisect_left/right 是同一模型的左右边界版本，再叠加 insort 插入维持有序"},{"id":"pre:luogu/P1739->luogu/P1241","source":"luogu/P1739","target":"luogu/P1241","type":"pre","directed":true,"reason":"B 复用 A 教的「栈维护未匹配左括号、右括号只能与之前未匹配左括号配对」这一步，再叠加本题特有规则（类型不符时两者都保留为未匹配）与扫描后按位补全括号输出。"},{"id":"pre:leetcodecn/valid-parentheses->luogu/P1241","source":"leetcodecn/valid-parentheses","target":"luogu/P1241","type":"pre","directed":true,"reason":"B 原样复用 A 的栈顶即最近未匹配左括号且右括号只与栈顶匹配这一步，再叠加未匹配括号原位补全的输出规则"},{"id":"pre:roj/1354->luogu/P1241","source":"roj/1354","target":"luogu/P1241","type":"pre","directed":true,"reason":"B 的配对规则就是复用 A 教的“右括号只能配栈顶那个同类左括号”，把 A 的判定式匹配（OK/Wrong）改成记录匹配位置，再对未匹配括号补全输出"},{"id":"pre:luogu/P3374->luogu/P2234","source":"luogu/P3374","target":"luogu/P2234","type":"pre","directed":true,"reason":"B 复用 A 教的「Fenwick 单点加、前缀和查询」这一步（计数树前缀和得 less 后用 kth 找前后继），再叠加 A 未教的离散化排名、kth 二进制倍增与重复值波动为零的处理。"},{"id":"pre:luogu/P5266->luogu/P3613","source":"luogu/P5266","target":"luogu/P3613","type":"pre","directed":true,"reason":"B 的寄包柜直接复用 A 教的字典 key→value 增删查建模（cell→item 两级字典稀疏存储），再叠加每柜独立字典与不共享引用的坑点"},{"id":"pre:roj/3067->luogu/P1032","source":"roj/3067","target":"luogu/P1032","type":"pre","directed":true,"reason":"B 的隐式图 BFS 直接沿用 A 教的「队列层数即最短距离、第一次到达即最近、入队即标记」这一判定：main.cpp 里 vis 判重后 step+1 入队、首次弹出目标串即返回步数，只是把 A 的网格格子换成字符串状态，并额外叠加每条规则全部出现位置的枚举、10 步截断与暴力对拍；A 的多源部分未被用到，故属 BFS 骨架的模板级复用。"},{"id":"pre:luogu/P3654->luogu/P1101","source":"luogu/P3654","target":"luogu/P1101","type":"pre","directed":true,"reason":"B 复用 A 教的「枚举起点+方向、检查固定长度连续段是否逐格满足」这一步（起点+方向唯一确定一次检查），再叠加 A 未教的 8 方向数组、首字母剪枝与命中位置打标记输出掩码。"},{"id":"pre:luogu/P2670->luogu/P1101","source":"luogu/P2670","target":"luogu/P1101","type":"pre","directed":true,"reason":"B 的逐字符核对直接复用 A 教的「八方向 (dx,dy) 方向数组 + 边界判断」这一步沿方向延伸 k 位，再叠加只从 y 出发的起点剪枝与 keep 布尔并集处理交叉。"},{"id":"pre:roj/1335->luogu/P1162","source":"roj/1335","target":"luogu/P1162","type":"pre","directed":true,"reason":"B 的解法二把 A 教的「扫描未访问格后栈式淹没整块、入栈即标记」连通块 flood fill 原样用在 BFS2 上，只把计数换成改写成 2，并在其上叠加正难则反判内外这一 A 未教的新步骤。"},{"id":"pre:roj/1329->luogu/P1162","source":"roj/1329","target":"luogu/P1162","type":"pre","directed":true,"reason":"A 教的『邻居在界内且不是 0 就改值并入队』的入队即标记 flood fill，被 B 解法二的 BFS2 原样用来把整片内部 0 区域改写成 2，B 只在此基础上叠加补边框/正难则反判内外。"},{"id":"pre:roj/1254->luogu/P1825","source":"roj/1254","target":"luogu/P1825","type":"pre","directed":true,"reason":"B 的内联瞬移 BFS 复用 A 教的网格 BFS 首次到达出口层号即答案这一步（main.cpp 里 dist 兼判重、四方向扩展、首次弹出终点返回），只是在展开邻居时多加了传送门端点替换"},{"id":"pre:luogu/T641741->luogu/P1825","source":"luogu/T641741","target":"luogu/P1825","type":"pre","directed":true,"reason":"B 的内联瞬移 BFS 复用 A 教的「网格 BFS 层数即最短距离、判重入队」这一步，把走一步加免费瞬移折叠成有效代价为 1 的边后照常按层计步，再叠加 A 未教的端点配对瞬移映射与恰好瞬移一次的判定。"},{"id":"pre:roj/1186->luogu/P1102","source":"roj/1186","target":"luogu/P1102","type":"pre","directed":true,"reason":"B 的 Counter 解法直接复用 A 教的哈希频率计数这一步，把每个值的出现次数用于计算 cnt[B]×cnt[B+C]，只是在计数基础上叠加数对聚合。"},{"id":"pre:roj/1241->luogu/P1024","source":"roj/1241","target":"luogu/P1024","type":"pre","directed":true,"reason":"B 先用「根间距至少为 1」扫描出 f(l)f(r)<0 的变号区间，再对每个区间照搬 A 教的「取中点看符号丢一半」的二分，只是从单根推广到三个根并额外处理整数根"},{"id":"pre:noi_openjudge/ch0111-02->luogu/P1024","source":"noi_openjudge/ch0111-02","target":"luogu/P1024","type":"pre","directed":true,"reason":"先用「变号区间内二分」求单个零点的入门模板，再把同一手法用到三次方程：扫描整数端点与开单位区间、对每个变号区间分别二分逼近三个实根。"},{"id":"pre:HDU/5124->luogu/P1884","source":"HDU/5124","target":"luogu/P1884","type":"pre","directed":true,"reason":"B 沿用 A 的扫描线端点加减事件与离散化压缩坐标这一步，把求最大单点覆盖扩成矩形并面积（Δx 乘活跃 y 区间并集长度）"},{"id":"pre:luogu/P2367->luogu/P2882","source":"luogu/P2367","target":"luogu/P2882","type":"pre","directed":true,"reason":"B 把 A 的区间修改端点差分标记思想用在区间翻转上：起点打 started[i]=1，扫描到 i-K 时异或消去，使每次翻转 O(1)"},{"id":"pre:luogu/P1147->luogu/P3029","source":"luogu/P1147","target":"luogu/P3029","type":"pre","directed":true,"reason":"B 复用 A 的同向双指针扩缩窗口调度机制，只是把和与目标的比较换成品种计数是否齐全再收缩左指针"},{"id":"pre:luogu/P1638->luogu/P3029","source":"luogu/P1638","target":"luogu/P3029","type":"pre","directed":true,"reason":"B 原样套用 A 的最短覆盖窗口右扩左缩步骤，只额外先按坐标排序再计数品种"},{"id":"pre:luogu/P2367->luogu/P3406","source":"luogu/P2367","target":"luogu/P3406","type":"pre","directed":true,"reason":"B 复用 A 的区间加差分（左端加、右端+1 减）再前缀和还原这一步，逐段统计使用次数后贪心取 min(count*A, count*B+C)"},{"id":"pre:roj/1294->luogu/P2392","source":"roj/1294","target":"luogu/P2392","type":"pre","directed":true,"reason":"B 的正式解 main.cpp 在 subject_time 里逐题倒序枚举时间 j，直接复用 A 教的「倒序枚举容量保证每件物品只用一次」这一步，只把 max 取最大价值换成 dp 可达标记；B 的新难度在互补子集归约到 total/2。"},{"id":"pre:luogu/U661986->luogu/P2392","source":"luogu/U661986","target":"luogu/P2392","type":"pre","directed":true,"reason":"B 的正式解 main.cpp（28-31 行）把每道题当体积、逆序枚举时间做 dp[j] |= dp[j-times[i]]，直接沿用 A 教的『容量倒序枚举保证每件物品只用一次』这一步，只把求最大价值换成可达性标记；B 新增的难度是用互补子集性质把分组化归为不超过 total/2 的子集和。"},{"id":"pre:noi_openjudge/ch0110-03->luogu/P1786","source":"noi_openjudge/ch0110-03","target":"luogu/P1786","type":"pre","directed":true,"reason":"B 复用 A 的负号改降序、多关键字压成一条元组排序键的写法，扩成职位高低、等级、输入顺序三级规则"},{"id":"common:luogu/P1104--luogu/P1786","source":"luogu/P1104","target":"luogu/P1786","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：B 复用 A 的把多级比较规则翻译成元组排序键（负号降序、输入顺序兜底）这一步，做先重排职位再全体输出的两阶段排序"},{"id":"common:luogu/P1093--luogu/P1786","source":"luogu/P1093","target":"luogu/P1786","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：B 沿用 A 的多规则压成排序键元组、降序取负升序用原值、末位输入序号兜底这一步，套在两段排序上再叠加职位重分配"},{"id":"pre:noi_openjudge/ch0108-11->luogu/P4924","source":"noi_openjudge/ch0108-11","target":"luogu/P4924","type":"pre","directed":true,"reason":"B 复用 A 的顺时针旋转行列对应步骤，把整图旋转换成奇数阶子矩阵按同一坐标映射旋转并复制写回"},{"id":"pre:noi_openjudge/ch0108-13->luogu/P4924","source":"noi_openjudge/ch0108-13","target":"luogu/P4924","type":"pre","directed":true,"reason":"B 的子矩阵旋转复用 A 教的『变换不能原地读写、先复制原数据生成新矩阵再写回』纪律，再叠加顺逆时针 90 度的下标映射"},{"id":"pre:roj/1127->luogu/P4924","source":"roj/1127","target":"luogu/P4924","type":"pre","directed":true,"reason":"B 的顺时针分支直接复用 A 教过的坐标映射 new[row][col]=old[size-1-col][row]，只是把它从整图一次旋转改造成子矩阵内、可反复调用的原地旋转"},{"id":"pre:luogu/P1157->luogu/P1036","source":"luogu/P1157","target":"luogu/P1036","type":"pre","directed":true,"reason":"B 的关键观察「组合不关心顺序、下标递增天然去重」正是 A 教的组合行内递增、不重不漏这一枚举形态，只是把库函数 combinations 换成带 last 下标与求和的组合 DFS 并叠加试除判素数"},{"id":"pre:luogu/P6437->luogu/P1036","source":"luogu/P6437","target":"luogu/P1036","type":"pre","directed":true,"reason":"B 的组合枚举 DFS 复用 A 教的下标严格递增保证每个组合只出现一次（不重不漏），推广到 k 层并加上界剪枝，再叠加选满 k 个后试除判素数"},{"id":"pre:roj/1317->luogu/P1036","source":"roj/1317","target":"luogu/P1036","type":"pre","directed":true,"reason":"B 的 main.cpp 组合枚举 DFS 复用了 A 的核心观察——用递增（下标大于 last）替代 vis 数组天然去重，并沿用 A 教的上界剪枝保证选满，仅在叶子额外加试除法判素数。"},{"id":"pre:luogu/P1428->luogu/P5741","source":"luogu/P1428","target":"luogu/P5741","type":"pre","directed":true,"reason":"B 的成对枚举复用 A 的「按下标对双重循环枚举（j<i）再逐对判断」这一步，只是条件从可爱度比较换成每科分差不超过 5 且总分差不超过 10，再叠加 A 未教的 is_close 函数封装与字典序输出顺序论证。"},{"id":"pre:noi_openjudge/ch0107-01->luogu/P1598","source":"noi_openjudge/ch0107-01","target":"luogu/P1598","type":"pre","directed":true,"reason":"B 复用 A 教的「逐字符分类计数」这一步（从数数字字符个数换成统计 26 个大写字母频次），再叠加 A 未教的按最高层数从上往下逐层输出星号的柱状图格式化。"},{"id":"pre:roj/1187->luogu/P1598","source":"roj/1187","target":"luogu/P1598","type":"pre","directed":true,"reason":"B 的 main.cpp/main.py 都直接复用 A 教的「26 长度计数数组按 ch-'a'/'A' 下标累加字母频次」并顺扫计数表求最大频次这一步，只是把 A 的「取次数最多字符」换成把最大频次当 max_height，再叠加逐层自上而下输出星号与行尾去空格的新流程。"},{"id":"pre:noi_openjudge/ch0108-12->luogu/P1205","source":"noi_openjudge/ch0108-12","target":"luogu/P1205","type":"pre","directed":true,"reason":"B 复用 A 教的「把各矩阵变换按下标构造出来、再按题目优先级逐个比较判定」这一步，只是把三种变换扩成旋转 90/180/270、水平反射及其复合共七种情况。"},{"id":"pre:noi_openjudge/ch0108-11->luogu/P1205","source":"noi_openjudge/ch0108-11","target":"luogu/P1205","type":"pre","directed":true,"reason":"B 的 rotate 函数复用 A 的「顺时针旋转 90 度 = 结果每行取原图一列、从下到上读取」这一步，并用它组合出 90/180/270 与反射后再旋转，再叠加 A 未教的水平反射与按编号顺序枚举取最小答案。"},{"id":"pre:roj/1127->luogu/P1205","source":"roj/1127","target":"luogu/P1205","type":"pre","directed":true,"reason":"B 的 rotate 函数直接复用 A 教的顺时针 90° 坐标映射（main.cpp 中 dst[j][n-1-i]=src[i][j] 与 A 的 b[i][j]=a[n-1-j][i] 等价）来生成旋转矩阵，只是再叠加水平反射与 7 种情况的枚举比较。"},{"id":"pre:luogu/P5722->luogu/P1009","source":"luogu/P5722","target":"luogu/P1009","type":"pre","directed":true,"reason":"B 直接复用 A 教的「从 1 到 n 循环逐项累加」这一步（answer += factorial），只是在累加变量外多维护一个递推的当前阶乘 factorial *= x，并靠 Python 大整数承载结果。"},{"id":"pre:noi_openjudge/ch0105-35->luogu/P1009","source":"noi_openjudge/ch0105-35","target":"luogu/P1009","type":"pre","directed":true,"reason":"B 沿用 A 的循环递推维护当前阶乘并每轮累加的写法，把倒数求和换成阶乘求和并用大整数支持 n<=50"},{"id":"pre:roj/1091->luogu/P1009","source":"roj/1091","target":"luogu/P1009","type":"pre","directed":true,"reason":"B 的解法直接复用 A 教的「递推维护当前阶乘、每轮先乘 i 再累加」这一具体步骤（B 的 main.py 就是 factorial *= x; answer += factorial），只是把 long long 换成 Python 大整数以支持 n<=50，方向从入门到普及-成立。"},{"id":"pre:luogu/P1106->usaco/1087","source":"luogu/P1106","target":"usaco/1087","type":"pre","directed":true,"reason":"B 复用 A 的栈顶大于当前即弹出的单调栈步骤，把删数改成判断刷漆层该在何处结束并计笔数"},{"id":"pre:roj/1191->usaco/1110","source":"roj/1191","target":"usaco/1110","type":"pre","directed":true,"reason":"两题都用「队列存网格上待处理的格子、入队即标记、每格至多处理一次」这一结构，A 里它保证流感 BFS 每个格子只入队一次，B 把它照搬成补牛队列并在取出时对已补位置直接跳过，只是把标记从患病格换成补牛格，再叠加舒适度判定与连锁补牛流程。"},{"id":"pre:luogu/P7910->usaco/1326","source":"luogu/P7910","target":"usaco/1326","type":"pre","directed":true,"reason":"B 复用 A 教的「把元素位置翻译成排序数组中的位次」这一步（old_pos/new_pos 都是排序位次，删除插入即位次移动），再叠加 A 未教的 Σi·b_i 贡献公式、位移区间和前缀和与单点修改的 O(log n) 增量计算。"},{"id":"pre:luogu/P2367->usaco/1373","source":"luogu/P2367","target":"usaco/1373","type":"pre","directed":true,"reason":"B 的满分做法先复用 A 的「相邻差分数组」这一步定义一阶差分 b_i=a_i-a_{i-1}，再差分出二阶差分把后缀阶梯操作压成单点修改并贪心从左修到右"},{"id":"pre:noi_openjudge/ch0103-17->usaco/1109","source":"noi_openjudge/ch0103-17","target":"usaco/1109","type":"pre","directed":true,"reason":"B 的有向面积校验解法（brute.cpp）复用 A 教的「鞋带有向面积的符号随顶点绕向变号」这一步，把 A 里用来取 abs 的符号信息反过来用作 CW/CCW 判定，主解法另叠加 A 未教的累计转角。"},{"id":"pre:roj/1593->luogu/P1896","source":"roj/1593","target":"luogu/P1896","type":"pre","directed":true,"reason":"B 的 main.cpp 里 `if ((cur & prev) || ((cur << 1) & prev) || ((cur >> 1) & prev)) continue;` 直接复用了 A 教的「把一行压成掩码、行间同列冲突用 `t & s == 0` 这一条位运算判定」的步骤，只在外面每人多带一个国王数维度并补上左上/右上两条冲突式，难度由普及+/提高- 升到普及+/提高，方向与台阶都成立。"},{"id":"pre:roj/3041->luogu/P3808","source":"roj/3041","target":"luogu/P3808","type":"pre","directed":true,"reason":"B 的 AC 自动机第一步直接复用 A 教的 Trie 插入与「终止结点累计结尾个数」：P3808 main.cpp 的 insert_pattern 与 end_count[u]++ 和 A 的 insert/cnt[结点] 完全同构，A 停在用路径累计计数回答前缀询问，B 在此之上再叠加 fail 指针与文本串扫描的 fail 链统计。"},{"id":"pre:roj/1455->luogu/P3808","source":"roj/1455","target":"luogu/P3808","type":"pre","directed":true,"reason":"B 的 AC 自动机扫描文本串时把 A 在 KMP 里教的「失配时沿 fail 链回退到最长相等真前后缀」直接搬到 Trie 上：当前状态沿 fail 链上溯统计终止节点（代码 while (p != 0 && end_count[p] != -1) p = fail_link[p]），只是把单模式串的失配链换成多模式串的树形 fail 链，再叠加建 Trie 与 BFS 补全转移。"},{"id":"pre:luogu/P1226->luogu/P3390","source":"luogu/P1226","target":"luogu/P3390","type":"pre","directed":true,"reason":"B 明确把 A 教的快速幂循环原样移植，仅把乘法换成满足结合律的矩阵乘法，res 与 base 的二进制拆指数骨架完全相同"},{"id":"pre:noi_openjudge/ch0108-09->luogu/P3390","source":"noi_openjudge/ch0108-09","target":"luogu/P3390","type":"pre","directed":true,"reason":"B 的矩阵快速幂每次相乘都调用 A 教的矩阵乘法（一行乘一列内积的三重循环），借结合律把普通快速幂的乘法替换成矩阵乘法"},{"id":"pre:roj/3000->luogu/P3390","source":"roj/3000","target":"luogu/P3390","type":"pre","directed":true,"reason":"B 的 matrix_power 完全沿用 A 教的指数按二进制拆分与 base=base*base 循环，只是把整数乘法替换为矩阵乘法、1 替换为单位矩阵"},{"id":"pre:luogu/P2362->luogu/P1020","source":"luogu/P2362","target":"luogu/P1020","type":"pre","directed":true,"reason":"B 的第一问直接复用 A 教的“以 i 结尾定义子序列长度、按前后元素大小关系从 j 转移”这一步（brute.cpp 的 if (a[j] >= a[i]) dp1[i]=max(dp1[i],dp1[j]+1)，正式解则写成对 -a[i] 求最长不下降子序列），只是把大小关系反向、再叠加二分维护尾值的 O(n log n) 优化与第二问的 Dilworth 结论。"},{"id":"pre:luogu/P1307->luogu/P1553","source":"luogu/P1307","target":"luogu/P1553","type":"pre","directed":true,"reason":"B 的四种形式处理都复用 A 教的「切片反转数字部分再去掉多余零」这一步，再叠加按分隔符分类讨论与小数部分去末尾零、空串保留 0 的新规则。"},{"id":"pre:noi_openjudge/ch0105-29->luogu/P1553","source":"noi_openjudge/ch0105-29","target":"luogu/P1553","type":"pre","directed":true,"reason":"B 的四种形式反转都复用 A 教的「切片反转数位、删除多余零、删空后补一个 0」这一步，再叠加按分隔符分类讨论与小数部分去末尾零的方向规则。"},{"id":"pre:roj/20001->luogu/P1553","source":"roj/20001","target":"luogu/P1553","type":"pre","directed":true,"reason":"B 的整数部分（含分数分子分母、百分数数字部分）就是照搬 A 教的 s[::-1] 后 lstrip('0') 化简前导零，只是外面套了一层按分隔符分类的流程"},{"id":"pre:luogu/P8605->luogu/P8817","source":"luogu/P8605","target":"luogu/P8817","type":"pre","directed":true,"reason":"B 复用 A 教的「固定中间一段、两端独立选」这一步分解（先定 B→C 再两侧各选一个），只是把邻居计数换成 BFS 可达矩阵+每侧 top3 分数候选枚举，再叠加 A 未教的可转车 k 次可达预处理与四景点互不相同的约束处理。"},{"id":"pre:luogu/P8855->luogu/P8820","source":"luogu/P8855","target":"luogu/P8820","type":"pre","directed":true,"reason":"B 的查询复用 A 的按 LCA 拆路径步骤，从两端向 LCA 收缩合并两侧状态，再叠加重链剖分与 min-plus 矩阵"},{"id":"pre:luogu/P3366->roj/1393","source":"luogu/P3366","target":"roj/1393","type":"pre","directed":true,"reason":"B 的可选边阶段完全复用了 A 教的按权排序加并查集判环的 Kruskal 选边步骤，只是先用并查集把必选边缩成连通块再对可选边跑同一流程"},{"id":"pre:luogu/P3374->luogu/P1972","source":"luogu/P3374","target":"luogu/P1972","type":"pre","directed":true,"reason":"B 复用 A 的树状数组单点修改加前缀差求和步骤，按右端点离线维护每色最后位置回答区间颜色数"},{"id":"pre:luogu/P1908->luogu/P1972","source":"luogu/P1908","target":"luogu/P1972","type":"pre","directed":true,"reason":"逆序对是最基础的二维偏序计数，掌握树状数组扫描一维统计另一维的模式"},{"id":"pre:luogu/P1494->luogu/P1903","source":"luogu/P1494","target":"luogu/P1903","type":"pre","directed":true,"reason":"B 的区间维护复用 A 的莫队骨架这一步：排序后双指针移动、每次只增删一个位置并 O(1) 更新统计量（B 维护不同颜色数），再叠加 A 未教的第三维时间 t 与修改的应用/撤销回退。"},{"id":"pre:luogu/P3397->luogu/P2822","source":"luogu/P3397","target":"luogu/P2822","type":"pre","directed":true,"reason":"B 用 A 教的二维前缀和递推预处理可整除位置表并 O(1) 读矩形计数；A 的四角差分部分未被用到，复用仅限前缀和步骤"},{"id":"pre:shumeng/CSP202104B->luogu/P2822","source":"shumeng/CSP202104B","target":"luogu/P2822","type":"pre","directed":true,"reason":"B 沿用 A 的二维前缀和递推与 O(1) 矩形查询回答每组询问，再叠加 Pascal 递推 mod k 标记可整除位置"},{"id":"pre:shumeng/CSP201412D->luogu/P1991","source":"shumeng/CSP201412D","target":"luogu/P1991","type":"pre","directed":true,"reason":"B 的 Kruskal 聚类完整复用 A 教的「边按权排序、并查集合并不同连通块、同块跳过」这一步，只是把连成一棵树的停止条件换成连通块数降到 S、答案取最后一次加入的边长。"},{"id":"pre:roj/1391->luogu/P1991","source":"roj/1391","target":"luogu/P1991","type":"pre","directed":true,"reason":"B 的 Kruskal 聚类直接复用 A 教的按边权升序扫描 + 并查集合并两端连通块这一步，只把停止条件从并满 n-1 条边改成连通块数降到 S；A 是普及的基础 MST 模板，B 为普及/提高- 且额外叠加完全图建边、欧氏距离与卫星电话的聚类建模。"},{"id":"pre:luogu/P3366->luogu/P1525","source":"luogu/P3366","target":"luogu/P1525","type":"pre","directed":true,"reason":"B 复用 A 教的「按权排序逐边处理、用并查集 find 判两端是否已连通」这一步做贪心取舍（第一次 find 同组处即答案），再叠加 A 未教的 2n 扩展域并查集表达异组约束与最大值最小化目标。"},{"id":"pre:roj/1547->luogu/P1250","source":"roj/1547","target":"luogu/P1250","type":"pre","directed":true,"reason":"B 直接复用 A 教的 lowbit 链操作（点加沿 i+=lowbit 上跳、区间和=两次前缀和相减）来实现 range_sum 查区间已种树数，再叠加 A 未教的按右端点排序、从右往左补树的贪心。"},{"id":"pre:roj/1535->luogu/P1250","source":"roj/1535","target":"luogu/P1250","type":"pre","directed":true,"reason":"B 的贪心每次都要问区间已有树数，直接复用 A 教的树状数组点加+两次前缀和相减的区间和维护，再叠加按右端点排序、从右补树的贪心流程"},{"id":"pre:roj/1324->luogu/P1250","source":"roj/1324","target":"luogu/P1250","type":"pre","directed":true,"reason":"B 复用 A 的按右端点排序并在区间未满足时取右端点的贪心选位，把“取右端点放一个点”扩展为“从 r 向左补足区间内缺少的树”。"},{"id":"pre:roj/1323->luogu/P2887","source":"roj/1323","target":"luogu/P2887","type":"pre","directed":true,"reason":"A 教的「结束时间最早优先」截止最紧贪心，被 B 的从奶牛出发解法迁移为「maxSPF 最小的奶牛最急」，按 maxSPF 升序依次分配最小可行 SPF。"},{"id":"pre:leetcodecn/sliding-window-maximum->luogu/P14638","source":"leetcodecn/sliding-window-maximum","target":"luogu/P14638","type":"pre","directed":true,"reason":"B 把 A 教的单调队列滑窗候选维护机制用在梯形区域的 x 侧候选上，配合 ST 表查询完成每轮线性扫描"},{"id":"pre:luogu/P2504->luogu/P14362","source":"luogu/P2504","target":"luogu/P14362","type":"pre","directed":true,"reason":"B 的每个子集解法复用 A 教的最小生成树求最小连接费用这一步，再叠加枚举乡镇子集与原图 MST 替换性质的交换论证"},{"id":"pre:luogu/P4047->luogu/P14362","source":"luogu/P4047","target":"luogu/P14362","type":"pre","directed":true,"reason":"B 沿用 A 的 Kruskal 并查集逐边合并建生成树骨架，枚举乡镇子集把修复边与建路边合并跑 Kruskal 再优化掉重复排序"},{"id":"pre:roj/3019->luogu/P11232","source":"roj/3019","target":"luogu/P11232","type":"pre","directed":true,"reason":"B 第二问求最少保留测速仪时直接复用 A 教的按右端点选点贪心（B 仅用分桶代替排序），A 是纯区间选点模板，B 在其上叠加了匀加速物理建模与二分求区间"},{"id":"pre:HDU/4911->luogu/P5149","source":"HDU/4911","target":"luogu/P5149","type":"pre","directed":true,"reason":"B 把相对顺序颠倒的不满值转化成逆序对后，直接用 A 教的归并过程计逆序对统计答案，再叠加姓名到原座位下标的哈希映射"},{"id":"pre:luogu/P3375->luogu/P3318","source":"luogu/P3375","target":"luogu/P3318","type":"pre","directed":true,"reason":"B 在把双旋转条件化为循环串匹配后，原样复用 A 教的 KMP 找模式串所有出现位置这一步，再叠加哈希查集合"},{"id":"pre:luogu/P4924->luogu/P3678","source":"luogu/P4924","target":"luogu/P3678","type":"pre","directed":true,"reason":"B 求旋转等价类时复用 A 教的「按新下标=旧下标映射生成旋转 90° 矩阵」这一步（0/90/180/270 各表示由同一映射复合而成），再叠加 A 未教的边框行列切窗口与最小表示编码去重。"},{"id":"pre:luogu/P1205->luogu/P3678","source":"luogu/P1205","target":"luogu/P3678","type":"pre","directed":true,"reason":"B 复用 A 的矩阵旋转函数这一步骤，对每个窗口枚举各角度旋转结果取字典序最小作规范表示再去重"},{"id":"pre:noi_openjudge/ch0112-09->luogu/P3678","source":"noi_openjudge/ch0112-09","target":"luogu/P3678","type":"pre","directed":true,"reason":"B 复用 A 的旋转实现步骤生成各角度旋转图案，取字典序最小编码作窗口的旋转等价类代表"},{"id":"pre:luogu/P1808->luogu/P8630","source":"luogu/P1808","target":"luogu/P8630","type":"pre","directed":true,"reason":"B 复用 A 教的「字母出现次数完全相同即同类、压成标准形比较」这一步（计数签名就是 A 的标准形的哈希版），再叠加 A 未教的长度 8 滑动窗口统计签名频次后统一应答查询。"},{"id":"pre:roj/3054->luogu/P8112","source":"roj/3054","target":"luogu/P8112","type":"pre","directed":true,"reason":"A 教的「把模式串与文本用分隔符拼成 B#A、一次 Z 函数求出每个后缀与模式串前缀的 LCP」被 B 原样复用为求 reach[i] 的第一步，B 只是在其上另加最少跳数贪心"},{"id":"pre:luogu/P1190->luogu/P8755","source":"luogu/P1190","target":"luogu/P8755","type":"pre","directed":true,"reason":"B 的任务管理复用 A 的「小根堆按结束时刻维护进行中任务、堆顶即最早结束者」这一步（新任务到达时弹出已结束任务回收资源），再叠加 A 未教的按机器分堆、算力余量判定与不足输出 -1。"},{"id":"pre:roj/3149->luogu/P4037","source":"roj/3149","target":"luogu/P4037","type":"pre","directed":true,"reason":"B 的 solve(u) 把 A 教的那一步——子树压成 dp 表后用分组背包逐儿子合并、容量维倒序分配——原样用于 f[u][j][c] 的合并（main.cpp 中 g[cost] 与 f[v][total*need][used] 的 cost 倒序循环），只是把名额维换成金币维并额外叠加「留 j 件给父亲」与枚举合成件数、森林合并。"},{"id":"pre:luogu/P2015->luogu/P4037","source":"luogu/P2015","target":"luogu/P4037","type":"pre","directed":true,"reason":"B 的 solve(u) 沿用 A 教的树上分组背包合并儿子（main.cpp 中 g[cost] 与 f[v][total*need][used] 逐儿子合并），并在 A“子树要留出连接父亲的那一份”上再加金币与件数维度，只多出枚举合成件数和森林合并。"},{"id":"pre:luogu/P2014->luogu/P4037","source":"luogu/P2014","target":"luogu/P4037","type":"pre","directed":true,"reason":"B 的 main.cpp 里 g[cost] 逐个儿子合并 f[v][total*need][used] 的那层循环，就是 A 教的「孩子子树作为物品组、在容量维上分配后合并」的树形分组背包，只是把名额维换成金币维，并额外叠加 f[u][留 j 件给父亲] 这一新维度。"},{"id":"pre:roj/3149->luogu/P4516","source":"roj/3149","target":"luogu/P4516","type":"pre","directed":true,"reason":"A 教的「子树压成 dp[v][j]（子树内选 j 个点）表、把预算在儿子之间切分并逐儿子合并两个名额」正是 B 实际使用的步骤：B 的 f[u][j][sel][cov] 同样以子树内设备数为维度，main.cpp 在 limit_u/limit_v 与 i+j<=k 下把已合并部分名额 i 与儿子名额 j 相加（tmp[i+j]+=ways_u*ways_v），B 只在其上叠加 sel/cov 覆盖状态与计数乘法。"},{"id":"pre:luogu/P2899->luogu/P4516","source":"luogu/P2899","target":"luogu/P4516","type":"pre","directed":true,"reason":"B 直接复用 A 的覆盖状态设计：B 的 cov 就是 A「被儿子覆盖」，A「等父亲覆盖」那一维被 B 推迟到父节点合并时用 cov_v||sel_u 裁定，只在其上叠加恰好选 k 个的树上背包计数。"},{"id":"pre:roj/3599->luogu/P4823","source":"roj/3599","target":"luogu/P4823","type":"pre","directed":true,"reason":"B 的第一步直接复用 A 教的邻项交换论证——同样只比较相邻两人、由局部换序不劣推出全局排序键（B 代码 cmp_dwarf 即按 a_i+b_i 排序），只是把排序键从 a_i×b_i 换成 a_i+b_i，之后再叠加 A 未教的前缀限制改写与大根堆「最多任务数」贪心。"},{"id":"pre:roj/3043->luogu/P4823","source":"roj/3043","target":"luogu/P4823","type":"pre","directed":true,"reason":"B 把 A 教的“先加入集合、前缀超限再用堆反悔删掉极值”的前缀容量贪心，套用到已逃走肩高前缀与截止限制上"},{"id":"pre:roj/3158->luogu/P5785","source":"roj/3158","target":"luogu/P5785","type":"pre","directed":true,"reason":"两题同属任务安排：A 第二步用费用提前计算推出 dp[i]=dp[j]+(sum_t[i]-sum_t[j])(sum_c[n]-sum_c[j])+s(sum_c[n]-sum_c[j])，B 原文第 47 行给出完全相同的式子，再叠单调队列维护下凸壳，正是 A 教的费用提前计算加凸壳两步。"},{"id":"pre:roj/1610->luogu/P5785","source":"roj/1610","target":"luogu/P5785","type":"pre","directed":true,"reason":"B 把转移里随 j 变化的部分写成一次函数并逐决策点建线，直接复用 A 教的「双单调 ⇒ 单调队列维护下凸壳、弹队首查询」这一步，代码里 bad() 交叉相乘弹队尾与 better_front() 弹队首同 A 的 is_shadow()/value() 结构一致；只在此外层叠加「把每批启动时间 s 摊到 sum_c[n]-sum_c[j]」的前缀和建模。"},{"id":"pre:roj/1610->luogu/P2120","source":"roj/1610","target":"luogu/P2120","type":"pre","directed":true,"reason":"B 的斜率优化解法直接复用 A 教的「双单调 ⇒ 单调队列维护下凸壳/凸包」这一步，把分段的二次代价换成前缀和写成的运输代价，模板完全一致"},{"id":"pre:roj/1610->luogu/P2900","source":"roj/1610","target":"luogu/P2900","type":"pre","directed":true,"reason":"B 的主解正是复用 A 教的「斜率与查询双单调 ⇒ 单调队列维护下凸壳」这一模板：A 用它在 dp[j]+(T[i]-T[j]-1-L)^2 上 O(N) 转移，B 把它套到 x_i 递增、y_{j+1} 递减的 dp[j]+x_i*y_{j+1} 上（main.cpp 的 q 队列弹队首/弹队尾），两端点斜率与查询同样单调；B 额外叠加的只是排序删被支配矩形、把分组化为连续分段 DP 这一步，故为台阶式前置。"},{"id":"pre:roj/3158->luogu/P2900","source":"roj/3158","target":"luogu/P2900","type":"pre","directed":true,"reason":"B 的 main.cpp 复用了 A 教的那一步「比较队首与第二点的边斜率、斜率不占优则永久弹出队首」：A 用它在斜率 k=t[i]+s 递增时淘汰队首，B 把同一判定搬到 x_i 递增的转移 dp[i]=min(dp[j]+x_i*y_{j+1}) 上（while(head<tail && slope(q[head],q[head+1])<=b[i].x) head++，队尾同样叉积判劣），只是在外面叠加了排序删被支配矩形、把分组化为连续分段 DP 的新流程。"},{"id":"pre:roj/3158->luogu/P8632","source":"roj/3158","target":"luogu/P8632","type":"pre","directed":true,"reason":"B 沿用 A 的整理法：把转移里与 j 有关的项拆成 f[j]-k·c[j] 式直线，再在对 d_i 的直线 dp_prev[j]+pre_dw[j]-d_i·pre_w[j] 上做斜率优化"},{"id":"pre:roj/1610->luogu/P8632","source":"roj/1610","target":"luogu/P8632","type":"pre","directed":true,"reason":"B 的 main.cpp 逐字复用了 A 教的「把 min 的 DP 转移展开成直线」这一步：都是把 dp[j] + cost(j+1,i) 整理成 i 的公共项加一条来自 j 的直线，再套 A 教的下凸壳队列（B 的 is_bad 与 A 的 is_shadow 同形，弹队头条件也同形），额外流程是把 1 层 dp 变成 dp[k][i] 的 3 层区间分段并按 d_i 递增查询。"},{"id":"pre:roj/1610->luogu/P5017","source":"roj/1610","target":"luogu/P5017","type":"pre","directed":true,"reason":"B 的 main.cpp 把 A 教的「转移展开成直线族、斜率与查询同向单调 ⇒ 单调队列维护下凸壳（队首弹、队尾弹、交叉相乘判死线）」整套照搬：A 用在 dp[j]+(T[i]-T[j]-1-L)^2 上 O(N) 转移，B 用在 (dp[y]+sum[y])-cnt[y]·x 上，只是先多做一层「平移时刻 + 按回程时刻 DP + y=x-m 延迟入队」的建模，故属台阶式前置。"},{"id":"pre:roj/3158->luogu/P5017","source":"roj/3158","target":"luogu/P5017","type":"pre","directed":true,"reason":"B 的斜率优化求解（main.cpp 中的 is_bad 弹尾与 line_value 弹首交叉相乘）就是 A 教的「把 j 相关部分拆成点、用下凸壳+单调队列在单调斜率查询」这一步的直接复用；额外叠加的是把题意平移成 a_i=t_i+m、按回程时刻设 dp[x] 并处理 y=x-m 才入队的新建模"},{"id":"pre:luogu/P2564->luogu/P3594","source":"luogu/P2564","target":"luogu/P3594","type":"pre","directed":true,"reason":"B 沿用 A 的滑动窗口右扩、条件满足即左缩并更新答案的双指针流程，再叠加长度 d 子段最大和的维护"},{"id":"pre:luogu/P1725->luogu/P3594","source":"luogu/P1725","target":"luogu/P3594","type":"pre","directed":true,"reason":"B 的双指针窗口里复用 A 的单调队列取窗口最值机制，队头维护当前窗口内长度恰好 d 的子段和最大值"},{"id":"pre:roj/1599->luogu/P3572","source":"roj/1599","target":"luogu/P3572","type":"pre","directed":true,"reason":"B 的 main.cpp 在窗口 [i-k,i-1] 上原样复用 A 教的单调队列三步：队首弹过期下标、队首取最优前驱、队尾弹掉更差候选；只是把窗口比较键从 h(j)=dp[j-1]-S_j 扩成「先比 dp 再比高度」两层，并额外叠加 q 组询问与 0/1 代价，A 的拆项步骤并未被复用。"},{"id":"pre:roj/1602->luogu/P3572","source":"roj/1602","target":"luogu/P3572","type":"pre","directed":true,"reason":"B 的 main.cpp 原样复用 A 教的三步滑动窗口单调队列（弹过期队首、取队首作最优前驱、按优先级单调插入），只是把窗口换成 [i-k,i-1]、把队内比较从「dp 最小」扩成「先比 dp 再比高度」，并在外层叠加多组询问。"},{"id":"pre:luogu/P1725->luogu/P3572","source":"luogu/P1725","target":"luogu/P3572","type":"pre","directed":true,"reason":"B 的 main.cpp 原样复用 A 教的滑动窗口单调队列三步（弹过期、取队头最优前驱、单调插入），只是把窗口换成 [i-k,i-1]、把队内比较从「dp 最小」扩成「先比 dp 再比高度」。"},{"id":"pre:luogu/U663797->luogu/P2569","source":"luogu/U663797","target":"luogu/P2569","type":"pre","directed":true,"reason":"B 的正解直接复用 A 教的这一步：A 把 dp[r+kv]=max(g[r+jv]-jw)+kw 的线性项提出 max 后，用单调队列维护递减的 g[r+jv]-jw 并取队首最优前驱；B 把同一手法套到 -j*AP_i+max(dp[pre][k]+k*AP_i)（卖出式同理），B 的 main.cpp 里弹出 q[head]<hold-as[day] 与 dp[pre][k]+k*ap[day] 的队尾淘汰和 A 的 main.cpp 三步完全同形，只是把容量下标换成持股数，额外叠加 A 未涉及的 pre=i-W-1 冷却期建模。"},{"id":"pre:luogu/P1725->luogu/P2569","source":"luogu/P1725","target":"luogu/P2569","type":"pre","directed":true,"reason":"B 的买入、卖出转移把 a_step 的滑动窗口单调队列原样套用，只是窗口从 P1725 的「跳来的前驱下标」换成「持股数 k 区间」。"},{"id":"pre:roj/1603->luogu/P2511","source":"roj/1603","target":"luogu/P2511","type":"pre","directed":true,"reason":"A 教的关键观察是「转移的合法前驱是一个随 i 单向右移的连续区间，只需增量维护窗口内的聚合值」，B 的计数 DP 正是在同一观察上把聚合值由 dp 最小值换成 dp[s-1][t] 之和、用双指针窗口增量维护，再叠加 A 未教的「固定上界后统计划分方案数」这一步。"},{"id":"pre:luogu/P1123->luogu/P2704","source":"luogu/P1123","target":"luogu/P2704","type":"pre","directed":true,"reason":"B 复用 A 的行状态压缩加位运算冲突判定再枚举兼容状态转移这一步，把冲突范围扩到两格并让状态记住最近两行"},{"id":"pre:luogu/P1896->luogu/P2704","source":"luogu/P1896","target":"luogu/P2704","type":"pre","directed":true,"reason":"B 直接沿用 A 教的「把一行压成 bitmask、先预处理单行合法状态、再按行做位运算冲突检查的状压转移」这一整套步骤，只是把单行约束从相邻一格放宽为距离 1/2，并把转移窗口由上一行扩到前两行"},{"id":"pre:luogu/P1879->luogu/P2704","source":"luogu/P1879","target":"luogu/P2704","type":"pre","directed":true,"reason":"B 直接沿用 A 教的两步：先枚举满足 ok_self 的单行合法态，再用 cur&pre1、cur&pre2 的同列不相容检查筛掉冲突行，只是把「只依赖上一行」扩成「依赖最近两行」，并把计数改成 +popcount 取 max。"},{"id":"pre:luogu/P1123->luogu/P3888","source":"luogu/P1123","target":"luogu/P3888","type":"pre","directed":true,"reason":"B 复用 A 的「每行选择压成二进制状态、枚举相邻行状态做按行 DP」这一步状压建模，只是窗口从相邻两行扩到 up/mid/down 三行并叠加覆盖检查与双目标取 min"},{"id":"pre:luogu/P1879->luogu/P3888","source":"luogu/P1879","target":"luogu/P3888","type":"pre","directed":true,"reason":"两题都用同一个关键观察（限制只在上下左右，故可把每行压成二进制状态、按行枚举相邻行状态转移），B 直接沿用 A 教的行压状态 + 两行滚动骨架，只是把兼容性判定换成三行覆盖判定的 row_ok，并叠加 (总代价, 油库数) 双关键字取 min；A 的代码里是 if (cur & last) continue 后 dp[now][cur] += dp[pre][last]，B 是同形循环里改判 row_ok 后再累加 cost，属模板级复用加台阶。"},{"id":"pre:luogu/P1896->luogu/P3888","source":"luogu/P1896","target":"luogu/P3888","type":"pre","directed":true,"reason":"B 的行覆盖判定 row_ok 复用 A 教的一行编码为 bitmask + 移位后与相邻行做按位与/或这两步，只是把冲突判定换成三行覆盖判定"},{"id":"pre:roj/3153->luogu/P8756","source":"roj/3153","target":"luogu/P8756","type":"pre","directed":true,"reason":"B 把 A 教的关键观察「行间攻击只往前看两行、状态只需带最近两行掩码」搬到列方向，落实为 dp[col][pre2][pre1][used] 的前两列轮廓转移"},{"id":"pre:luogu/P1879->luogu/P8756","source":"luogu/P1879","target":"luogu/P8756","type":"pre","directed":true,"reason":"B 直接复用 A 教的「把一层压成二进制状态、先预处理单层合法状态、再逐层枚举兼容状态做计数 DP」这一整套按层状压步骤，只是层从行变成列、兼容表从 cur&pre==0 换成 ok1/ok2 两张表，并按马的数量多开一维 used。"},{"id":"pre:luogu/P1896->luogu/P8756","source":"luogu/P1896","target":"luogu/P8756","type":"pre","directed":true,"reason":"B 直接复用 A 教的『一行/一列压成 bitmask + 用 popcount 累加已放置数量、逐行/逐列计数转移』这一状态设计（B 的 main.cpp 逐列转移时 dp[nxt][p1][cur][nused]=…+bit_cnt[cur]），只是把 A 的单行前驱扩成前两列轮廓、把同行相邻限制换成马的跳列攻击限制，属加维度加约束的台阶式叠加。"},{"id":"pre:luogu/P2914->luogu/P8733","source":"luogu/P2914","target":"luogu/P8733","type":"pre","directed":true,"reason":"B 沿用 A 的按距离阈值连欧氏边建图步骤（D 折进补油限制）先 Floyd 求任意两点最短代价，再叠加 TSP 状压 DP"},{"id":"pre:luogu/P1294->luogu/P8733","source":"luogu/P1294","target":"luogu/P8733","type":"pre","directed":true,"reason":"B 的 TSP 状压直接复用 A 教过的 dp[mask][u]（已访问点集 + 当前终点）状态与「v 不在 mask 时 dp[mask|1<<v][v] 由 dp[mask][u] 扩展」的转移，只是先跑 Floyd 把补油限制折进距离矩阵，再把 A 的取 max 改成取 min 并补一条回总部 1 的收尾边，代码 main.cpp 第 58-73 行与 A 的 main.cpp 第 31-45 行同构。"},{"id":"pre:roj/3006->luogu/P7859","source":"roj/3006","target":"luogu/P7859","type":"pre","directed":true,"reason":"A 教过的「mask 第 i 位标记元素是否入选、枚举 2^n 个掩码」被 B 正解直接用作原料子集的表示，B 只在这层之上叠加一步按位与检查冲突对并计数，属典型台阶式复用。"},{"id":"pre:roj/3148->luogu/P2899","source":"roj/3148","target":"luogu/P2899","type":"pre","directed":true,"reason":"B 的最小支配集三状态 DP 把 A 教的「父节点状态锁定儿子可选档」直接推广：A 中 v 参会则儿子都不能参会、只能取不选档，B 中 dp[u][2] 表示 u 不放塔等父亲覆盖，此时儿子被禁止等 u 覆盖、只能取放塔或被儿子覆盖，即 dp[u][2]=Σmin(dp[v][0],dp[v][1])，在此外层再叠加第三档状态与「至少一个儿子放塔」的修正。"},{"id":"pre:roj/3149->luogu/P3177","source":"roj/3149","target":"luogu/P3177","type":"pre","directed":true,"reason":"B 的树形背包复用 A 教的「子树压成 dp 表后逐儿子枚举分给儿子的名额做分组背包合并」这一步：B 的 dp[u][a+b]=max(dp[u][a+b], dp[u][a]+dp[v][b]+add) 与 A 的 dp[v][j+k]=max(dp[v][j+k], dp[v][j]+dp[u][k]) 同型（main.cpp 的 a/b 双层循环即该合并），A 用「选课程得学分」教模板，B 只在其上叠加「边两侧黑/白同色点对数 × 边权」的贡献与 sz 维，难度从提高升到提高+/省选-，构成台阶。"},{"id":"pre:luogu/P2014->luogu/P3177","source":"luogu/P2014","target":"luogu/P3177","type":"pre","directed":true,"reason":"B 直接沿用 A 的 dp[u][j]=子树内选 j 个 + 把名额分给孩子子树的树上背包合并（dp[u][a+b]=dp[u][a]+dp[v][b]+add），只是在每一步叠加了边的黑/白点对贡献，A 是纯模板、B 是带贡献权值的同型合并。"},{"id":"pre:luogu/P1892->luogu/P7846","source":"luogu/P1892","target":"luogu/P7846","type":"pre","directed":true,"reason":"B 处理 t=2 相等约束时复用 A 教的「并查集 union 把受同一约束的点合并成块、按代表元/块再处理」这一步，再叠加 t=0 不同色森林染色计数与二分侧放 1/2 求最小和。"},{"id":"pre:luogu/P1955->luogu/P7846","source":"luogu/P1955","target":"luogu/P7846","type":"pre","directed":true,"reason":"B 复用 A 教的「先把相等关系并查集合并成等价类、再在类间处理不等约束」这两步分解（t=2 缩块后 t=0 只在块间做不同色限制），再叠加 A 未教的块权计数与森林 proper coloring 计数 R*(R-1)^(s-1)。"},{"id":"pre:luogu/P1330->luogu/P7846","source":"luogu/P1330","target":"luogu/P7846","type":"pre","directed":true,"reason":"B 的第二问直接复用 A 教的「连通块二分图染色后按侧计数、把较小侧作为被选侧」这一步（较小侧放值 2 得 A+B+min(A,B)），再叠加并查集缩 t=2 块与森林 proper coloring 计数 R(R-1)^{s-1}。"},{"id":"pre:roj/3148->luogu/P4084","source":"roj/3148","target":"luogu/P4084","type":"pre","directed":true,"reason":"B 的 dp[u][c] 把 A 教的「父节点取某状态则儿子被禁止取同一状态」升级为 3 色同色排除求和后连乘，完成了从 0/1 两态取 max 到有限多态计数的迁移"},{"id":"pre:luogu/P1352->luogu/P4084","source":"luogu/P1352","target":"luogu/P4084","type":"pre","directed":true,"reason":"B 的 dp[u][c] 直接沿用 A 教的「父节点状态禁止孩子取同一状态」这一步，只是把 0/1 两态扩成 3 色、并把取 max 换成同色排除后求和再连乘。"},{"id":"pre:roj/3148->luogu/P3047","source":"roj/3148","target":"luogu/P3047","type":"pre","directed":true,"reason":"B 的第一遍把 A 教的「儿子先于父亲」自底向上递推原样用在 down_dp 上，只是把 A 的两档 0/1 状态换成 0..K 的距离维，再用换根补齐父亲方向"},{"id":"pre:roj/3019->luogu/P1937","source":"roj/3019","target":"luogu/P1937","type":"pre","directed":true,"reason":"B 直接复用了 A 教的按右端点升序贪心顺序，先按右端点排序再逐请求判定，只是把『选最少点』换成用线段树维护区间剩余容量。"},{"id":"pre:luogu/P2564->luogu/P1712","source":"luogu/P2564","target":"luogu/P1712","type":"pre","directed":true,"reason":"B 复用 A 教的「排序后同向双指针滑动窗口、右端扩张左端在窗口合法时尽量收缩」这一步（每个元素至多被两指针各经过一次），再叠加 A 未教的长度窗口枚举目标、离散化与线段树维护区间覆盖最大值判合法。"},{"id":"pre:HDU/1199->luogu/P1712","source":"HDU/1199","target":"luogu/P1712","type":"pre","directed":true,"reason":"B 把端点离散化后在压缩下标上用线段树做区间加减维护覆盖次数最大值，复用 A 教的离散化压缩坐标 + 线段树区间修改，再叠加长度排序双指针滑窗枚举最短窗口"},{"id":"pre:luogu/P1253->luogu/P1607","source":"luogu/P1253","target":"luogu/P1607","type":"pre","directed":true,"reason":"B 用 A 的「线段树区间加+区间最大值查询」这一步维护每段路占用量决定接几头牛，A 的双懒标记赋值复合并未用到，复用限于基础区间操作"},{"id":"pre:luogu/P1816->luogu/P2471","source":"luogu/P1816","target":"luogu/P2471","type":"pre","directed":true,"reason":"B 把 A 教的 ST 表区间最值查询原样用作核心一步，再叠加 lower_bound 定位年份与四类分类讨论"},{"id":"pre:roj/1544->luogu/P2471","source":"roj/1544","target":"luogu/P2471","type":"pre","directed":true,"reason":"B 的 main.cpp 里 build_sparse_table/query_max 与 A 完全相同（st[k][i]=max(st[k-1][i],st[k-1][i+half]) 加 k=floor(log2(len)) 的两块重叠取 max），把 A 教的「两块可重叠 2 幂区间求 max」直接用作查询 (Y,X) 中间已知年份最大降雨量这一步，B 在此外层再叠加 lower_bound 定位与四类分类讨论。"},{"id":"pre:roj/1541->luogu/P2471","source":"roj/1541","target":"luogu/P2471","type":"pre","directed":true,"reason":"B 的正解把 A 教的重叠两段 ST 表查询原样搬来求 (Y,X) 中间已知年份的最大降雨量，在此之上再叠加年份存在性判定与四种分类讨论。"},{"id":"pre:leetcodecn/daily-temperatures->luogu/P4393","source":"leetcodecn/daily-temperatures","target":"luogu/P4393","type":"pre","directed":true,"reason":"B 的贡献法结算直接复用 A 教的「单调递减栈、遇更大值弹栈结算」这一步，且正是 A 弹栈语义（新栈顶=左侧第一个不小于它的值）给出 min(L,R) 的另一侧，再叠加全局最大值外每元素消失一次的贪心论证。"},{"id":"pre:luogu/P1614->luogu/P3353","source":"luogu/P1614","target":"luogu/P3353","type":"pre","directed":true,"reason":"B 复用 A 的滑动窗口逐右端点推进并维护最值步骤，把窗口和改成前缀和 O(1) 计算并聚合同坐标星星"},{"id":"pre:noi_openjudge/ch0106-06->luogu/P1276","source":"noi_openjudge/ch0106-06","target":"luogu/P1276","type":"pre","directed":true,"reason":"B 的模拟解法复用 A 的逐位置标记步骤，把布尔标记扩成老树/空地/树苗三态并统计被砍树苗"},{"id":"pre:luogu/P1918->luogu/P3879","source":"luogu/P1918","target":"luogu/P3879","type":"pre","directed":true,"reason":"B 复用 A 教的「先预处理反向映射、询问时直接查表」这一步（把一对一映射扩成一对多的倒排索引），再叠加 A 未教的篇内 set 去重、编号天然有序免排序与 Trie 备选实现。"},{"id":"pre:luogu/P1481->luogu/P3879","source":"luogu/P1481","target":"luogu/P3879","type":"pre","directed":true,"reason":"B 的 C++ 解法复用 A 的 Trie 建模这一步（逐字符插入、终结节点挂元数据）：A 在终结点标 # 并沿路计数前缀，B 在终结点挂 belong 文章编号列表。注意 B 是单词精确查找，并未使用前缀匹配；倒排索引与篇内 set 去重是 B 叠加的新步骤。"},{"id":"pre:roj/3015->luogu/P6186","source":"roj/3015","target":"luogu/P6186","type":"pre","directed":true,"reason":"B 的初始 inv_cnt[x] 正是复用 A 的树状数组从左往右插入统计左边更大元素这一步，再叠加冒泡轮数的 max(c[x]-k,0) 推导与两棵 BIT 在线维护。"},{"id":"pre:luogu/P3374->luogu/P5200","source":"luogu/P3374","target":"luogu/P5200","type":"pre","directed":true,"reason":"B 用 A 教的 Fenwick 点加 + 前缀和查询维护值域出现次数（sum(a[i]) 得有序集合里比它小的个数），再叠加最长严格递增后缀观察与逐步插入的步数公式"},{"id":"pre:luogu/P2058->luogu/P2564","source":"luogu/P2058","target":"luogu/P2564","type":"pre","directed":true,"reason":"B 的最短覆盖区间双指针直接复用 A 教的「窗口内按种类计数、计数归零/从零变一时同步增减不同种类数」这一步，再叠加按坐标排序打平输入与合法就收缩左端的双指针推进。"},{"id":"pre:luogu/P3029->luogu/P2564","source":"luogu/P3029","target":"luogu/P2564","type":"pre","directed":true,"reason":"B 的最短彩带步骤逐字复用 A 的最短覆盖滑窗：覆盖全部种类后右移左端点收缩到刚好缺一种、每次收缩更新最短长度，再叠加按颜色分组输入与坐标差长度定义"},{"id":"pre:noi_openjudge/ch0112-08->luogu/P2636","source":"noi_openjudge/ch0112-08","target":"luogu/P2636","type":"pre","directed":true,"reason":"B 的三种逆操作之一直接复用 A 教的维吉尼亚解密步骤（密钥长度不足时按取模循环使用、按位移量前后平移模 26），再叠加从后往前逐层撤销与栅栏、QWE 两个新逆函数。"},{"id":"pre:roj/1402->luogu/P2636","source":"roj/1402","target":"luogu/P2636","type":"pre","directed":true,"reason":"B 的维吉尼亚逆变换直接套用 A 教的密钥循环取位（代码 key[i % key_len]）与按偏移反向平移解密，只是把它放进倒序撤销 N 层加密、并叠加栅栏与 QWE 两种 A 未教的新逆变换。"},{"id":"pre:roj/3598->luogu/P2636","source":"roj/3598","target":"luogu/P2636","type":"pre","directed":true,"reason":"B 把 A 教的维吉尼亚逐位模 26 减法（解密即向前平移）连同密钥循环复用直接写成 decode_vigenere 这一层逆变换，再在外层叠加栅栏、QWE 与倒序撤销"},{"id":"pre:leetcodecn/daily-temperatures->luogu/P2897","source":"leetcodecn/daily-temperatures","target":"luogu/P2897","type":"pre","directed":true,"reason":"B 建最大笛卡尔树复用 A 的单调栈「保存未解决候选、遇到更高元素逐个弹出」这一步（弹出的低平台挂到新节点下），再叠加 A 未教的笛卡尔树建模与灌水淹没时刻递推。"},{"id":"pre:leetcodecn/subarray-sum-equals-k->luogu/P2843","source":"leetcodecn/subarray-sum-equals-k","target":"luogu/P2843","type":"pre","directed":true,"reason":"B 把 A 的前缀和+哈希查询历史前缀模型迁移到 k 维差分状态 state(x)=state(y)，用哈希找最远相等状态求最长区间"},{"id":"pre:luogu/P1821->luogu/P2296","source":"luogu/P1821","target":"luogu/P2296","type":"pre","directed":true,"reason":"B 的合法性预处理直接复用 A 的反图技巧这一步：把「原图里谁能到 t」翻成「反图里从 t 出发能搜到谁」，再叠加 A 未教的安全点筛选（出边全都能到 t）与安全子图上的 BFS 最短路。"},{"id":"pre:luogu/P1725->luogu/P3522","source":"luogu/P1725","target":"luogu/P3522","type":"pre","directed":true,"reason":"B 的最长合法区间步骤复用 A 教的单调队列滑动窗口取最值（队首弹过期、队首即窗内最值），这里换成维护窗口内最小 bad_pos，再叠加冲突对转化与树状数组求 bad_pos"},{"id":"pre:luogu/P2251->luogu/P3088","source":"luogu/P2251","target":"luogu/P3088","type":"pre","directed":true,"reason":"B 沿用 A 的单调队列窗口最值原语（队尾淘汰更差候选、队首过期弹出、队首即最值）左右各扫一遍求两侧窗口最大高度"},{"id":"pre:luogu/P1886->luogu/P3088","source":"luogu/P1886","target":"luogu/P3088","type":"pre","directed":true,"reason":"B 两遍套用 A 的单调队列模板弹过期队首、维护队尾单调，分别求左右 D 窗口最大高度"},{"id":"pre:roj/1598->luogu/P3088","source":"roj/1598","target":"luogu/P3088","type":"pre","directed":true,"reason":"B 直接复用 A 教的单调队列滑动窗口极值模板（队首弹出过期、队尾淘汰更差候选），把求窗口最小前缀和换成求左右两侧窗口最大高度，再叠加排序与两倍高度判断"},{"id":"pre:leetcodecn/daily-temperatures->luogu/P9485","source":"leetcodecn/daily-temperatures","target":"luogu/P9485","type":"pre","directed":true,"reason":"B 的第 3 步原样复用 A 教的「单调栈求每个元素左右第一个更高/不低于的位置」这一步来确定每个挡板的影响区间，再叠加 A 未教的前后缀最高值预处理与改低一个挡板的积水增量计算。"},{"id":"pre:luogu/P1293->luogu/P4998","source":"luogu/P1293","target":"luogu/P4998","type":"pre","directed":true,"reason":"B 复用 A 的中位数最小化绝对距离和判定（A 用于选会场，B 用同一判定确定 f(x)=Σ|x-a_i| 的最小值平台），再叠加平台两侧单调、多站互不干扰与取前 k 小归并"},{"id":"pre:luogu/P4779->luogu/P3956","source":"luogu/P4779","target":"luogu/P3956","type":"pre","directed":true,"reason":"B 把格子扩成 (x,y,color,magic) 状态后，直接复用 A 教的堆优化 Dijkstra 定出各状态最小花费，再叠加 0/1/2 三类边权转移与魔法不能连续使用的状态约束。"},{"id":"pre:luogu/P3368->luogu/P1083","source":"luogu/P3368","target":"luogu/P1083","type":"pre","directed":true,"reason":"B 的可行性检查复用 A 教的「区间加塌缩成差分两端修改、前缀和还原」这一步（前 k 份订单的区间借教室一次差分累加后前缀和即每日总量），再叠加 A 未教的二分答案与按天判超限的单调性判定。"},{"id":"pre:roj/1673->luogu/P2697","source":"roj/1673","target":"luogu/P2697","type":"pre","directed":true,"reason":"B 直接复用 A 教的“记录值首次出现位置、再次出现时用旧位置得出答案”这一关键观察：A 用它配出同色的两个编号，B 把值换成前缀和，用同一张首次位置表算出最长零和区间。"},{"id":"pre:luogu/P2957->luogu/P1019","source":"luogu/P2957","target":"luogu/P1019","type":"pre","directed":true,"reason":"B 的 best_overlap 预处理直接复用 A 教的枚举重合长度、比较一串后 len 位与另一串前 len 位的判定，再叠加最小正重叠贪心证明与 DFS 接龙"},{"id":"pre:luogu/P1414->luogu/P6298","source":"luogu/P1414","target":"luogu/P6298","type":"pre","directed":true,"reason":"B 沿用 A 的枚举倍数统计每个约数覆盖个数这一步得 cnt[t] 并算 C(cnt[t],k)，再叠加容斥把倍数方案反转成 gcd 恰好为 t"},{"id":"pre:luogu/P1226->luogu/P3197","source":"luogu/P1226","target":"luogu/P3197","type":"pre","directed":true,"reason":"B 用补集计数得到 m^n − m·(m−1)^(n−1) 后，两次幂运算直接调用 A 教的模快速幂（指数二进制拆分逐位平方取模），再叠加正难则反的计数转化"},{"id":"pre:roj/3000->luogu/P3197","source":"roj/3000","target":"luogu/P3197","type":"pre","directed":true,"reason":"A 教的「指数按二进制拆分 + 底数逐位平方」快速幂，正是 B 在用补集计数得到 m^n - m*(m-1)^(n-1) 后必须调用的同一步幂运算，B 的 main.cpp 里 quick_pow 与 A 的 mod_pow 同构。"},{"id":"pre:roj/1638->luogu/P1292","source":"roj/1638","target":"luogu/P1292","type":"pre","directed":true,"reason":"B 把倒酒抽象成一元同余方程 b·y≡g(mod a) 后，直接复用 A 教的「约去 gcd、再乘模逆元取最小非负解」这一具体步骤，只是把求最小跳跃次数换成求最小正整数 y。"},{"id":"pre:roj/3601->luogu/P1292","source":"roj/3601","target":"luogu/P1292","type":"pre","directed":true,"reason":"B 把倒酒建模成 b·y≡g (mod a) 并除以 g 后，直接复用 A 教的扩展欧几里得求逆元步骤，再取模折回最小正整数解得到 y，只是多了一层把倒酒化为同余的建模"},{"id":"pre:roj/1640->luogu/P2833","source":"roj/1640","target":"luogu/P2833","type":"pre","directed":true,"reason":"B 直接复用 A 教的「用 exgcd 求特解、以 g 整除判有解」这一步，只是把单个未知量 t 扩成通解参数 k 并在 k 的区间上计数"},{"id":"pre:roj/3601->luogu/P2833","source":"roj/3601","target":"luogu/P2833","type":"pre","directed":true,"reason":"A 把 ax≡1(mod b) 化成 ax+by=1 并用扩展欧几里得求出特解，B 的 exgcd 解法直接复用这一步求 ax+by=-c 的一组特解，随后才叠加通解参数化与区间求交"},{"id":"pre:roj/1631->luogu/P2833","source":"roj/1631","target":"luogu/P2833","type":"pre","directed":true,"reason":"A 教的 exgcd 求特解与 g|c 判有解被 B 直接复用，B 在同样的通解族上叠加把 x、y 范围转成参数 k 区间求交的计数流程"},{"id":"pre:roj/3601->luogu/P5656","source":"roj/3601","target":"luogu/P5656","type":"pre","directed":true,"reason":"B 直接复用 A 教的扩展欧几里得系数同步维护算法求 ax0+by0=d 的特解，再叠加通解与不等式分类，A 是 c=1 的特例"},{"id":"pre:roj/1631->luogu/P5656","source":"roj/1631","target":"luogu/P5656","type":"pre","directed":true,"reason":"B 的 exgcd 解 ax+by=c 直接复用 A 教的「先判 gcd 是否整除右端、再把 exgcd 特解整体乘 c/g」这一步（main.cpp 中 c%d!=0 判无解、mul=c/d 求特解）；A 只把它用于单个线性同余方程，B 在此之上叠加通解 x=x0+k·b/d、y=y0-k·a/d 与正整数解 k 区间。"},{"id":"pre:roj/3601->luogu/P1516","source":"roj/3601","target":"luogu/P1516","type":"pre","directed":true,"reason":"B 直接复用 A 教的「同余方程改写成不定方程再用扩展欧几里得求一组解并取模」这一步，只是把模数从互质的 b 推广到 gcd(m-n,L) 未必为 1 的一般情形"},{"id":"pre:luogu/P4779->luogu/P2939","source":"luogu/P4779","target":"luogu/P2939","type":"pre","directed":true,"reason":"B 在改造次数分层的状态图上原样复用 A 的堆优化 Dijkstra 松弛步骤求最短时间"},{"id":"pre:roj/1382->luogu/P2939","source":"roj/1382","target":"luogu/P2939","type":"pre","directed":true,"reason":"B 直接复用 A 的判据“边权全正/非负 ⟹ 用堆优化 Dijkstra 定型”，把同一算法原样套到 (点, 已用改造数) 状态图上，只额外叠加了改造机会这一状态维与 0 权转移"},{"id":"pre:luogu/P1796->luogu/P4822","source":"luogu/P1796","target":"luogu/P4822","type":"pre","directed":true,"reason":"B 把『已用卡数』展开成分层状态 dist[u][used]，复用 A 的「分层图逐层 min 转移」这一步最短路递推（层内走原价、跨层走半价），再叠加 Dijkstra 堆优化"},{"id":"pre:luogu/P4779->luogu/P4822","source":"luogu/P4779","target":"luogu/P4822","type":"pre","directed":true,"reason":"B 把 A 教的「堆优化 Dijkstra（含弹出跳过过期条目）」整体搬到分层状态 (u,used) 的状态图上，再叠加 A 未教的用卡次数进状态分层与 w/2 转移。"},{"id":"pre:roj/1382->luogu/P4822","source":"roj/1382","target":"luogu/P4822","type":"pre","directed":true,"reason":"B 的 main.cpp 把 A 教的堆优化 Dijkstra（弹出即定型 + 用 done 惰性删除过期堆条目）原样搬到 (城市, 已用卡数) 状态图上，只把一维 done 扩成 vis[u][used]，A 未教的分层状态建模是 B 叠加的新流程，故 A 是 B 的前置台阶。"},{"id":"pre:roj/1494->luogu/P2966","source":"roj/1494","target":"luogu/P2966","type":"pre","directed":true,"reason":"两题都把 Floyd 维护的 dist 直接当作『只允许某一批点当中转』的受限最短路：A 用它探测候选环，B 在每轮加入新中转点后用 dist 更新 dist[i][j]+最大点权 的答案，属同一套阶段不变量用法，差别只是 B 的加入顺序由点权排序决定、并对 max 点权额外取一次最大值"},{"id":"pre:luogu/P4779->luogu/P1186","source":"luogu/P4779","target":"luogu/P1186","type":"pre","directed":true,"reason":"B 的正式做法把 A 教的堆优化 Dijkstra 当引擎反复调用（先求一条最短路并记录前驱，再逐条禁用最短路边重跑），再叠加只枚举最短路边的观察与回溯路径"},{"id":"pre:roj/1382->luogu/P1186","source":"roj/1382","target":"luogu/P1186","type":"pre","directed":true,"reason":"B 的整个解法就是把 A 教的堆优化 Dijkstra 模板当作子过程反复调用：第一次跑出 1 到 N 的最短路并记录前驱点/前驱边，回溯出路径边集后逐条禁用重跑同一 Dijkstra 取最大值；A 负责的 Dijkstra 定型与惰性删除是 B 每一步实际执行的一步，B 在其上叠加的是路径恢复与枚举删边的额外流程。"},{"id":"pre:luogu/P4779->luogu/P2914","source":"luogu/P4779","target":"luogu/P2914","type":"pre","directed":true,"reason":"B 直接复用 A 的 Dijkstra 核心步骤「每次取当前距离最小的状态、松弛所有出边」（B 因 n≤1000 改用朴素数组版），再叠加 A 未教的 0 权旧电线与距离不超过 M 才连欧氏边的建图、乘 1000 取整输出。"},{"id":"pre:luogu/P6770->luogu/P2914","source":"luogu/P6770","target":"luogu/P2914","type":"pre","directed":true,"reason":"B 复用 A 教的「非负权图上跑 Dijkstra 求单源最短路」这一步（cost 矩阵上求 1→n 最短路），再叠加 A 未教的 0 权旧电线与距离不超过 M 才能拉线的双状态建图。"},{"id":"pre:roj/3211->luogu/P1522","source":"roj/3211","target":"luogu/P1522","type":"pre","directed":true,"reason":"B 的 main.cpp 直接复用 A 教过的“Floyd 求全源最短路后取最大即直径”这一步，算出每个连通块的最大直径 old_diameter，可见于代码中先 Floyd 再统计 farthest_dist 取最大；新范式只是在此之上加了枚举跨块新边、用 farthest[i]+dist(i,j)+farthest[j] 更新最小直径，因此 A 是 B 的严格前置台阶"},{"id":"pre:luogu/P4779->luogu/P2176","source":"luogu/P4779","target":"luogu/P2176","type":"pre","directed":true,"reason":"B 复用 A 的「堆优化 Dijkstra 求单源最短路」这一步作为求解引擎（第一次求原最短路、候选边加倍后再跑），再叠加只枚举最短路上的边这一剪枝观察"},{"id":"pre:roj/1382->luogu/P2176","source":"roj/1382","target":"luogu/P2176","type":"pre","directed":true,"reason":"B 把 A 教的堆优化 Dijkstra 当黑盒模板调用：第一次跑时在同一次松弛里顺带记录前驱边以回溯最短路，随后对这条路上的每条边重跑同一模板取增量"},{"id":"pre:luogu/P1339->luogu/P3003","source":"luogu/P1339","target":"luogu/P3003","type":"pre","directed":true,"reason":"B 复用 A 教的「从源点跑一次 Dijkstra 得到 dist 数组」这一步（分别以 PB、PA1 为源各跑一次），再叠加 A 未教的两种送货顺序分解与三段距离组合取 min。"},{"id":"pre:roj/1376->luogu/P3003","source":"roj/1376","target":"luogu/P3003","type":"pre","directed":true,"reason":"B 把 A 教的「堆优化 Dijkstra 用出边松弛得到到所有点 dist」原样当子程序（main.cpp 与 A 的堆+vis+松弛结构一致），先 PB 一次取出到两送货点的距离、再从 PA1 一次取出 PA1-PA2 距离，值即 dist 数组，最后才叠加 A 未教的「只有两种送货顺序」枚举取 min。"},{"id":"pre:roj/1381->luogu/P3003","source":"roj/1381","target":"luogu/P3003","type":"pre","directed":true,"reason":"B 的直接复用 A 教的堆优化 Dijkstra——main.cpp 的 dijkstra() 就是 A 的小根堆按 dist 出队模板，B 只是把它调用两次求出 PB、PA1 到两送货点的距离，再叠加「只有两种送货顺序取较小值」这一步额外流程。"},{"id":"pre:luogu/P2951->luogu/P6770","source":"luogu/P2951","target":"luogu/P6770","type":"pre","directed":true,"reason":"B 复用 A 教的「多对一距离需求只需一次单源最短路、不必全源」这一步（B 再用无向图对称性 dist(cow,1)=dist(1,cow) 换源），只是边权非负后从 BFS 换成 Dijkstra，再叠加逐头奶牛判 dist<=M。"},{"id":"pre:roj/1342->luogu/P2888","source":"roj/1342","target":"luogu/P2888","type":"pre","directed":true,"reason":"B 保留 A 教的三重循环以 k 作中转点松弛全源距离矩阵这一步骤，只把转移从加法改成取 max 求最小瓶颈路"},{"id":"pre:luogu/P1339->luogu/P1821","source":"luogu/P1339","target":"luogu/P1821","type":"pre","directed":true,"reason":"B 复用 A 的堆优化 Dijkstra 单源最短路步骤，正图反图各跑一次得到所有牛的往返最短路"},{"id":"pre:roj/2051->luogu/P1821","source":"roj/2051","target":"luogu/P1821","type":"pre","directed":true,"reason":"B 复用 A 教的“换向”：把求所有 i 到 x 的最短路改成从 x 出发的一次单源最短路，只是无向图对称直接用原图，有向图须再建反图跑一次"},{"id":"pre:roj/2037->luogu/P1821","source":"roj/2037","target":"luogu/P1821","type":"pre","directed":true,"reason":"B 把 A 教的「源点换成固定终点、一次单源最短路拿到所有点到该点距离」这一步搬到有向图上，靠反图实现全源去程，再与原图回程相加取最大"},{"id":"pre:luogu/P1339->luogu/P2935","source":"luogu/P1339","target":"luogu/P2935","type":"pre","directed":true,"reason":"B 把 A 教的单源堆优化 Dijkstra 当子程序对每个喜欢牧场各跑一次，利用无向图对称性 dist(u,v)=dist(v,u) 累加出各候选点距离和，再扫描取最小。"},{"id":"pre:roj/1376->luogu/P2935","source":"roj/1376","target":"luogu/P2935","type":"pre","directed":true,"reason":"B 的正解把 A 教的「从单源跑堆优化 Dijkstra、用出边松弛得到到所有点的 dist」原样当作子程序，对每个喜欢的牧场调用一次同一函数把距离累加，再叠加无向图对称换向与编号最小 tie-break；A 只教这一次松弛模板，B 是它的多次调用加统计。"},{"id":"pre:roj/2037->luogu/P2935","source":"roj/2037","target":"luogu/P2935","type":"pre","directed":true,"reason":"B 直接复用 A 教的无向图距离对称观察，把 dist(候选点,喜欢点) 翻转成 dist(喜欢点,候选点)，从而不必枚举候选点、改成从每个喜欢点跑一次 Dijkstra 再累加，只多了平均→求和与多源累加这层流程"},{"id":"pre:luogu/P3258->luogu/P3833","source":"luogu/P3258","target":"luogu/P3833","type":"pre","directed":true,"reason":"B 把 A 教的树上四点差分标记步骤用于路径加操作，再叠加 DFS 序与树状数组查询子树和"},{"id":"pre:roj/1553->luogu/P3833","source":"roj/1553","target":"luogu/P3833","type":"pre","directed":true,"reason":"B 的 add_path 直接复用 A 教的「先求 LCA、再在端点处打差分标记让路径加在 LCA 抵消」这一步，只是把 A 的边差分 diff[w]-=2 改写成点差分 diff[p]-=d、diff[parent(p)]-=d，再叠加子树和展开与两棵树状数组。"},{"id":"pre:luogu/P2307->luogu/P9235","source":"luogu/P2307","target":"luogu/P9235","type":"pre","directed":true,"reason":"B 的最大生成森林构建复用 A 教的「并查集 find 判两端连通性、不同才合并」这一步（Kruskal 逐边判连通块合并），再叠加 A 未教的边权降序取瓶颈与倍增维护路径最小边权。"},{"id":"pre:luogu/P8805->luogu/P9235","source":"luogu/P8805","target":"luogu/P9235","type":"pre","directed":true,"reason":"B 明确在 A 教的倍增 LCA 祖先表与提层上跳查询上扩展 min_edge 表，迁移的是同一套倍增跳层步骤"},{"id":"pre:luogu/P5903->luogu/P9235","source":"luogu/P5903","target":"luogu/P9235","type":"pre","directed":true,"reason":"B 查询第一步「把深的点提到同一层」正是执行 A 教的「k 拆二进制沿 up 表逐位跳 k 级祖先」这一步（这里 k 即深度差），只是沿跳更新最小边权，再叠加 A 未教的同步上跳收尾与最大生成森林建树。"},{"id":"pre:luogu/P5903->luogu/P4427","source":"luogu/P5903","target":"luogu/P4427","type":"pre","directed":true,"reason":"B 的前缀和拆分公式依赖 p=lca(x,y)，求 p 时复用 A 教的倍增表 up[x][j] 预处理与按二进制拆分上跳步骤，B 在此之上叠加 depth^k 前缀和预处理与容斥拆路径。"},{"id":"pre:luogu/P8855->luogu/P4427","source":"luogu/P8855","target":"luogu/P4427","type":"pre","directed":true,"reason":"B 的路径和计算复用 A 的「用 LCA 把路径拆成根到点的组合」这一步：设 p=lca(x,y) 后按根前缀和容斥并把多减的 depth[p]^k 加回，再叠加 A 未教的 depth^k 前缀和预处理与 k≤50 批量幂。"},{"id":"pre:luogu/P8805->luogu/P4427","source":"luogu/P8805","target":"luogu/P4427","type":"pre","directed":true,"reason":"B 原样复用 A 的路径和拆成根到点前缀和再加回 LCA 点的公式，只是把点权从度数换成 depth 的 k 次幂"},{"id":"pre:luogu/P3884->luogu/P8805","source":"luogu/P3884","target":"luogu/P8805","type":"pre","directed":true,"reason":"B 把 A 教的 LCA 拆路径为两条到根链再修正重复段用在点权和上，减两次根到 p 段后加回 deg[p] 保留 LCA 点"},{"id":"pre:HDU/3836->luogu/P5676","source":"HDU/3836","target":"luogu/P5676","type":"pre","directed":true,"reason":"B 复用 A 的 SCC 判定步骤：约数兴奋值与 e_i 同 SCC 即游戏可玩两次，叠加兴奋值图建模与约数枚举"},{"id":"pre:roj/3197->luogu/P5676","source":"roj/3197","target":"luogu/P5676","type":"pre","directed":true,"reason":"A 教的关键观察「互相可达的点合并为同一强连通分量（SCC），同分量内点可整体处理」被 B 直接用作判定条件：B 把兴奋值当状态建值图后，游戏 i 能玩两次当且仅当 e_i 与某个整除 w_i 的值 x 同 SCC（main.cpp 的 scc_id[div_nodes[j]] == my_scc；brute.cpp 亦沿用 A 的两遍 Kosaraju 缩点判非平凡分量），B 额外叠加的只是「后续可选游戏只取决于当前兴奋值」的值图建模与约数枚举这一层。"},{"id":"pre:luogu/P1656->luogu/P7687","source":"luogu/P1656","target":"luogu/P7687","type":"pre","directed":true,"reason":"B 的候选边枚举直接复用 A 的 Tarjan 桥判定这一步（low[v]>dfn[u] 找出全部桥，只有桥才可能成为答案），再叠加 A 未教的子树内 A/B 服务计数与 total-sub 缺服务判定。"},{"id":"pre:luogu/P3388->luogu/P7687","source":"luogu/P3388","target":"luogu/P7687","type":"pre","directed":true,"reason":"B 把 A 的 DFS 树 low/dfn 子树能否绕过回祖先的判定迁移到桥判定（low[v]>dfn[u]），再叠加桥两侧 sub_a/sub_b 服务计数"},{"id":"pre:HDU/3836->luogu/P1653","source":"HDU/3836","target":"luogu/P1653","type":"pre","directed":true,"reason":"B 的最后一步直接套用 A 的「缩点成 DAG 后数入度 0 的源分量与出度 0 的汇分量、答案取 max(P,Q)」这一结论，只叠加网格建模与同高连通块即 SCC 的观察"},{"id":"pre:roj/3197->luogu/P1653","source":"roj/3197","target":"luogu/P1653","type":"pre","directed":true,"reason":"A 详细证明的「缩点 DAG 中每个源分量要补入边、每个汇分量要补出边，一条边只能同时补一对，故 answer=max(源,汇)，单分量特判 0」这一具体结论被 B 原样用作最后一步：B 的 main.cpp 同样用 indeg/outdeg 标记加 max(source_cnt, sink_cnt)，并保留 comp_cnt==1 输出 0 的特判，B 额外叠加的只是网格建模与「同高连通块即 SCC」的观察；B 的 brute.cpp 还直接沿用 A 的 Kosaraju 缩点度数流程。"},{"id":"pre:luogu/P1656->luogu/P8435","source":"luogu/P1656","target":"luogu/P8435","type":"pre","directed":true,"reason":"B 复用 A 的 dfn/low 判定框架并原样沿用记边号只跳反向边的重边处理，只是把判桥不等式换成判割点后弹栈划点双"},{"id":"pre:POJ/1523->luogu/P8435","source":"POJ/1523","target":"luogu/P8435","type":"pre","directed":true,"reason":"B 的点双构造直接复用 A 方法二的 v-BCC 分解步骤：low[v]>=dfn[u] 时把点栈弹到 v 再把 u 并入同一 BCC，且都用「割点属于多个点双」的性质，B 再叠加重边按边号跳反向边与显式栈防爆栈。"},{"id":"pre:luogu/P3388->luogu/P8435","source":"luogu/P3388","target":"luogu/P8435","type":"pre","directed":true,"reason":"B 的点双分解以 A 教的 low[v]>=dfn[u] 割点判据为触发点（子树无法绕过 u 回祖先），据此弹点栈到 v 再加 u 得到一个点双，A 的判定是 B 的实际一步"},{"id":"pre:luogu/P3388->luogu/P8436","source":"luogu/P3388","target":"luogu/P8436","type":"pre","directed":true,"reason":"B 把 A 教的 low/dfn 子树绕回判据从割点判定改写成桥判定（>= 变 >），再删桥二次 DFS 划出边双"},{"id":"pre:luogu/P1656->luogu/P8436","source":"luogu/P1656","target":"luogu/P8436","type":"pre","directed":true,"reason":"B 求边双的第一步直接复用 A 教的 Tarjan 判桥（low[v]>dfn[u] 即桥、记录进入边只跳反向边处理重边），再在其上叠加删桥后第二遍 DFS 收集边双与非递归防爆栈。"},{"id":"pre:luogu/P1265->luogu/P9709","source":"luogu/P1265","target":"luogu/P9709","type":"pre","directed":true,"reason":"B 的主解复用 A 教的 Prim 流程（每次取最近未访问点接入、顺带更新其余点的最优接入距离），只是把欧氏距离换成 BFS 求骑士步数 dist(u,v)+1，并叠加串行攻占转化为 MST 的建模。"},{"id":"pre:luogu/P2504->luogu/P9709","source":"luogu/P2504","target":"luogu/P9709","type":"pre","directed":true,"reason":"B 复用 A 的「把点间距离作边权建完全图、用 Prim O(n^2) 求最小生成树」这一步建模与求解，只是答案从 MST 最大边换成 MST 总边权（串行攻占总天数），并叠加马步 BFS 现算边权"},{"id":"pre:luogu/P2212->luogu/P9709","source":"luogu/P2212","target":"luogu/P9709","type":"pre","directed":true,"reason":"B 沿用 A 的 Prim 逐次选 dist 最小未访问点并更新其它点接入代价的主循环，把更新动作换成从新城市 BFS 出 dist(u,v)+1"},{"id":"pre:shumeng/CSP201412D->luogu/P1195","source":"shumeng/CSP201412D","target":"luogu/P1195","type":"pre","directed":true,"reason":"B 直接套用 A 的 Kruskal 选边步骤，只是终点从连成一块改成连通块数降到 K 时立即停止"},{"id":"pre:roj/1346->luogu/P1195","source":"roj/1346","target":"luogu/P1195","type":"pre","directed":true,"reason":"B 的 Kruskal 直接复用 A 教的并查集模板：用 find 判断边两端是否同属一块，不同才合并计数连通块"},{"id":"pre:roj/1391->luogu/P1195","source":"roj/1391","target":"luogu/P1195","type":"pre","directed":true,"reason":"B 的 main.cpp 直接复用 A 教的 Kruskal 骨架——边按权升序用并查集判环、两端不同块才合并（unite 失败即 continue），只是把终止条件从连成 1 棵树改成停在 K 个连通块并补一个 No Answer 判断，属于在 A 的并查集判环步骤上叠加额外流程。"},{"id":"pre:luogu/P2097->luogu/P2330","source":"luogu/P2097","target":"luogu/P2330","type":"pre","directed":true,"reason":"B 的 Kruskal 复用 A 教的并查集维护连通块并合并这一步，再叠加按边权升序加边得到最大边权最小的生成树"},{"id":"pre:luogu/P3366->luogu/P2504","source":"luogu/P3366","target":"luogu/P2504","type":"pre","directed":true,"reason":"B 复用 A 的最小生成树构造步骤（B 用 Prim 实现）并取树上最大边作为连通全图的最小跳距"},{"id":"pre:luogu/P2330->luogu/P2504","source":"luogu/P2330","target":"luogu/P2504","type":"pre","directed":true,"reason":"B 判定最小可行跳距 D 就是 MST 的最大边长，正是 A 教的最小瓶颈生成树判断（任何方案最大边权不可能更小、MST 已做到），只是 B 用 Prim 替代 Kruskal 并叠加平方比较与猴子计数"},{"id":"pre:luogu/P1111->luogu/P2504","source":"luogu/P1111","target":"luogu/P2504","type":"pre","directed":true,"reason":"B 判定最小可行跳距 D 就是 MST 的最大边长，正是 A 教的『最早连通时刻等于生成树中最后修复那条边（MST 最大边权）』这一步，再叠加平方距离比较与猴子计数"},{"id":"pre:luogu/P1111->luogu/P2916","source":"luogu/P1111","target":"luogu/P2916","type":"pre","directed":true,"reason":"B 复用 A 的并查集按边权从小到大加边建最小生成树这一步，先把点权谈话成本折进边权再求 MST 总代价"},{"id":"pre:luogu/P2330->luogu/P2916","source":"luogu/P2330","target":"luogu/P2916","type":"pre","directed":true,"reason":"B 复用 A 的 Kruskal 求最小生成树这一步，先用点权转边权的建模把总代价化成新边权和再跑 MST"},{"id":"pre:luogu/P3366->luogu/P2916","source":"luogu/P3366","target":"luogu/P2916","type":"pre","directed":true,"reason":"B 把边权改写成 2L+C_u+C_v 后直接复用 A 教的 Kruskal 选边步骤（排序、并查集判两端异块才合并累加），再叠加树上每边走两次与点权分摊到边的代价改写"},{"id":"pre:luogu/P2307->luogu/P9666","source":"luogu/P2307","target":"luogu/P9666","type":"pre","directed":true,"reason":"B 沿用 A 的加边时 find 判两端已连通即成环否则合并的并查集步骤，找出第一条成环边再取森林唯一路径拼环"},{"id":"pre:luogu/P3366->luogu/P8710","source":"luogu/P3366","target":"luogu/P8710","type":"pre","directed":true,"reason":"B 复用 A 的并查集判连通与合并步骤，把每次真实合并事件记成合并树虚点再 DFS 下传消息标记"},{"id":"pre:luogu/P2307->luogu/P1196","source":"luogu/P2307","target":"luogu/P1196","type":"pre","directed":true,"reason":"B 在 A 的并查集 find/union 与路径压缩步骤上叠加距离维护，扩展成带权并查集求队列位置差"},{"id":"pre:roj/1389->luogu/P8686","source":"roj/1389","target":"luogu/P8686","type":"pre","directed":true,"reason":"B 的后继并查集直接复用 A 教的带路径压缩 find 求根骨架（main.cpp 的 find_root 含 fa[x]=find_root(fa[x])），只把「同家族求根」换成 fa[x]=find(x+1) 一路跳到下一个空位，B 的难点是叠加在这层 find 骨架上的后继建模与值域 V。"},{"id":"pre:roj/1346->luogu/P8686","source":"roj/1346","target":"luogu/P8686","type":"pre","directed":true,"reason":"B 把 A 教的「parent 数组 + find 沿父指针找根并路径压缩」原样用作后继并查集的求根（main.cpp 的 find_root 含 fa[x]=find_root(fa[x])），只把 A 的回答询问换成 fa[x]=find(x+1) 一路跳到下一个空位；A 只讲连通性模板，B 的难点是叠加在这层 find 骨架上的后继建模与取值域 V。"},{"id":"pre:luogu/P3366->luogu/P6121","source":"luogu/P3366","target":"luogu/P6121","type":"pre","directed":true,"reason":"B 的动态连通性维护复用 A 的「并查集维护连通块、不同集合就合并」这一步（components 计数随合并增减，==1 判全连通），再叠加 A 未教的倒序重开（逆过程）建模与答案倒回原顺序。"},{"id":"pre:HDU/1116->luogu/P7771","source":"HDU/1116","target":"luogu/P7771","type":"pre","directed":true,"reason":"B 先复用 A 教的有向欧拉路存在性判定（连通 + 恰一点出度=入度+1 为起点、恰一点入度=出度+1 为终点），再叠加 Hierholzer 回溯放点与贪心走小编号出边求字典序最小"},{"id":"pre:roj/3196->luogu/P7771","source":"roj/3196","target":"luogu/P7771","type":"pre","directed":true,"reason":"B 直接复用 A 教出的「出边用光即钉到回路末尾、答案逆序产生」这一迭代 Hierholzer 实现，只是额外叠加连通性/度数判定与出边升序排序来求字典序最小"},{"id":"pre:roj/1528->luogu/P7771","source":"roj/1528","target":"luogu/P7771","type":"pre","directed":true,"reason":"B 的 check_connected/find_start 直接复用 A 教的非零度点弱连通+入出度差合法这一有向欧拉路存在判定，再叠加 Hierholzer 处理字典序最小。"},{"id":"pre:luogu/P4017->luogu/P7113","source":"luogu/P4017","target":"luogu/P7113","type":"pre","directed":true,"reason":"B 复用 A 的拓扑序沿边累计传播步骤，把计数加法换成分数均分并精确通分约分"},{"id":"pre:luogu/P3183->luogu/P7113","source":"luogu/P3183","target":"luogu/P7113","type":"pre","directed":true,"reason":"B 复用 A 的 DAG 拓扑序顺推骨架（入度0起点置初值、逐点沿出边推给后继、出度0汇点汇总），把整数计数换成分数均分"},{"id":"pre:roj/3095->luogu/P7113","source":"roj/3095","target":"luogu/P7113","type":"pre","directed":true,"reason":"A 教的具体转移是「把量按出边数均分给每个后继」——出边带走 1/出度的系数（f[u]=(Σc+Σf[v])/出度），B 的 main.cpp 用 `each = water[u] / outdeg[u]; water[v] = water[v] + each` 原样复用这一均分转移，只把 A 的逆拓扑序期望聚合换成正拓扑序分数流量传播，并叠加精确分数类、多接收口与汇点求和。"},{"id":"pre:luogu/P3183->luogu/P1038","source":"luogu/P3183","target":"luogu/P1038","type":"pre","directed":true,"reason":"B 沿用 A 的按拓扑序逐点沿出边把值累加到后继的传播步骤，把路径计数换成阈值兴奋信号传播并只输出出度为 0 的点"},{"id":"pre:luogu/P3074->luogu/P1038","source":"luogu/P3074","target":"luogu/P1038","type":"pre","directed":true,"reason":"B 复用 A 的 Kahn 拓扑流程（入度零入队、弹点、后继减入度）按拓扑序传播神经元信号"},{"id":"pre:luogu/P2712->luogu/P1038","source":"luogu/P2712","target":"luogu/P1038","type":"pre","directed":true,"reason":"B 的传播模拟直接建立在 A 教的 Kahn 步骤「入度为 0 才能处理、处理后去掉出边使后继入度减一」的队列拓扑序上，只是把删除语义换成沿边累加 C[u]*W[u][v] 信号"},{"id":"pre:roj/1280->luogu/P6145","source":"roj/1280","target":"luogu/P6145","type":"pre","directed":true,"reason":"B 的正式解复用 A 教的『先确定无环顺序、再沿该顺序用 dp[u]+w 转移』这一步：A 用高度升序当拓扑序、按序填表转移，B 换成 Kahn 拓扑排序后逐边做最长路松弛并把 min 改成 max；Kahn 模板与下界初值 S[i] 是额外流程。"},{"id":"pre:roj/1352->luogu/P6145","source":"roj/1352","target":"luogu/P6145","type":"pre","directed":true,"reason":"B 把 A 教的『沿拓扑序用前驱值加边权取 max 松弛后继』原样搬用，只把权重从固定的 1 换成任意 x、初值从 100 换成下界 S[i]，因此 A 是 B 的模板台阶。"},{"id":"pre:luogu/P1359->luogu/P6145","source":"luogu/P1359","target":"luogu/P6145","type":"pre","directed":true,"reason":"B 直接复用 A 教的『DAG 上沿无环顺序用 dp[u]+w 松弛后继 dp[v]』这一步：A 靠编号天然有序，B 换成 Kahn 拓扑序求顺序，再把 min 松弛改成长路 max 并加下界初值 S[i] 作为额外流程。"},{"id":"pre:roj/1352->luogu/P2712","source":"roj/1352","target":"luogu/P2712","type":"pre","directed":true,"reason":"B 的 main.cpp 里 TopologicalSort::kahn_prune() 直接复用 A 教的这一步 Kahn 循环——「出队遍历后继、入度减 1、归零再入队」并用删点数（A 用出队计数 done 判环，B 用 removed 得 n-removed），只是把 A 的「反向建边 + pay[v]=max(pay[u])+1 定薪递推」换成位置到摄像头的建模与剩余计数。"},{"id":"pre:luogu/P6145->luogu/P8893","source":"luogu/P6145","target":"luogu/P8893","type":"pre","directed":true,"reason":"B 把 A 的拓扑序 max 转移步骤改写到规则节点图上，用前提计数归零触发 day[v]=max(day[u])+1 的最早天数传播"},{"id":"pre:luogu/P3183->luogu/P8893","source":"luogu/P3183","target":"luogu/P8893","type":"pre","directed":true,"reason":"B 把规则转成节点后复用 A 的「按拓扑序沿边传播递推」这一步（day[v]=max(day[u])+1 对应 dp[v]+=dp[u]），再叠加规则超边建模与未满足前提数、前提最大完成天数的维护"},{"id":"pre:luogu/P3074->luogu/P8893","source":"luogu/P3074","target":"luogu/P8893","type":"pre","directed":true,"reason":"B 复用 A 教的「拓扑序上取前驱 max 再加常数得最早完成时间」这一步（max(day[u])+1 与 f(i)=max(f(pre))+T[i] 同形，按依赖序单调传播），再叠加 A 未教的题集→规则→目标题超边建模与前提计数归零才触发的判定。"},{"id":"pre:roj/3144->luogu/P1776","source":"roj/3144","target":"luogu/P1776","type":"pre","directed":true,"reason":"A 教的把 C_i 枚硬币二进制拆成 1,2,4,… 组当 0/1 物品这一步，正是 B 把 m_i 件宝物拆成若干个包再做 0/1 背包所实际复用的核心；B 只是把可达性 DP 换成取最大值的一维 01 背包。"},{"id":"pre:luogu/P2347->luogu/P1776","source":"luogu/P2347","target":"luogu/P1776","type":"pre","directed":true,"reason":"B 的二进制拆分正是 A 教的「把有件数限制的物品展开成 0/1 物品后做一维 01 背包」这一步的升级：A 逐枚展开、价值退化成布尔可达，B 用 1,2,4…打包把展开数量从 O(m_i) 压到 O(log m_i)，转移仍是有件数限制物品转 0/1 物品的同一模式，B 另外叠加的只是拆分组数与组内重量价值计算。"},{"id":"pre:luogu/P1734->luogu/P1776","source":"luogu/P1734","target":"luogu/P1776","type":"pre","directed":true,"reason":"B 把每个二进制拆分包当 0/1 物品后，直接复用 A 教的「内层容量倒序枚举、每件至多选一次」的一维 01 背包转移（main.cpp 的 for(j=W;j>=wv;j--) dp[j]=max(dp[j],dp[j-wv]+vv)），B 额外叠加的只是拆分出的包数控制与包内重量价值计算。"},{"id":"pre:luogu/P1175->luogu/P8815","source":"luogu/P1175","target":"luogu/P8815","type":"pre","directed":true,"reason":"B 的第一阶段直接复用 A 的「运算符栈按优先级与括号把中缀转后缀」这一步调度场转换，再叠加由后缀建语法树与短路语义的迭代求值"},{"id":"pre:luogu/P4779->luogu/P9751","source":"luogu/P4779","target":"luogu/P9751","type":"pre","directed":true,"reason":"B 沿用 A 的堆优化 Dijkstra 主循环与过期堆项跳过，把节点状态扩成(点,时刻 mod k)状态图，再叠加入口时刻整体推迟观察"},{"id":"pre:roj/1522->luogu/P8604","source":"roj/1522","target":"luogu/P8604","type":"pre","directed":true,"reason":"B 的建圆方树步骤直接接着 A 的割点判据写：main.cpp 第 28 行仍是 if (low[v] >= dfn[u])，只是 A 在此处标记 is_cut，B 在此处弹出边栈收点双并新建方点；dfn/low 的维护方式与 A 完全一致，B 在此之上叠加圆方树路径计数这一新流程，方向为 A(普及+/提高-) → B(普及+/提高)，满足台阶性。"},{"id":"pre:roj/1525->luogu/P8604","source":"roj/1525","target":"luogu/P8604","type":"pre","directed":true,"reason":"B 的圆方树解法复用 A 教的 low(c)>=dfn(v) 割点判据，只是把该判据作为弹栈建点双的触发条件，再在圆方树路径上数割点"},{"id":"pre:luogu/P1775->luogu/P1622","source":"luogu/P1775","target":"luogu/P1622","type":"pre","directed":true,"reason":"B 沿用 A 的区间 dp[l][r] 枚举分割点、子区间独立合并的区间 DP 骨架，把石子合并的最后断点换成第一个释放的囚犯"},{"id":"pre:luogu/P1775->luogu/P3146","source":"luogu/P1775","target":"luogu/P3146","type":"pre","directed":true,"reason":"B 复用 A 的区间 DP 枚举最后一次断点把区间拆成两段转移这一具体步骤，只是把代价和换成两边相等才合并的取值规则"},{"id":"pre:roj/3549->luogu/P5322","source":"roj/3549","target":"luogu/P5322","type":"pre","directed":true,"reason":"B 的主解（main.cpp：先 ndp = dp 冻结上一组状态，再对每个台阶用 dp[money-cost] 转移）复用了 A 教的「一维容量倒序/旧状态转移做分组背包、每组至多选一个选项」这一步，只是把家族买法换成排序合并得到的阈值台阶，并叠加阈值建模"},{"id":"pre:luogu/U661995->luogu/P5322","source":"luogu/U661995","target":"luogu/P5322","type":"pre","directed":true,"reason":"A 教的分组背包用 previous = dp 冻结上一组状态、组内物品统一从 previous 转移以保证组内最多选一个；B 把每座城堡的台阶方案当成一组的互斥选项，main.cpp 先 ndp = dp 再对每个台阶用 dp[money - cost] 转移，正是同一份旧状态隔离手法，额外只加了排序合并阈值台阶的建模。"},{"id":"pre:roj/3549->luogu/P5365","source":"roj/3549","target":"luogu/P5365","type":"pre","directed":true,"reason":"B 的 main.cpp 先把 dp 备份成 ndp、再对每个英雄用 x 从旧状态 dp[cost] 转移，正是 A 教的『组内选项只能从上一组状态转移、不能叠加』这一步；A 靠容量倒序保证，B 换成 ndp 备份保证，只是把累加收益改成乘法收益并叠加阈值截断。"},{"id":"pre:luogu/P1510->luogu/P5365","source":"luogu/P1510","target":"luogu/P5365","type":"pre","directed":true,"reason":"B 复用 A 教的阈值截断：A 把体积超过 v 的 dp 截断到 v、再线性扫首个 dp≥v 求最小代价，B 把展示方式数与 M 取 min 截断、再扫首个 dp≥M 求最小花费，只是把累加收益换成分组背包的乘法收益。"},{"id":"pre:luogu/U661995->luogu/P5365","source":"luogu/U661995","target":"luogu/P5365","type":"pre","directed":true,"reason":"B 的 main.cpp 第 35-45 行先把 dp 备份成 ndp，再对每个英雄枚举 x≥2 从旧状态 dp[cost] 乘出 ndp[nc]，正是 A 教的『分组背包用 previous 备份冻结上一组状态、组内选项统一从旧状态转移』这一步，用来防止同一英雄重复买皮肤；B 只在其上叠加乘法收益与截断到 M，难度 普及+/提高- → 普及+/提高，方向正确。"},{"id":"pre:roj/3144->luogu/P6771","source":"roj/3144","target":"luogu/P6771","type":"pre","directed":true,"reason":"B 的 main.cpp 直接复用 A 教的布尔可达背包状态设计（dp_h 表示高度 h 是否可达），在其上叠加按 a_i 升序排序与 used 数组限制每种方块数量两步新流程。"},{"id":"pre:luogu/P2347->luogu/P6771","source":"luogu/P2347","target":"luogu/P6771","type":"pre","directed":true,"reason":"B 直接沿用了 A 教的“布尔 dp 表示某个重量/高度是否可达”的可达性背包状态设计（main.cpp 中 dp[j] 转移），再叠加按 a_i 排序与 used 数组限制每种方块数量这两步新流程，因此是 A 基础上的台阶题。"},{"id":"pre:roj/1294->luogu/P2904","source":"roj/1294","target":"luogu/P2904","type":"pre","directed":true,"reason":"B 的一维容量 dp 结构与 A 教的「用 dp 按容量归并选择、做 max 背包转移」是同一步，只是把物品固定成一趟运 k 头、max 换 min；但 A 只教到 0-1 背包的倒序枚举（每件至多一次），B 的真实难度之一「重复使用故为完全背包（正序枚举）」正是 A 未教的相反规则，A 的这一步在 B 中只是骨架，因此属模板级复用。"},{"id":"pre:luogu/U661986->luogu/P2904","source":"luogu/U661986","target":"luogu/P2904","type":"pre","directed":true,"reason":"B 的完全背包主循环直接沿用 A 的按容量维度一维 dp 转移结构，只把 max 改 min、把物品重量换成趟次 k，A 只教到 01 背包（倒序，见 problems/luogu/U661986/index.md:59-61），可重复使用物品、预处理 trip_cost[k] 以及最后减 M 都是 A 未涉及的新增内容，属模板级复用"},{"id":"pre:luogu/U661988->luogu/P2904","source":"luogu/U661988","target":"luogu/P2904","type":"pre","directed":true,"reason":"B 把 A 教的完全背包容量正序枚举原样用在 `for (int j = i; j <= n; j++)` 上，从而让「一趟运 k 头牛」可重复使用，再叠加 trip_cost 预处理与末尾减 M"},{"id":"pre:roj/1294->luogu/P2946","source":"roj/1294","target":"luogu/P2946","type":"pre","directed":true,"reason":"B 直接沿用 A 教的『一维数组按累积量作下标、每件物品只用一次的 0/1 转移』这一步（B 原文也明确把一头奶牛看成一个只能选一次的物品），只把容量下标换成余数下标、把 max 换成方案数累加，再叠加取模与空集去重，难度从 入门 升到 普及-，台阶性成立。"},{"id":"pre:luogu/U661986->luogu/P2946","source":"luogu/U661986","target":"luogu/P2946","type":"pre","directed":true,"reason":"B 直接复用 A 教的『只保留最优状态、不关心前面具体选了哪些物品』这一最优子结构观察，把它从背包切容量视角迁移成按余数设状态的 01 计数 DP。"},{"id":"pre:roj/1294->luogu/P1679","source":"roj/1294","target":"luogu/P1679","type":"pre","directed":true,"reason":"B 把 A 教的容量维枚举方向判定反用：由 A 的倒序改为正序枚举，使 dp[j-w] 取到本轮已更新值，从而让每个四次方数无限复用，再叠加 min 项数与 INF 初始化。"},{"id":"pre:luogu/U661986->luogu/P1679","source":"luogu/U661986","target":"luogu/P1679","type":"pre","directed":true,"reason":"B 直接复用 A 讲透的「一维 dp 容量维枚举方向决定物品使用次数」这一判定：A 用倒序保证每件只选一次，B 的完全背包把同一容量维转移改成正序让 dp[j-w] 用本轮新值以实现无限复用，再叠加 min 项数目标、INF 初始化与四次方数枚举。"},{"id":"pre:luogu/U661988->luogu/P1679","source":"luogu/U661988","target":"luogu/P1679","type":"pre","directed":true,"reason":"B 直接沿用 A 教的一维完全背包骨架（外层枚举物品、内层容量正序枚举以允许同一对象重复使用），只把目标从 max 价值改成 min 项数并把初始化改为 INF，B 的 main.cpp 也确为同型一维正序 dp。"},{"id":"pre:roj/1294->luogu/P2430","source":"roj/1294","target":"luogu/P2430","type":"pre","directed":true,"reason":"B 的 main.cpp 第 34-36 行原样复用 A 教的容量倒序枚举转移，只把容量换成时间、物品换成题目，A 未涉及的是耗时按水平倍率换算的建模环节"},{"id":"pre:luogu/U661986->luogu/P2430","source":"luogu/U661986","target":"luogu/P2430","type":"pre","directed":true,"reason":"B 的 main.cpp 第 34-36 行直接套用 A 教的『容量倒序枚举保证每件物品只选一次』这一步，只把容量维换成时间维 t、物品换成题目；B 额外叠加的、A 完全未涉及的是把知识点耗时乘水平倍率换算成 cost[i] 的物品建模环节，难度由 入门 升到 普及-，台阶性与方向都成立。"},{"id":"pre:luogu/B4141->luogu/P1926","source":"luogu/B4141","target":"luogu/P1926","type":"pre","directed":true,"reason":"B 的作业子问题复用 A 教的 0/1 背包倒序枚举保证每件一次，只是状态从容量换成分数并改成求最少时间"},{"id":"pre:roj/1267->luogu/P1926","source":"roj/1267","target":"luogu/P1926","type":"pre","directed":true,"reason":"B 第一段直接复用 A 教的 0/1 背包（每项作业只选一次、倒序转移）来求达到分数 k 的最少时间，再叠加剩余时间贪心刷题这一新步骤。"},{"id":"pre:roj/1294->luogu/P1164","source":"roj/1294","target":"luogu/P1164","type":"pre","directed":true,"reason":"B 的计数 0/1 背包（main.cpp:24 注释“倒序枚举金额，保证每种菜最多只选一次”）原样沿用 A 教的倒序枚举保证物品至多选一次这一步，只把转移由 max 换成 dp[j]+=dp[j-a[i]]、dp[0] 由 0 改成 1，在同一骨架上叠加“恰好花完”的计数语义"},{"id":"pre:luogu/U661986->luogu/P1164","source":"luogu/U661986","target":"luogu/P1164","type":"pre","directed":true,"reason":"B 的 main.cpp:23-27（以及 main-rainboy.cpp:25-31、main.py:7-9）原样复用 A 教的『容量倒序枚举保证每件物品只选一次』这一步，只把转移从 max(dp[c],dp[c-v]+w) 换成 dp[j]+=dp[j-price]、初始化 dp[0] 从 0 改成 1，得到恰好花完 M 元的方案数；B 原文也明确以“和求最大价值的 01 背包比”为起点，属于在同一 0/1 背包骨架上叠加计数语义，难度从 入门 升到 普及-，台阶性成立。"},{"id":"pre:roj/1260->luogu/P1233","source":"roj/1260","target":"luogu/P1233","type":"pre","directed":true,"reason":"B 的「最少的不升链覆盖数等于最长严格上升子序列长度」直接复用 A 第二问用过的 Dilworth 步骤，A 未教的只是先按长宽降序排序把二维条件降成一维。"},{"id":"pre:luogu/P2362->luogu/P1233","source":"luogu/P2362","target":"luogu/P1233","type":"pre","directed":true,"reason":"B 的 main.cpp 沿用 A 教的「dp[i] 记为以 i 结尾的 LIS 长度、用 dp[j]+1 转移」这一步，只把不下降条件换成严格上升 a[j].w<a[i].w，并在前面叠加 A 未教的排序降维和 Dilworth 最少不升链覆盖，因而比普及- 的 A 更难。"},{"id":"pre:roj/1298->luogu/P1140","source":"roj/1298","target":"luogu/P1140","type":"pre","directed":true,"reason":"B 的 DP 直接沿用 A 教的三前置状态转移结构（左上/上/左），只是把「替换、删 a_i、删 b_j 取 min」换成「字符对字符、字符对空位、空位对字符加打分取 max」，B 的 main.cpp 三条 max 转移与 brute.cpp 的三种递归分支都可指认该步骤。"},{"id":"pre:roj/1276->luogu/P1140","source":"roj/1276","target":"luogu/P1140","type":"pre","directed":true,"reason":"A 教的「三选一收尾前缀 DP」被 B 原样复用：删除 a_i 对应 f(i-1,j) 这一步在 B 里就是 a[i] 对 '-' 的 dp[i-1][j]+score(a[i],'-')，B 只是把取 min 改成取 max 并配上打分表，状态定义与三条前驱完全相同。"},{"id":"pre:luogu/P1177->luogu/P1103","source":"luogu/P1177","target":"luogu/P1103","type":"pre","directed":true,"reason":"B 复用 A 的「按关键字把序列排序」这一步作为 DP 前的预处理，之后的 dp[i][j] 递推是 A 未教的新内容，复用仅限排序这一基础操作"},{"id":"pre:luogu/P8218->luogu/P1569","source":"luogu/P8218","target":"luogu/P1569","type":"pre","directed":true,"reason":"B 的 DP 转移合法性判断复用 A 的前缀和差求区间和这一步：sum(j+1..i)=pre[i]-pre[j] 正是 prefix[r]-prefix[l-1] 取 l=j+1，再叠加 A 未教的前缀划分 DP 与 dp[j] 可达性约束。"},{"id":"pre:noi_openjudge/ch0111-08->luogu/P1808","source":"noi_openjudge/ch0111-08","target":"luogu/P1808","type":"pre","directed":true,"reason":"B 沿用 A 的把元素放入集合自动去重这一步数不同标准形个数，先给单词内部字母排序归一再入集合"},{"id":"pre:noi_openjudge/ch0110-09->luogu/P1808","source":"noi_openjudge/ch0110-09","target":"luogu/P1808","type":"pre","directed":true,"reason":"B 复用 A 教的「set 自动去重、按排序结果作标准形」这一步（单词排序后入 set 数集合大小），再叠加 A 未教的字母异位词标准形与数类别数的目标。"},{"id":"common:luogu/P1059--luogu/P1808","source":"luogu/P1059","target":"luogu/P1808","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：B 复用 A 教的「用 set 自动去重、以排序结果作标准形」这一步（把单词排序后插入 set 数集合大小），再叠加 A 未教的字母异位词标准形与输出类别数的目标。"},{"id":"pre:noi_openjudge/ch0110-10->luogu/P1603","source":"noi_openjudge/ch0110-10","target":"luogu/P1603","type":"pre","directed":true,"reason":"B 沿用 A 的 split 分词后对字符串升序排序这一步拼出最小密码，再叠加单词转数字平方取模与前导零删除"},{"id":"pre:luogu/P5715->luogu/P1603","source":"luogu/P5715","target":"luogu/P1603","type":"pre","directed":true,"reason":"B 的拼接最小化复用 A 的「升序排序后按序输出」这一步：两位字符串升序排序再依次拼接，再叠加 A 未教的单词到数字映射、平方取模两位补零格式化与整体前导零删除。"},{"id":"pre:noi_openjudge/ch0107-17->luogu/P1308","source":"noi_openjudge/ch0107-17","target":"luogu/P1308","type":"pre","directed":true,"reason":"B 复用 A 教的「统一 lower() 规范化实现忽略大小写」这一步做不区分大小写匹配，再叠加 A 未教的两端补空格判完整单词边界与 find/count 定位统计。"},{"id":"pre:noi_openjudge/ch0107-18->luogu/P1308","source":"noi_openjudge/ch0107-18","target":"luogu/P1308","type":"pre","directed":true,"reason":"B 的单词统计用 A 教的子串存在性判定做完整单词匹配（两端补空格后查找 \" word \" 子串并计数与取首现位置），再叠加大小写归一与边界补空格技巧"},{"id":"pre:noi_openjudge/ch0107-16->luogu/P1308","source":"noi_openjudge/ch0107-16","target":"luogu/P1308","type":"pre","directed":true,"reason":"B 复用 A 的转小写规范化步骤，在统一大小写后再做完整单词的补空格查找匹配"},{"id":"pre:noi_openjudge/ch0108-17->luogu/P2919","source":"noi_openjudge/ch0108-17","target":"luogu/P2919","type":"pre","directed":true,"reason":"B 沿用 A 的未访问点洪水填充整块并标记的连通块计数外层流程，只把四连通同字符换成同高度 8 连通并叠加更高邻居判定"},{"id":"pre:roj/1216->luogu/P2919","source":"roj/1216","target":"luogu/P2919","type":"pre","directed":true,"reason":"B 复用 A 教的网格洪水填充与入队即标记避免重复计数的具体做法，把连通块从黑砖扩到同高度平台并加判更高邻居"},{"id":"pre:roj/1249->luogu/P2919","source":"roj/1249","target":"luogu/P2919","type":"pre","directed":true,"reason":"B 的解法直接复用了 A 教的 8 连通洪泛填充：把 A 中「同字符 W 即同片」换成「同高度即同平台」，用同一套迭代扩展搜出整块再去判断有无更高邻居。"},{"id":"pre:roj/1593->luogu/P1123","source":"roj/1593","target":"luogu/P1123","type":"pre","directed":true,"reason":"B 直接复用 A 的合法行掩码横向不相邻判定 `(mask&(mask<<1))==0` 与逐行按位相容转移，只在其上补两条对角位移约束。"},{"id":"pre:roj/1335->luogu/P1332","source":"roj/1335","target":"luogu/P1332","type":"pre","directed":true,"reason":"B 的多源 BFS 直接沿用 A 教的网格四邻扩展模板（越界判定 + 未访问才入队并即时标记），只把单个入栈起点换成全部感染源"},{"id":"pre:roj/1255->luogu/P1332","source":"roj/1255","target":"luogu/P1332","type":"pre","directed":true,"reason":"B 复用 A 教的无权网格 BFS 首次访问即最短距离这一观察，只是把单源改成所有感染源同时入队，得到每格最早感染时间"},{"id":"pre:luogu/P1135->luogu/P3395","source":"luogu/P1135","target":"luogu/P3395","type":"pre","directed":true,"reason":"B 的关键论证正是 A 教的「同一状态第二次到达没有帮助、首访即最短」，用它把 (x,y,t) 时间维状态压缩成每格只记首访时间，再叠加 block_time 最早落障时间预处理与入格时刻约束。"},{"id":"pre:roj/1255->luogu/P1443","source":"roj/1255","target":"luogu/P1443","type":"pre","directed":true,"reason":"B 的正式解直接复用 A 教的网格无权图 BFS 按层扩展、以 -1 兼作访问标记、首次入队即最短步数这一步，只是把四方向换成马步 8 方向、单终点扩成全部格子"},{"id":"pre:noi_openjudge/ch0110-04->luogu/P7910","source":"noi_openjudge/ch0110-04","target":"luogu/P7910","type":"pre","directed":true,"reason":"B 复用 A 的多关键字排序键步骤，用 (值,下标) 二元组复合键排序把稳定插入排序改写成排名查询"},{"id":"pre:noi_openjudge/ch0104-17->luogu/P8831","source":"noi_openjudge/ch0104-17","target":"luogu/P8831","type":"pre","directed":true,"reason":"B 的格里高利历段复用 A 的闰年判定规则这一步（能被 4 整除且非 100 倍数、或被 400 整除，折算成闰年计数式 n/4-n/100+n/400），再叠加 A 未教的儒略历 4 倍数规则、1582 年 10 月跳日与三段式天数统计。"},{"id":"pre:luogu/P1047->luogu/P2692","source":"luogu/P1047","target":"luogu/P2692","type":"pre","directed":true,"reason":"B 在行、列两个方向分别复用 A 教的一维区间差分（left 处加一、right+1 处减一、前缀和统计覆盖），再叠加行/列覆盖的容斥 R×M+C×N−R×C"},{"id":"pre:roj/1102->luogu/P1097","source":"roj/1102","target":"luogu/P1097","type":"pre","directed":true,"reason":"B 的正解（main.cpp）先排序再用 A 教的『线性扫描 + 计数器统计等于目标值的个数』这一步：把固定目标 m 换成当前段值 cur，靠 a[i]==cur 时 cnt++ 统计每段出现次数，只是额外叠加了排序让相同数字连续与换段输出。"},{"id":"pre:roj/1069->luogu/P8813","source":"roj/1069","target":"luogu/P8813","type":"pre","directed":true,"reason":"B 的朴素解与正式解都直接复用 A 教的「result 初始化 1、循环乘 a」累乘求幂骨架，只是在此基础上叠加每次乘前判断是否超上界的截断，属于同一循环的加强版"},{"id":"pre:luogu/P1075->luogu/P1029","source":"luogu/P1075","target":"luogu/P1029","type":"pre","directed":true,"reason":"B 的正式做法复用 A 教的「从 2 试除到 sqrt(n) 分解质因子」这一步来数不同质因子个数 k，再叠加 gcd/lcm 换元 P=x0·a、质因子块去向得 2^k 的组合论证。"},{"id":"pre:roj/1620->luogu/P1217","source":"roj/1620","target":"luogu/P1217","type":"pre","directed":true,"reason":"A 教的「从 2 试除到 ⌊√x⌋ 并用 math.isqrt 取精确上界」这一步，正是 B 的 is_prime 用来判素的同一步（range(3, isqrt(x)+1) 试除），B 只是在其上叠加 A 未教的回文构造（偶数位回文除 11 外可被 11 整除）与区间过滤；难度从普及- 升到普及/提高-。"},{"id":"pre:noi_openjudge/ch0107-01->luogu/P1619","source":"noi_openjudge/ch0107-01","target":"luogu/P1619","type":"pre","directed":true,"reason":"B 的 extract_digits() 复用 A 教的「逐字符 isdigit 筛出数字字符」这一步（提取即计数筛选的取材版），再叠加 A 未教的按原序拼接大整数、素数判定与质因数分解。"},{"id":"pre:roj/1129->luogu/P1619","source":"roj/1129","target":"luogu/P1619","type":"pre","directed":true,"reason":"A 教的「整行读入后逐字符判断 ASCII 数字字符」在 B 的 extract_digits 里被原样复用（ch>='0'&&ch<='9' 收集），B 只在其上叠加去前导零、字符串比界、试除判素与质因数分解，难度从入门升到普及-。"},{"id":"pre:roj/1210->luogu/P9750","source":"roj/1210","target":"luogu/P9750","type":"pre","directed":true,"reason":"A 教的从 2 试除并用 while 除尽统计指数的分解法，正是 B 解法一提取平方因子时对 Δ 试除分解的原样实现，B 只是额外按指数奇偶拆出 s²·r"},{"id":"pre:roj/3013->luogu/P1293","source":"roj/3013","target":"luogu/P1293","type":"pre","directed":true,"reason":"B 把 A 教的“距离和最小点在中位数”这一选址结论直接迁移过来，只是把每个点的人数当作权重推广成带权中位数。"},{"id":"pre:roj/1150->luogu/P1851","source":"roj/1150","target":"luogu/P1851","type":"pre","directed":true,"reason":"B 的 sum_proper_divisors 直接复用 A 教的因子成对平方根试除求约数和，只是把完全数判定换成互查真约数和。"},{"id":"common:luogu/P1571--luogu/P1678","source":"luogu/P1571","target":"luogu/P1678","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：B 把 A 教的排序数组 lower_bound 分界点模型用作第一步，再叠加前驱比较求最近分数线"},{"id":"common:atcoder/abc248_d--luogu/P1678","source":"atcoder/abc248_d","target":"luogu/P1678","type":"common","directed":false,"reason":"同为在有序数组中使用二分查找定位目标值的应用。"},{"id":"pre:luogu/P1047->luogu/P2367","source":"luogu/P1047","target":"luogu/P2367","type":"pre","directed":true,"reason":"B 沿用 A 的区间加只改 l 与 r+1 两端、最后前缀和还原的差分步骤处理批量成绩修改"},{"id":"pre:luogu/P9094->luogu/P5542","source":"luogu/P9094","target":"luogu/P5542","type":"pre","directed":true,"reason":"B 复用 A 教的「区间整体加塌缩成差分边界点修改、最后前缀和还原」这两步，从一维的两端点推广到半开矩形的四角 (+1,-1,-1,+1)（B 原文即“只改四个边界点…再做一次二维前缀和还原”），再叠加 A 未教的 x2/y2 不加 1 的半开边界处理与恰好 K 层判定。"},{"id":"pre:luogu/P2367->luogu/P5542","source":"luogu/P2367","target":"luogu/P5542","type":"pre","directed":true,"reason":"B 把 A 的区间加改差分端点步骤推广到二维矩形四点差分，最后同样用前缀和统一还原"},{"id":"pre:luogu/P3397->luogu/P5542","source":"luogu/P3397","target":"luogu/P5542","type":"pre","directed":true,"reason":"B 复用 A 的二维差分四角容斥修改加二维前缀和还原，只按半开区间把 x2+1、y2+1 换成 x2、y2 再统计恰好 K 层"},{"id":"pre:roj/1360->luogu/P3068","source":"roj/1360","target":"luogu/P3068","type":"pre","directed":true,"reason":"A 教的『出队后枚举后继并入队、每状态只入队一次』队列扩展步骤，被 B 直接用于邀请传播：出队奶牛后扩展所属组、把新触发的牛入队，从而每牛每关系只处理一次；B 只是在此基础上叠加 rest_count 触发条件，故属模板级复用。"},{"id":"pre:luogu/P1614->luogu/P1147","source":"luogu/P1614","target":"luogu/P1147","type":"pre","directed":true,"reason":"B 复用 A 的窗口和滑动维护步骤并把固定长窗改成和目标比较后双指针扩缩"},{"id":"pre:luogu/P1106->luogu/P1901","source":"luogu/P1106","target":"luogu/P1901","type":"pre","directed":true,"reason":"B 复用 A 的单调栈弹掉被当前元素压制的候选这一机制，从删数贪心换到左右最近更高站的求解"},{"id":"pre:leetcodecn/valid-parentheses->luogu/P4387","source":"leetcodecn/valid-parentheses","target":"luogu/P4387","type":"pre","directed":true,"reason":"B 复用 A 的「压入/按栈顶条件弹出/以栈空收尾」这一栈判定流程，只是把括号配对规则换成 pushed→poped 的出栈序列匹配"},{"id":"pre:luogu/P1739->luogu/P3056","source":"luogu/P1739","target":"luogu/P3056","type":"pre","directed":true,"reason":"B 原样复用 A 的 balance 计数与 balance<0 判定步骤，在其上叠加最少翻转的贪心决策"},{"id":"pre:luogu/P1147->luogu/P2058","source":"luogu/P1147","target":"luogu/P2058","type":"pre","directed":true,"reason":"B 的 24 小时窗口沿用 A 教的同向双指针滑动窗口（过期的从左端移出、新的从右端加入、两端只前进故线性），再叠加队列存乘客与国籍计数维护不同国家数"},{"id":"pre:luogu/P1638->luogu/P2058","source":"luogu/P1638","target":"luogu/P2058","type":"pre","directed":true,"reason":"B 的不同国家计数复用 A 的窗口种类计数这一步：进窗计数 0→1 时种类数加一、出窗计数归零时减一，配合滑动窗口淘汰过期元素，再叠加 A 未教的按 86400 秒时间过期与逐船批量加入。"},{"id":"pre:roj/3582->luogu/P2058","source":"roj/3582","target":"luogu/P2058","type":"pre","directed":true,"reason":"B 的 main.py 用 deque 保存窗口乘客，过期时 while 从队头 popleft，正是 A 教的「队头 popleft 淘汰最早进入的元素」这一步，只是把淘汰条件从容量满换成时间超过 86400 秒，并额外叠加 Counter 统计不同国籍的计数层。"},{"id":"pre:luogu/P1114->luogu/P14359","source":"luogu/P1114","target":"luogu/P14359","type":"pre","directed":true,"reason":"B 把 A 的『前缀量相等⇔区间为零、哈希记录历史前缀』迁移为前缀异或：条件化为 pre[l-1]=pre[r]⊕k 后查哈希表，再叠加按结束最早贪心与时间戳清空"},{"id":"pre:luogu/P2697->luogu/P14359","source":"luogu/P2697","target":"luogu/P14359","type":"pre","directed":true,"reason":"B 把 A 教的「区间条件化成两个前缀值的关系再查前缀值出现情况」这一步从加法和=0 推广到异或和=k（哈希查 pre[i]^k 是否出现），再叠加按结束最早贪心分段与时间戳清空。"},{"id":"pre:luogu/P1803->luogu/P14359","source":"luogu/P1803","target":"luogu/P14359","type":"pre","directed":true,"reason":"经典区间贪心，P14359 的核心子问题就是最大不相交区间数量"},{"id":"common:luogu/P4014--luogu/P4015","source":"luogu/P4014","target":"luogu/P4015","type":"common","directed":false,"reason":"同为基本费用流供需匹配建模题，P4014 为二分图完美匹配，P4015 为多仓库多商店的一般运输模型。"},{"id":"pre:roj/1168->OpenJ_Bailian/2506","source":"roj/1168","target":"OpenJ_Bailian/2506","type":"pre","directed":true,"reason":"A 教的「逆序存位、逐位相加传递进位」高精度加法，正是 B 文章为 2×n 铺砖递推 f(n)=f(n-1)+2f(n-2) 的爆范围结果指定的 BigInt Add 写法，B 在此之上叠加线性 DP 递推与逐行读入输出，故为模板级前置。"},{"id":"pre:luogu/P3033->luogu/P2774","source":"luogu/P3033","target":"luogu/P2774","type":"pre","directed":true,"reason":"B 把 A 的「最大独立集 = 总数 - 最小点覆盖」归约升级为加权版「最大权独立集 = 总权 - 最小权点覆盖」再用最小割实现，König 归约这一步直接来自 A，外加网格黑白染色与源汇建图"},{"id":"pre:HDU/3549->luogu/P2774","source":"HDU/3549","target":"luogu/P2774","type":"pre","directed":true,"reason":"先掌握最大流 = 最小割的基本概念，再理解网格图黑白染色的最小割建图。"},{"id":"common:luogu/P2774--luogu/P3749","source":"luogu/P2774","target":"luogu/P3749","type":"common","directed":false,"reason":"同为最小割建模题，都使用 S-正权-T-负权的建图结构，对比不同应用场景。"},{"id":"common:OpenJ_Bailian/4135--luogu/P1182","source":"OpenJ_Bailian/4135","target":"luogu/P1182","type":"common","directed":false,"reason":"同为二分最大段和+贪心分段判定：4135 按月度上限分段计数，P1182 按段和上限分段计数，check 结构可互相套用。"},{"id":"pre:luogu/P2367->luogu/P2879","source":"luogu/P2367","target":"luogu/P2879","type":"pre","directed":true,"reason":"B 把 A 教的「区间整体加减塌缩成差分两端修改、最后前缀和还原」整步照搬（区间减 1 即 D[L]-=1、D[R+1]+=1），再叠加 A 未教的视线关系转化成区间压制与重叠累加、关系去重。"},{"id":"pre:luogu/P1129->luogu/P2765","source":"luogu/P1129","target":"luogu/P2765","type":"pre","directed":true,"reason":"B 的匈牙利增量匹配复用 A 的 DFS 增广这一步：目标被占就递归让对方腾出位置、每次成功增广匹配数加一，再叠加 A 未教的最小路径覆盖定理（柱子数=球数-匹配数）与逐球加点停机判断。"},{"id":"pre:luogu/P2764->luogu/P2765","source":"luogu/P2764","target":"luogu/P2765","type":"pre","directed":true,"reason":"先掌握静态 DAG 最小路径覆盖的拆点法和公式 n - 最大匹配，再处理动态加点增广。"},{"id":"pre:roj/3661->luogu/P5049","source":"roj/3661","target":"luogu/P5049","type":"pre","directed":true,"reason":"B 的 O(N) 正解复用了 A 教的「邻接表升序、每次走最小编号未访问邻居即字典序最小先根序」这一步（代码里 sort(adj[i]) 后按序 dfs），只是把 A 的『枚举环上 c 条边各跑一次贪心取最小』的 O(n^2) 框架换成在遍历中动态断环边，并额外叠加找环标记 on_ring 与回溯备胎比较；A 正是 B 题解里提到的 n^2 基础做法。"},{"id":"pre:shumeng/CSP201509D->HDU/3072","source":"shumeng/CSP201509D","target":"HDU/3072","type":"pre","directed":true,"reason":"B 复用 A 教的「互相可达⇔同一强连通分量、Tarjan 求 SCC」这一步把分支内费用清零并缩点成 DAG，再在其上叠加每个非根 SCC 选最小进入边的 DAG 贪心。"},{"id":"pre:HDU/3836->HDU/3639","source":"HDU/3836","target":"HDU/3639","type":"pre","directed":true,"reason":"B 复用 A 的 SCC 缩点成 DAG 后找入度为 0 分量这一步，反转图后正是这些汇点做 DFS 统计支持人数决定赢家"},{"id":"pre:roj/3197->HDU/3639","source":"roj/3197","target":"HDU/3639","type":"pre","directed":true,"reason":"B 的 Tarjan 缩点后用 dag_indeg[su]++ 统计出度、再以 dag_indeg[i]==0 判定原图汇点，正是 A 教的那一步：缩点后每个汇分量（SCC）都没有出边；B 把它用作反向 DAG 的入度为 0 起点（'只有入度为0的点（原图的汇点）才可能是最终赢家'）再叠加从这些分量 DFS 求点权和。"},{"id":"pre:luogu/P2341->HDU/3639","source":"luogu/P2341","target":"HDU/3639","type":"pre","directed":true,"reason":"需要先理解缩点后汇点和可达性分析，再统计反向 DAG 可达人数。"},{"id":"pre:shumeng/CSP201509D->HDU/3836","source":"shumeng/CSP201509D","target":"HDU/3836","type":"pre","directed":true,"reason":"B 复用 A 教的「Tarjan 求强连通分量、把互相可达的点归为同一 SCC」这一步并在其上缩点成 DAG，再叠加 A 未教的入度 0/出度 0 统计与 max(P,Q) 加边结论。"},{"id":"pre:luogu/P1341->OpenJ_Bailian/1392","source":"luogu/P1341","target":"OpenJ_Bailian/1392","type":"pre","directed":true,"reason":"B 复用 A 的 Hierholzer 回溯记录得逆序路径与贪心走最小边得字典序最小这两步，用在 De Bruijn 图上构造最小序列"},{"id":"pre:luogu/P7771->OpenJ_Bailian/1392","source":"luogu/P7771","target":"OpenJ_Bailian/1392","type":"pre","directed":true,"reason":"B 复用 A 的「Hierholzer 后序入栈、按出边贪心顺序走、逆序产出」这一步求字典序最小欧拉序列，只是把有向图出边升序换成 De Bruijn 图上优先走 0 边"},{"id":"pre:HDU/1878->OpenJ_Bailian/1392","source":"HDU/1878","target":"OpenJ_Bailian/1392","type":"pre","directed":true,"reason":"先掌握欧拉回路判定条件，再理解 De Bruijn 序列的隐式图建模和 Hierholzer 构造。"},{"id":"pre:HDU/1878->OpenJ_Bailian/1300","source":"HDU/1878","target":"OpenJ_Bailian/1300","type":"pre","directed":true,"reason":"先掌握无向图欧拉回路的基本度数条件，再处理含指定起点终点的路径判定。"},{"id":"pre:OpenJ_Bailian/1300->OpenJ_Bailian/1041","source":"OpenJ_Bailian/1300","target":"OpenJ_Bailian/1041","type":"pre","directed":true,"reason":"B 的可行性判定复用 A 的欧拉回路判定这一步（所有点度数为偶数、度数大于 0 的点连通），再叠加 A 未教的 Hierholzer 后序入栈构造与邻接表按街道编号排序求字典序最小。"},{"id":"pre:roj/3196->OpenJ_Bailian/1041","source":"roj/3196","target":"OpenJ_Bailian/1041","type":"pre","directed":true,"reason":"A 讲的 Hierholzer 死胡同回溯、后序入栈并逆序输出正是 B 的主算法，B 只是在其上叠加边编号排序以取字典序最小，属模板级台阶递进。"},{"id":"pre:HDU/1878->OpenJ_Bailian/1041","source":"HDU/1878","target":"OpenJ_Bailian/1041","type":"pre","directed":true,"reason":"先掌握欧拉回路判定条件，再学 Hierholzer 算法构造字典序最小的欧拉回路。"},{"id":"common:OpenJ_Bailian/1041--luogu/P1341","source":"OpenJ_Bailian/1041","target":"luogu/P1341","type":"common","directed":false,"reason":"同为 Hierholzer 算法求最小字典序欧拉路，HDU 1878 是无向图模板。"},{"id":"pre:luogu/P6145->POJ/3018","source":"luogu/P6145","target":"POJ/3018","type":"pre","directed":true,"reason":"B 的最长链 DP 复用 A 的按拓扑序 max 转移：A 显式建图 Kahn 求序，B 用维度排序得到同一偏序线性序（自述相当于 topsort 省了建图）再做 dp[i]=max(dp[j])+1，再叠加排序不等式维度转化"},{"id":"pre:luogu/P3074->POJ/3018","source":"luogu/P3074","target":"POJ/3018","type":"pre","directed":true,"reason":"B 把 A 的拓扑序 DP 最长路步骤换成按维度排序得到的偏序序，再做同型的最长链转移"},{"id":"pre:luogu/P1233->POJ/3018","source":"luogu/P1233","target":"POJ/3018","type":"pre","directed":true,"reason":"B 复用 A 教的「先排序让偏序约束化成下标先后、转移只指向已处理元素，再做 dp[i]=1+max dp[j] 最长链 DP」这一步（两处转移式同形），只是把长度/宽度两键比较推广成 d 维逐维比较并叠加礼物锚定初值。"},{"id":"pre:OpenJ_Bailian/1300->HDU/5883","source":"OpenJ_Bailian/1300","target":"HDU/5883","type":"pre","directed":true,"reason":"B 复用 A 的「连通性+度数奇偶分类判定欧拉路/回路」这一步（全偶为回路、恰两奇为路径），再叠加按起点终点两种情形枚举异或和最大化"},{"id":"pre:roj/1528->HDU/5883","source":"roj/1528","target":"HDU/5883","type":"pre","directed":true,"reason":"B 的 HDU 5883 复用 A 教的「非零度点弱连通」判定：并查集合并所有边端点后统计度>0 的根节点数，>1 输出 Impossible，与 A 的连通条件判定是同一步；B 再叠加 A 未教的异或贡献奇偶分析与回路起点贪心。"},{"id":"pre:HDU/1878->HDU/5883","source":"HDU/1878","target":"HDU/5883","type":"pre","directed":true,"reason":"先掌握无向图欧拉回路/路径的度数判定条件，再处理异或贡献的计算。"},{"id":"pre:POJ/2976->luogu/P3199","source":"POJ/2976","target":"luogu/P3199","type":"pre","directed":true,"reason":"先理解分数规划中 D(x) = w_i - x 的变换思想，再结合 DFS-SPFA 判负环求最小均值环。"},{"id":"pre:luogu/P1642->luogu/P4322","source":"luogu/P1642","target":"luogu/P4322","type":"pre","directed":true,"reason":"先掌握分数规划 + 树形背包 DP 的基本组合，再学 sz 优化避免 O(NK²) 超时。"},{"id":"pre:luogu/P1832->luogu/P2918","source":"luogu/P1832","target":"luogu/P2918","type":"pre","directed":true,"reason":"B 复用 A 的完全背包正序容量枚举保证物品可重复使用，再叠加把超过 h 的状态压缩到 dp[h] 的至少型截断技巧"},{"id":"pre:luogu/P1679->luogu/P2918","source":"luogu/P1679","target":"luogu/P2918","type":"pre","directed":true,"reason":"B 的完全背包正序转移复用 A 教的正序枚举允许同一物品反复使用这一步，再叠加至少型处理（超过 h 的重量统一压到 dp[h]）"},{"id":"pre:roj/2033->luogu/P2918","source":"roj/2033","target":"luogu/P2918","type":"pre","directed":true,"reason":"A 教的完全背包内层金额正序枚举让同一物品可重复选，B 在 main.cpp 的内层正序转移上叠加把超过 h 的重量压到 dp[h] 的至少型处理，因此 A 的正序枚举是 B 解法的实际一步。"},{"id":"common:POJ/2976--luogu/P4951","source":"POJ/2976","target":"luogu/P4951","type":"common","directed":false,"reason":"同为 01 分数规划：二分比值 x 并把分式判定转为 Σ(参数+x·参数) 的线性判定；差别只在验证结构（2976 取最大 n-k 项求和，P4951 求最小生成树）。"},{"id":"common:POJ/2976--luogu/P3199","source":"POJ/2976","target":"luogu/P3199","type":"common","directed":false,"reason":"同为分数规划问题，POJ 2976 用贪心排序验证，P3199 用 SPFA 判负环验证，对比不同检验手段。"},{"id":"pre:roj/3149->luogu/P1642","source":"roj/3149","target":"luogu/P1642","type":"pre","directed":true,"reason":"B 的二分答案内层判定直接复用 A 的树上分组背包（dp[u][j] 在 u 子树选 j 个必含 u 的最大权值和、儿子间分组背包合并、j 倒序），再叠加 01 分数规划二分与重定义权值 w−x·c"},{"id":"pre:luogu/P2014->luogu/P1642","source":"luogu/P2014","target":"luogu/P1642","type":"pre","directed":true,"reason":"先掌握树形背包 DP 的基本转移，再结合分数规划处理树上连通块选点。"},{"id":"pre:POJ/2976->luogu/P1642","source":"POJ/2976","target":"luogu/P1642","type":"pre","directed":true,"reason":"先掌握 01 分数规划中的 D(x) = a_i - x*b_i 变换，再应用到树形 DP 的验证中。"},{"id":"common:luogu/P1642--luogu/P4322","source":"luogu/P1642","target":"luogu/P4322","type":"common","directed":false,"reason":"同为分数规划 + 树形背包的组合问题，P4322 数据更大需要 sz 优化避免 TLE。"},{"id":"pre:luogu/P1775->OpenJ_Bailian/1651","source":"luogu/P1775","target":"OpenJ_Bailian/1651","type":"pre","directed":true,"reason":"B 直接复用 A 教的「区间 DP 倒推最后一步、枚举断点 k、按区间长度递增递推」这一步（dp[i][j]=min_k dp[i][k]+dp[k][j]+cost），只是把最后一次合并换成了最后一张被取走的卡片并改计 A[i]*A[k]*A[j]。"},{"id":"pre:luogu/P2032->luogu/P2627","source":"luogu/P2032","target":"luogu/P2627","type":"pre","directed":true,"reason":"B 沿用 A 的单调队列窗口最值原语，把候选换成 dp[j-1]-S[j] 做 DP 优化，再叠加断点建模与初始断点入队"},{"id":"pre:luogu/P1440->luogu/P2627","source":"luogu/P1440","target":"luogu/P2627","type":"pre","directed":true,"reason":"B 把 A 的单调队列取窗内最值步骤用于 dp[j-1]-S[j] 的滑窗最大值，逐断点转移求最大效率和"},{"id":"pre:luogu/P1714->luogu/P2627","source":"luogu/P1714","target":"luogu/P2627","type":"pre","directed":true,"reason":"B 的转移 dp[i]=S[i]+max(dp[j-1]-S[j]) 复用 A 教的「递增单调队列维护滑窗内前缀式最值、先删过期队头再算答案」这一步，只是队列里存的量从前缀和换成 dp[j-1]-S[j] 并由求最小换成求最大"},{"id":"pre:HDU/2586->luogu/P4281","source":"HDU/2586","target":"luogu/P4281","type":"pre","directed":true,"reason":"B 求三个点的集合点后算 Cost=dist(x,P)+dist(y,P)+dist(z,P) 时复用 A 教的倍增 LCA 预处理与树上距离公式 d[u]+d[v]-2d[lca]，B 在此之上叠加三点两两 LCA 取最深的集合点结论。"},{"id":"pre:HDU/2586->HDU/2874","source":"HDU/2586","target":"HDU/2874","type":"pre","directed":true,"reason":"B 明言是 A 的变种升级版，在判定同树后原样套用 A 的 LCA 距离公式这一步"},{"id":"pre:luogu/P2330->luogu/P4047","source":"luogu/P2330","target":"luogu/P4047","type":"pre","directed":true,"reason":"B 复用 A 的「按边权从小到大 Kruskal+并查集合并」这一步扫描流程，只是合并到集合数等于 k 就停，第一条跨部落边就是最近的两个部落距离"},{"id":"pre:luogu/P1195->luogu/P4047","source":"luogu/P1195","target":"luogu/P4047","type":"pre","directed":true,"reason":"B 的部落划分直接复用 A 教的 K 块即停 Kruskal（并查集从小到大加边、连通块数降到 k 就停），再把最后那条跨集合边解释为最近部落距离并叠加平方距离比较"},{"id":"pre:luogu/P3366->luogu/P4047","source":"luogu/P3366","target":"luogu/P4047","type":"pre","directed":true,"reason":"B 复用 A 的 Kruskal 并查集加边循环，把终止条件从连通改成集合数等于 k，遇到的第一条跨集合边即答案"},{"id":"pre:shumeng/CSP201412D->luogu/P2872","source":"shumeng/CSP201412D","target":"luogu/P2872","type":"pre","directed":true,"reason":"B 的主解把 A 教的「Kruskal 排序逐边、并查集判两端不同连通块才选边」整步搬用，再叠加 A 未教的完全图隐式建边与已有道路先合并的初始化。"},{"id":"pre:roj/1346->luogu/P2872","source":"roj/1346","target":"luogu/P2872","type":"pre","directed":true,"reason":"B 把 A 教的「parent 数组 + find 到根、根相同即同块、union 合并两块」原样用于 Kruskal：先 union 合并已有免费道路的端点，再对排好序的候选边用根是否相同判断能否连入并累加长度；A 只做连通块划分与询问比较，B 在此基础上叠加完全图取距离与边排序。"},{"id":"pre:roj/1391->luogu/P2872","source":"roj/1391","target":"luogu/P2872","type":"pre","directed":true,"reason":"B 把 A 教的 Kruskal 判环步骤照搬到主解：先并查集合并已有道路，再对完全图候选边按长度排序，只保留连接两个不同连通块的边累加长度"},{"id":"pre:luogu/P2097->luogu/P3366","source":"luogu/P2097","target":"luogu/P3366","type":"pre","directed":true,"reason":"B 的成环判断复用 A 的「并查集维护连通块、find 比较代表元判断是否同块」这一步（不同连通块才把边加入生成树），再叠加 A 未教的边权排序贪心与选满 n-1 条判定不连通。"},{"id":"pre:roj/1346->luogu/P3366","source":"roj/1346","target":"luogu/P3366","type":"pre","directed":true,"reason":"B 直接把 A 教的代表元判定（find 相同即同块）用作是否成环的判断，并把 A 用于建块的 union 换成按边权依次合并两端，A 是纯并查集模板，B 在其上叠加排序与选边流程，难度差 2 级"},{"id":"pre:luogu/P2692->HDU/5124","source":"luogu/P2692","target":"HDU/5124","type":"pre","directed":true,"reason":"B 的覆盖层数统计复用 A 的一维差分区间覆盖这一步：L 处 +1、R+1 处 −1，前缀和还原每个点的覆盖次数，再叠加 A 未教的坐标离散化与扫描取最大覆盖层数。"},{"id":"pre:luogu/P9094->HDU/5124","source":"luogu/P9094","target":"HDU/5124","type":"pre","directed":true,"reason":"B 的最大重叠区间统计复用 A 教的差分区间加两端点修改（l 处加 1、r+1 处减 1、前缀和还原），再叠加离散化 rank 映射与扫描求最大覆盖层数"},{"id":"pre:luogu/P2367->HDU/5124","source":"luogu/P2367","target":"HDU/5124","type":"pre","directed":true,"reason":"B 原样复用 A 的区间加改 L 与 r+1 差分端点步骤，再叠加离散化处理 10^9 的大坐标"},{"id":"pre:HDU/5124->HDU/1199","source":"HDU/5124","target":"HDU/1199","type":"pre","directed":true,"reason":"B 原样复用 A 教的「离散化把端点和 R+1 一起收进关键点表」这一步（存 a 与 b+1 才能划出精确边界），并进一步用相邻关键点构成基本区间，再叠加 A 未教的线段树懒标记染色与基本区间合并求最长白色段。"},{"id":"pre:shumeng/CSP202104B->HDU/6514","source":"shumeng/CSP202104B","target":"HDU/6514","type":"pre","directed":true,"reason":"B 沿用 A 的二维前缀和递推与四角容斥矩形求和查被监控格子数，再叠加二维差分批量标记监控矩形"},{"id":"pre:luogu/P3397->HDU/6514","source":"luogu/P3397","target":"HDU/6514","type":"pre","directed":true,"reason":"B 的构建覆盖图步骤完整复用 A 的二维差分四角容斥修改与二维前缀和还原，再叠加 0/1 覆盖矩阵的二次前缀和与 RealCount==Area 判定"},{"id":"pre:luogu/P9094->HDU/6273","source":"luogu/P9094","target":"HDU/6273","type":"pre","directed":true,"reason":"B 复用 A 的按类别分别开差分数组做区间加（l 加 1、r+1 减 1）再前缀和还原这一步，把三种颜色换成 2 与 3 的质因子指数"},{"id":"pre:luogu/P1090->HDU/2527","source":"luogu/P1090","target":"HDU/2527","type":"pre","directed":true,"reason":"B 的哈夫曼编码总长直接复用 A 的每次合并最小两堆并累加合并代价步骤，WPL 就是合并代价总和"},{"id":"pre:luogu/P9094->luogu/P3948","source":"luogu/P9094","target":"luogu/P3948","type":"pre","directed":true,"reason":"B 的阶段一区间加完全复用 A 教的差分端点标记 diff[L]+=X、diff[R+1]-=X 再前缀和还原这一步，又把同一前缀和思想用到答案数组 ans_sum[R]-ans_sum[L-1] 做 O(1) 大量查询。"},{"id":"pre:luogu/P8218->luogu/P3948","source":"luogu/P8218","target":"luogu/P3948","type":"pre","directed":true,"reason":"B 的离线海量查询阶段复用 A 教的前缀和作差：先标记每个位置是否合格再建 ans_sum，直接用 ans_sum[R]-ans_sum[L-1] O(1) 回答"},{"id":"pre:luogu/P2367->luogu/P3948","source":"luogu/P2367","target":"luogu/P3948","type":"pre","directed":true,"reason":"B 的 A L R X 区间加直接复用 A 的「区间加只动 diff[l] 与 diff[r+1] 两个边界点再前缀和还原」这一步，再叠加 A 未教的少量 Q 查询暴力还原与答案前缀和 O(1) 回答 Final。"},{"id":"pre:luogu/P1090->POJ/1521","source":"luogu/P1090","target":"POJ/1521","type":"pre","directed":true,"reason":"B 求哈夫曼带权路径长直接复用 A 教的「每次弹出两个最小值合并、把和累加进答案再压回堆、直到只剩一堆」这一步，再叠加字符频率统计与定长 8 位压缩率对比、单字符特判。"},{"id":"pre:roj/1358->luogu/P1175","source":"roj/1358","target":"luogu/P1175","type":"pre","directed":true,"reason":"B 的中缀转后缀直接沿用 A 教的运算符栈优先级比较弹栈步骤，只在同一规则上追加 ^ 右结合特判与逐步模拟输出。"},{"id":"pre:roj/1356->luogu/P1175","source":"roj/1356","target":"luogu/P1175","type":"pre","directed":true,"reason":"B 的第一步中缀转后缀直接沿用 A 教的“栈顶优先级不低于当前就弹出”的调度场弹栈规则，只在 ^ 上追加右结合特判，额外新增的是逐层维护后缀序列并输出每一步。"},{"id":"pre:luogu/P3225->POJ/2942","source":"luogu/P3225","target":"POJ/2942","type":"pre","directed":true,"reason":"B 复用 A 教的 Tarjan 求 v-BCC 分解这一具体步骤，把补图按割点切成 BCC 后再逐块二分图染色判奇环"},{"id":"pre:POJ/1523->POJ/2942","source":"POJ/1523","target":"POJ/2942","type":"pre","directed":true,"reason":"需要先掌握 v-BCC 分解和边入栈的 Tarjan 写法，再结合二分图染色处理奇环传染定理。"},{"id":"pre:luogu/P2746->luogu/P2272","source":"luogu/P2746","target":"luogu/P2272","type":"pre","directed":true,"reason":"B 复用 A 教的「SCC 缩点成 DAG、只保留跨分量的边」这一步（id[u]!=id[v] 才加新边），再叠加 A 未教的缩点边去重、DAG 上加权最长路与方案数计数（g[v] 等长累加）。"},{"id":"pre:luogu/P2863->luogu/P2272","source":"luogu/P2863","target":"luogu/P2272","type":"pre","directed":true,"reason":"B 的算法第一步原样沿用 A 教的 Tarjan 求 SCC 并记录每分量 size，再把 size[i] 作为缩点 DAG 最长路计数的点权"},{"id":"pre:luogu/P3387->luogu/P2272","source":"luogu/P3387","target":"luogu/P2272","type":"pre","directed":true,"reason":"需要先掌握缩点后在 DAG 上做路径 DP，再扩展到最长半连通子图计数。"},{"id":"pre:roj/1522->POJ/1523","source":"roj/1522","target":"POJ/1523","type":"pre","directed":true,"reason":"B 的方法一直接复用 A 教过的非根割点判据 low[v]>=dfn[u]，并把它从“判定是否割点”升级为“每命中一个孩子就给删除后的子网数 subnet_cnt 加 1”，本质是对同一步的计数扩展"},{"id":"pre:POJ/1144->POJ/1523","source":"POJ/1144","target":"POJ/1523","type":"pre","directed":true,"reason":"更基础的割点模板题，先掌握 Tarjan 求割点后再学习 v-BCC 分解视角。"},{"id":"pre:HDU/1269->luogu/P2746","source":"HDU/1269","target":"luogu/P2746","type":"pre","directed":true,"reason":"B 的第一步原样复用 A 教的 Tarjan 求强连通分量，再叠加缩点成 DAG 后统计入度出度为 0 分量的新步骤"},{"id":"pre:luogu/P3387->luogu/P2746","source":"luogu/P3387","target":"luogu/P2746","type":"pre","directed":true,"reason":"需要先会 Tarjan 缩点并把原图变成 DAG，再统计源点和汇点。"},{"id":"pre:HDU/1269->luogu/P2341","source":"HDU/1269","target":"luogu/P2341","type":"pre","directed":true,"reason":"B 的算法流程第一步就是 A 教的 Tarjan 求出所有强连通分量，再在其上缩点建 DAG 并统计出度为 0 的分量找明星奶牛"},{"id":"pre:luogu/P2863->luogu/P2341","source":"luogu/P2863","target":"luogu/P2341","type":"pre","directed":true,"reason":"需要先掌握统计 SCC 大小，再在缩点 DAG 上分析出度为 0 的分量。"},{"id":"common:POJ/2186--luogu/P2341","source":"POJ/2186","target":"luogu/P2341","type":"common","directed":false,"reason":"同为受欢迎的牛模型，缩点后判断唯一汇点 SCC。"},{"id":"pre:luogu/P6145->luogu/P3387","source":"luogu/P6145","target":"luogu/P3387","type":"pre","directed":true,"reason":"B 缩点后的 DAG 直接复用 A 教的拓扑排序加按拓扑序 max 松弛转移来求最大权值和路径"},{"id":"pre:HDU/1269->luogu/P3387","source":"HDU/1269","target":"luogu/P3387","type":"pre","directed":true,"reason":"先掌握 Tarjan 求 SCC，再理解缩点后在 DAG 上做 DP。"},{"id":"pre:luogu/P2251->luogu/P2629","source":"luogu/P2251","target":"luogu/P2629","type":"pre","directed":true,"reason":"B 把条件转成「窗口内最小前缀和 ≥ 基准前缀」后直接复用 A 的「单调队列维护滑动窗口最小值、队首即窗口最值」这一步逐个判定起点 k，额外叠加断环成链与前缀和条件转化"},{"id":"pre:luogu/P1886->luogu/P2629","source":"luogu/P1886","target":"luogu/P2629","type":"pre","directed":true,"reason":"B 的滑动窗口最小前缀直接复用 A 的单调队列模板：s[i] 比队尾小就弹队尾、队头下标滑出窗口就弹队头、队头即窗口最小值，再叠加 A 未教的断环成链与 S[head]>=S[k-1] 的起点合法性判定。"},{"id":"pre:luogu/P1714->luogu/P2629","source":"luogu/P1714","target":"luogu/P2629","type":"pre","directed":true,"reason":"先掌握前缀和 + 单调队列求滑动窗口最小值，再处理断环成链 + 窗口验证的组合。"},{"id":"pre:luogu/P1714->luogu/P1725","source":"luogu/P1714","target":"luogu/P1725","type":"pre","directed":true,"reason":"B 的转移优化复用 A 教的「单调队列维护滑动窗口最值、先删除过期队头再加入新候选」这一步（dp[i]=A[i]+max(dp[j]) 的窗口最值就靠它），再叠加 A 未教的跳跃模型与「答案可从 i+R>N 直接跳出」的非固定终点处理。"},{"id":"pre:luogu/P1440->luogu/P1725","source":"luogu/P1440","target":"luogu/P1725","type":"pre","directed":true,"reason":"B 的转移优化复用 A 的滑动窗口单调队列取最值这一步（过期前驱弹出后队头即窗口内 dp 最大位置），把 O(nR) 前驱枚举压成 O(n)，再叠加 A 未教的 dp 转移设计与最后一步可越过 N 的答案收集。"},{"id":"pre:luogu/P2032->luogu/P1725","source":"luogu/P2032","target":"luogu/P1725","type":"pre","directed":true,"reason":"B 把 A 教的「单调队列维护滑动窗口最大值」这一步直接用在 DP 转移 dp[i]=A[i]+max(dp[j]) 的前驱窗口上，再叠加最后一步可越过 N 时用 dp[i] 更新答案的边界处理。"},{"id":"pre:luogu/P2251->luogu/P1714","source":"luogu/P2251","target":"luogu/P1714","type":"pre","directed":true,"reason":"B 把 A 教的滑窗最小值单调队列原样套在前缀和数组上，用当前前缀减窗口最小前缀得限长最大子段和"},{"id":"pre:luogu/P1886->luogu/P1714","source":"luogu/P1886","target":"luogu/P1714","type":"pre","directed":true,"reason":"B 的限长最大子段和直接复用 A 的「单调队列维护滑动窗口最值、删过期队头后队头即答案」这一步，只是队列里存的是前缀和并以当前前缀减窗口最小前缀得到答案"},{"id":"pre:luogu/P2032->luogu/P1714","source":"luogu/P2032","target":"luogu/P1714","type":"pre","directed":true,"reason":"先掌握单调队列维护滑动窗口最值。"},{"id":"pre:luogu/P2251->luogu/P2032","source":"luogu/P2251","target":"luogu/P2032","type":"pre","directed":true,"reason":"B 复用 A 教的「单调队列滑窗最值：队尾劣者永久淘汰、队首过期弹出、队首即最值」这一步（B 原文同样是“更早过期且更小可以永久删除”），只是从窗口最小值换成最大值。"},{"id":"common:luogu/P1714--luogu/P2032","source":"luogu/P1714","target":"luogu/P2032","type":"common","directed":false,"reason":"同为单调队列维护滑动窗口的应用。"},{"id":"pre:luogu/P1714->luogu/P1419","source":"luogu/P1714","target":"luogu/P1419","type":"pre","directed":true,"reason":"B 的二分答案检验把均值判定化成 b[i]=a[i]-est 后，逐 i 取 [i-T,i-S] 窗口内最小前缀做 S[i]-min 判定，直接复用 A 教的「单调队列取滑动窗口最小前缀和」这一步，外面再套二分答案与数组平移。"},{"id":"common:atcoder/abc248_d--luogu/P1571","source":"atcoder/abc248_d","target":"luogu/P1571","type":"common","directed":false,"reason":"同难度同型题（M5 自 pre 移入；master 重新定级后两者同档）：B 的查询原样复用 A 教的「lower_bound 找第一个 >=x 的位置、再判该位置值是否等于 x」这一步来划定数值 X 的块边界并判存在，再叠加 A 未教的 (数值,下标) 双关键字排序与在块内二次二分下标求 rr-ll 计数。"},{"id":"pre:luogu/P1047->luogu/P9094","source":"luogu/P1047","target":"luogu/P9094","type":"pre","directed":true,"reason":"B 复用 A 的区间加在 l 记 +1、r+1 记 -1 再前缀和统一还原的差分步骤，扩成黄蓝红各一个差分数组后按出现与否判绿色"},{"id":"pre:roj/3196->luogu/P1127","source":"roj/3196","target":"luogu/P1127","type":"pre","directed":true,"reason":"B 的 main.py 直接复用 A 教的迭代 Hierholzer：走不动就出栈并把边加入结果，最后反转后序，只是把点换成单词并加字典序取尾部边"},{"id":"pre:roj/1528->luogu/P1127","source":"roj/1528","target":"luogu/P1127","type":"pre","directed":true,"reason":"B 直接复用 A 教的『单词降维成首尾字母有向边 + 各字母 out-in 只取 0/±1 且 ±1 至多各一个』作存在性判定与起点定位（代码里同样扫描 difference==1/-1），再叠加有序 Hierholzer 与降序取边输出字典序最小词链。"},{"id":"pre:HDU/1116->luogu/P1127","source":"HDU/1116","target":"luogu/P1127","type":"pre","directed":true,"reason":"先掌握有向图欧拉路径的入度/出度判定条件，再学 Hierholzer 算法输出字典序最小路径。"},{"id":"pre:HDU/1116->luogu/P1341","source":"HDU/1116","target":"luogu/P1341","type":"pre","directed":true,"reason":"B 沿用 A 的把字母对建成图点边、问题化为欧拉路的建模步骤，把有向判定换成无向奇度点条件并叠加 Hierholzer 字典序构造"},{"id":"pre:OpenJ_Bailian/1300->luogu/P1341","source":"OpenJ_Bailian/1300","target":"luogu/P1341","type":"pre","directed":true,"reason":"B 原样复用 A 教的「无向图全偶度=欧拉回路、恰两个奇点=欧拉路且奇点为起终点」这套奇偶判据（B 的判别条件逐条对应 A 的情况 A/B），再叠加 A 未教的字母对建边、字典序起点选择与 Hierholzer 构造。"},{"id":"pre:HDU/1878->luogu/P1341","source":"HDU/1878","target":"luogu/P1341","type":"pre","directed":true,"reason":"先掌握欧拉回路度数判定条件，再学 Hierholzer 算法构造最小字典序路径，注意自环处理。"},{"id":"pre:roj/1479->roj/1485","source":"roj/1479","target":"roj/1485","type":"pre","directed":true,"reason":"同专题的 AC 自动机入门题（Keywords Search），先掌握 Trie + fail 的建法与多模式匹配扫描"},{"id":"pre:roj/1481->roj/1485","source":"roj/1481","target":"roj/1485","type":"pre","directed":true,"reason":"同专题的 AC 自动机应用题（Censoring），已熟悉转移表补全与 dead 沿 fail 链下传后可读本题的 DP 建模"},{"id":"pre:roj/1479->roj/1481","source":"roj/1479","target":"roj/1481","type":"pre","directed":true,"reason":"同专题的 AC 自动机入门题（Keywords Search），先掌握 Trie + fail 的建法与扫描"},{"id":"pre:luogu/P3808->roj/1481","source":"luogu/P3808","target":"roj/1481","type":"pre","directed":true,"reason":"AC 自动机简单版模板题，掌握 fail 指针建立后方可理解本题的转移表补全"},{"id":"common:luogu/P3121--roj/1481","source":"luogu/P3121","target":"roj/1481","type":"common","directed":false,"reason":"同一模型的英文原题 Censoring Gold，同样的 AC 自动机 + 栈回退解法"},{"id":"common:luogu/P2375--roj/1481","source":"luogu/P2375","target":"roj/1481","type":"common","directed":false,"reason":"同样依赖 fail 链结构的 AC 自动机题，可对照 fail 树上的计数思路"},{"id":"pre:luogu/P2440->roj/19999","source":"luogu/P2440","target":"roj/19999","type":"pre","directed":true,"reason":"先掌握二分答案与计数判定的写法，再处理把判定式变形为两个序列比较、排序后双指针计数求第 k 大的问题。"},{"id":"pre:luogu/P2347->luogu/P8742","source":"luogu/P2347","target":"luogu/P8742","type":"pre","directed":true,"reason":"B 的 main.cpp 直接沿用 A 教的「由已有可达状态加砝码重量推出新可达状态」这一可行性转移，只把可达量从总重量换成带偏移量的两侧差值，并新增负下标偏移与正差值统计"},{"id":"pre:noi_openjudge/ch0107-34->leetcodecn/palindrome-partitioning","source":"noi_openjudge/ch0107-34","target":"leetcodecn/palindrome-partitioning","type":"pre","directed":true,"reason":"B 的 main.cpp 直接复用 A 教的“两端相等且中间回文”区间 DP 递推，先预处理出 pal[i][j] 表 O(1) 判段，再在回溯枚举分割点时查表剪枝"},{"id":"pre:luogu/P2440->luogu/P1462","source":"luogu/P2440","target":"luogu/P1462","type":"pre","directed":true,"reason":"先掌握二分答案 + 单调 check 模板，再把 check 换成受限最短路（Dijkstra 判断可达性）。"},{"id":"pre:luogu/P2949->luogu/P4053","source":"luogu/P2949","target":"luogu/P4053","type":"pre","directed":true,"reason":"单位时间版本的反悔贪心，按截止时间排序后用小根堆删利润最小的任务"},{"id":"pre:luogu/P3375->luogu/P4391","source":"luogu/P3375","target":"luogu/P4391","type":"pre","directed":true,"reason":"KMP 模板题：先掌握 nxt 数组（最长 border）的本质，才能理解最小正周期公式"},{"id":"common:noi_openjudge/ch0111-10--usaco/1038","source":"noi_openjudge/ch0111-10","target":"usaco/1038","type":"common","directed":false,"reason":"同为最大化最小值+贪心扫描判定：跳房子统计需移走的石头数，1038 放置奶牛，判定过程可互相套用。"},{"id":"common:luogu/P2678--usaco/1038","source":"luogu/P2678","target":"usaco/1038","type":"common","directed":false,"reason":"同为最大化最小值+贪心扫描判定：P2678 二分跳跃距离统计移石数，1038 二分间距放置奶牛，判定均为单遍贪心。"},{"id":"common:luogu/P1824--usaco/1038","source":"luogu/P1824","target":"usaco/1038","type":"common","directed":false,"reason":"同型题：都是二分最小间距+从左到右尽量靠左放牛的贪心判定，唯一差别是牛舍为点还是区间。"},{"id":"pre:luogu/P1725->luogu/P5665","source":"luogu/P1725","target":"luogu/P5665","type":"pre","directed":true,"reason":"B 的满分解法把 A 教的单调队列取最优前驱原样搬到划分 DP 上：P1725 用队列在滑动窗口内取 dp 最大前驱，P5665 把「上一段段和不超过最后一段」化成 key(t)<=c[i] 的单调性判据后，同样靠队头取最优、队尾弹掉更差候选来 O(1) 得到最靠后的可行断点，只在 A 之上多出正数序列贪心与 key 的推导。"},{"id":"pre:luogu/P1722->luogu/P5658","source":"luogu/P1722","target":"luogu/P5658","type":"pre","directed":true,"reason":"B 的两条解法线都站在 A 教的「任意前缀不失衡即合法括号串」这一步上：brute.cpp 的 is_valid 直接照该判据扫描（balance<0 即非法），main.cpp 则把同一条前缀差非负判据实现成未匹配左括号栈（stack_top[fa]!=0 才配对），再叠加 A 未教的树上 total_count[u]=total_count[f_u]+end_count[u] 拆解、end_count[u]=end_count[f_p]+1 配对观察与 stack_next 共享栈。"},{"id":"pre:luogu/P2340->luogu/P4832","source":"luogu/P2340","target":"luogu/P4832","type":"pre","directed":true,"reason":"B 的差值背包直接复用 A 教的负数下标技巧：状态下标是可正可负的差值/智商和，靠偏移量换成非负下标，状态值存另一个量（A 存情商和、B 存总长度），且都按转移方向的正负决定倒序或正序枚举（B 的 main.cpp 里 diff>=0 倒序、diff<0 正序与 A 完全同型）；B 额外叠加的是 sin²+cos²=1 推出的配平建模与答案取一半。"},{"id":"pre:luogu/P2120->luogu/P6302","source":"luogu/P2120","target":"luogu/P6302","type":"pre","directed":true,"reason":"B 的主解法沿用 A 的“决策点对应一条线、查询点单调所以用单调队列维护凸包”这一步，只是把它从单序列复制到每个站点并按时间扫描插入/查询。"},{"id":"pre:luogu/P3195->luogu/P6302","source":"luogu/P3195","target":"luogu/P6302","type":"pre","directed":true,"reason":"B 把 A 教的“展开平方后转成关于查询量的一次函数、再用单调队列维护下凸壳”这一步，从单条全局凸壳搬到每个站点各自的凸壳上，并叠加按时间先插入后查询的新流程"},{"id":"pre:luogu/P5785->luogu/P6302","source":"luogu/P5785","target":"luogu/P6302","type":"pre","directed":true,"reason":"B 直接复用 A 教的“把 DP 转移中对固定 j 的线性部分看成直线、用单调队列维护下凸壳”这一斜率优化步骤，只是从单条队列扩成按站点分组、按时间扫描插入与查询，并把判劣/取最优的 bad()、better_front() 写成同一套叉积写法。"},{"id":"pre:luogu/P3195->luogu/P4056","source":"luogu/P3195","target":"luogu/P4056","type":"pre","directed":true,"reason":"B 先按 A 的做法把平方代价展开成斜率优化形式（`把平方项展开`，index.md:47），再把每个历史点 `f_j(x) = 2x_j * x + (dp(j) - x_j^2 - y_j^2)` 当作直线逐行维护凸壳查询，main.cpp 的 Hull::bad/query 就是 A 教的同一套凸壳弹头弹尾机制，只是把一维链上单凸壳扩成历史列、前缀、同列三层凸壳。"},{"id":"pre:luogu/P2120->luogu/P4056","source":"luogu/P2120","target":"luogu/P4056","type":"pre","directed":true,"reason":"P2120 教的“决策点写成直线 + 单调队列维护凸包做直线查询”被 P4056 直接复用：B 把每个历史点按固定 y_j 写成直线 f_j(x)=2x_j·x+(dp(j)-x_j^2-y_j^2)，对每一行用同一套凸包做单调查询，只是额外叠加了二维偏序的两层凸包拆分"},{"id":"pre:luogu/P5785->luogu/P4056","source":"luogu/P5785","target":"luogu/P4056","type":"pre","directed":true,"reason":"B 的按列扫描解法把 A 教的“决策点看成直线、用单调队列维护下凸壳做单调查询”直接搬过来：每个历史点按固定 y_j 写成直线 f_j(x)=2x_j*x+(dp(j)-x_j^2-y_j^2)，对每一行用同一套凸包查询，只是额外叠加了二维偏序的两层凸包拆分与同列内转移"},{"id":"pre:luogu/P2196->luogu/P1685","source":"luogu/P2196","target":"luogu/P1685","type":"pre","directed":true,"reason":"B 的关键前提「题目给的是 DAG，所以可以在拓扑序上做 DP」正是复用 A 教的「编号顺序就是拓扑序、按拓扑序用前驱更新后继」这一步，B 在这级台阶上把单个最大值 f 扩成 cnt/sum 两个量并补上高精度（main.cpp:165-166 沿边累加），不是重新发现拓扑序。"},{"id":"pre:luogu/P1807->luogu/P1685","source":"luogu/P1807","target":"luogu/P1685","type":"pre","directed":true,"reason":"B 正式解（main.cpp:165-166）沿边做 cnt[v]+=cnt[u]、sum[v]+=sum[u]+cnt[u]*w 的第 83-84 行，正是 A 教的『按拓扑序用前驱 state[u] 松弛后继』这一步；B 只是把 A 的单个最大值 distance 扩成 cnt/sum 两个量，并叠加路径条数计数、t0 坐船项与高精度实现。"},{"id":"pre:luogu/P4017->luogu/P1685","source":"luogu/P4017","target":"luogu/P1685","type":"pre","directed":true,"reason":"B 的花销转移 sum[v]+=sum[u]+cnt[u]*w 完全建立在 A 教过的拓扑序路径条数 DP 之上，cnt[v]+=cnt[u] 就是 ways[v]+=ways[u]，cnt[t] 同时充当答案里的坐船次数，A 的计数步骤是 B 主解里实际用到的一步"},{"id":"pre:luogu/P1359->luogu/P3074","source":"luogu/P1359","target":"luogu/P3074","type":"pre","directed":true,"reason":"B 的正式解法把 A 教的那一步「在 DAG 的有效点序上对每条出边做 dp[后继]=opt(dp[后继],dp[当前]+权值)」整步搬过来：A 里这个序就是天然编号顺序、取 min 求最少租金，B 用 Kahn 求出拓扑序、取 max 求最长依赖链，新增的只是求序手段。"},{"id":"pre:luogu/P8687->luogu/P1294","source":"luogu/P8687","target":"luogu/P1294","type":"pre","directed":true,"reason":"B 沿用 A 教的「把已处理的集合压成二进制 mask、用按位或把新元素并入集合」这一步，把覆盖口味换成已访问点并加一维当前终点，由集合覆盖最少包数变成最长简单路径。"},{"id":"pre:luogu/U661993->luogu/P5662","source":"luogu/U661993","target":"luogu/P5662","type":"pre","directed":true,"reason":"B 每天的收益计算（main.cpp:30-32 `for (cap = cost; cap <= money; ++cap)`）就是把 A 教的『完全背包正序枚举容量、让 dp[j-v] 取本轮新值从而同一物品可无限次选取』复用一次，重量取今日价、价值取两日差价；B 额外叠加的是 A 未讲的相邻两日差价观察与 `money += dp[money]` 滚动更新。"},{"id":"pre:luogu/U613136->luogu/P5662","source":"luogu/U613136","target":"luogu/P5662","type":"pre","directed":true,"reason":"先理解相邻两天价格差为正才值得买卖的股票模型，再把单物品交易扩展为每天一次完全背包。"},{"id":"pre:luogu/U661986->luogu/P2725","source":"luogu/U661986","target":"luogu/P2725","type":"pre","directed":true,"reason":"A 教的一维背包状态设计（dp[容量] 只留最优值、dp[x-v]+代价 转移、按枚举方向控制物品可用次数）正是 B 完全背包解法的第一步：B 把它换成 min 与 +1 求每个面值的最少邮票数，再叠加扫描最长连续前缀这一新步骤。"},{"id":"pre:luogu/U661988->luogu/P2725","source":"luogu/U661988","target":"luogu/P2725","type":"pre","directed":true,"reason":"A 教的『容量正序枚举让同一物品无限复用』被 B 的原样搬进 main.cpp 的 for (x = v; x <= limit; x++) dp[x] = min(dp[x], dp[x-v]+1)，只把 max/+价值 换成 min/+1 张数；B 在此完全背包模板之上再叠加 limit = k*max_a 与扫描最长连续可达前缀这一额外流程。"},{"id":"pre:luogu/P1734->luogu/P2340","source":"luogu/P1734","target":"luogu/P2340","type":"pre","directed":true,"reason":"B 主解在智商和状态上复用 A 教的「01 背包按转移方向倒序枚举、保证每件物品至多选一次」这一步（main.cpp 中 iq>=0 分支 for(s=hi;s>=lo;s--) 与 A 的 for(j=s;j>=i;--j) 同型，iq<0 分支才改为正序），并在其上叠加 A 未教的偏移量处理负智商和与双非负约束的收尾。"},{"id":"pre:luogu/P8742->luogu/P2340","source":"luogu/P8742","target":"luogu/P2340","type":"pre","directed":true,"reason":"B 直接复用 A 教的「用偏移量把负数状态映射成非负下标」这一步：A 把差值 [-sum,sum] 平移 sum 得 offset=sum、dp[offset] 为初值，B 把智商和平移 OFFSET（main.cpp 里 OFFSET=400000、dp[OFFSET]=0），只是把可行性 dp 换成最大值 dp，并额外处理 iq 为负时正序枚举。"},{"id":"pre:luogu/U663298->luogu/P1509","source":"luogu/U663298","target":"luogu/P1509","type":"pre","directed":true,"reason":"B 的正式解 main.cpp 用 `for (int b = r; b >= rp[i]; b--)` 逐维复用 A 教的倒序枚举容量以保证每件物品只选一次，只是从一维 dp[c] 扩到钱与人品两维 dp[a][b]。"},{"id":"pre:luogu/U661994->luogu/P1509","source":"luogu/U661994","target":"luogu/P1509","type":"pre","directed":true,"reason":"A 教的『二维费用 01 背包 dp[j][k] 两维均倒序枚举，用 dp[j-v][k-w] 更新』这一整步被 B 的 main.cpp 原样搬进钱与人品两维（for(a=m..rmb[i]) for(b=r..rp[i]) 用 dp[a-rmb][b-rp] 更新 dp[a][b]），B 只是把 max 价值换成 (cnt,time) 优先级 pair 叠加，属台阶式叠加，难度 普及- → 普及/提高- 方向正确。"},{"id":"pre:luogu/P1048->luogu/P1509","source":"luogu/P1048","target":"luogu/P1509","type":"pre","directed":true,"reason":"B 的正式解（main.cpp 双层倒序循环）直接把 A 教的“倒序枚举容量以保证每个物品只选一次”从一维容量扩成钱与人品两维，B 在此基础上只多出“先比 cnt 再比 time”的 pair 比较。"},{"id":"pre:luogu/U661993->luogu/P1853","source":"luogu/U661993","target":"luogu/P1853","type":"pre","directed":true,"reason":"A 讲透并给出代码的「完全背包正序枚举容量即可重复取用同一物品」这一步，被 B 的 main.cpp（第 37 行 for (int j = bond.cost_unit; j <= max_unit; j++)）直接用作一年内求最大利息的完全背包主体；B 在此之上额外叠加了 A 未涉及的按 1000 元压缩预算与按年滚动本金的流程，难度 普及+/提高- → 普及+/提高，台阶方向正确。"},{"id":"pre:leetcodecn/perfect-squares->luogu/P1853","source":"leetcodecn/perfect-squares","target":"luogu/P1853","type":"pre","directed":true,"reason":"B 的某一年解法直接复用 A 的完全背包 dp（容量维 + 可重复物品），只是从最少物品数改成最大收益并在年份上滚动"},{"id":"pre:luogu/U661986->luogu/P1832","source":"luogu/U661986","target":"luogu/P1832","type":"pre","directed":true,"reason":"B 直接沿用 A 的“外层枚举物品、内层枚举容量”一维 dp 状态设计，只把 A 的 max 转移改成正序累加，并借用 A 的“枚举顺序决定物品能否重复使用”这一结论说明正序让素数可重复选，属于在同一背包模板上叠加素数筛与组合计数流程。"},{"id":"pre:luogu/U661988->luogu/P1832","source":"luogu/U661988","target":"luogu/P1832","type":"pre","directed":true,"reason":"B 把 A 教的「容量正序枚举即允许物品无限次使用」当成完全背包内层骨架，直接套在筛出的素数集合上（main.cpp 内层 for(j=p;j<=n;++j) 正序），只把 A 的 max 转移换成 dp[j]+=dp[j-p] 的计数转移，并叠加埃氏筛与组合计数（外层物品、内层容量的顺序论证）这一层。"},{"id":"pre:luogu/P1164->luogu/P2347","source":"luogu/P1164","target":"luogu/P2347","type":"pre","directed":true,"reason":"B（main-rainboy.cpp 分支）把 A 教的「每件最多选一次、容量倒序枚举」这一 0/1 背包转移直接复用：先把每枚砝码逐枚展开成 0/1 物品，再对 j 从 sum 到 w 倒序做 dp[j]|=dp[j-w]，只把 A 的方案数累加换成布尔可达性并最后统计可达重量数。"},{"id":"pre:luogu/P2871->luogu/P2347","source":"luogu/P2871","target":"luogu/P2347","type":"pre","directed":true,"reason":"B 的 main-rainboy.cpp 分支把 A 教的「每件物品只能选一次，所以容量必须倒序枚举」原样套到逐枚展开的砝码上（for j = sum..w[i]），只把 max 值转移改成布尔可达 dp[j] |= dp[j-w[i]]，并在此基础上新增逐枚展开与可达重量计数。"},{"id":"pre:luogu/U663295->luogu/P2347","source":"luogu/U663295","target":"luogu/P2347","type":"pre","directed":true,"reason":"B 的 main-rainboy.cpp 把 A 教的「布尔可达性 dp + 容量倒序枚举（每件最多选一次）」原样用作其可达性背包主体：A 是 dp[c]=dp[c]||dp[c-v] 单问 V 是否可达，B 先叠加 A 未教的「每种多枚砝码逐枚展开成 0/1 物品」这一层，再用同一倒序转移 dp[j]|=dp[j-w] 把可达性从单点扩到所有重量并计数（bitset 版 dp|=dp<<w 是同一转移的位并行写法），属台阶式叠加而非同难度孪生。"},{"id":"pre:luogu/P2639->luogu/P1734","source":"luogu/P2639","target":"luogu/P1734","type":"pre","directed":true,"reason":"B 主解把 A 教的「一维 dp 内层容量倒序枚举、每件物品最多选一次」原样搬入自己的 main.cpp（for(j=s;j>=i;--j) dp[j]=max(dp[j],dp[j-i]+val[i])），只是先补上真约数和预处理这一步；A 的 dp[j] 状态含义从「容量 j 内最大重量」换成「总和 j 内最大约数和」，模板完全一致。"},{"id":"pre:luogu/P1060->luogu/P1734","source":"luogu/P1060","target":"luogu/P1734","type":"pre","directed":true,"reason":"B 的主解把 A 教的「内层容量倒序枚举、每件至多选一次」一维 0/1 背包原样搬用（main.cpp 的 for (j=s; j>=i; --j) dp[j]=max(dp[j],dp[j-i]+val[i])），只多加真约数和预处理这层题意翻译。"},{"id":"pre:luogu/P2871->luogu/P1734","source":"luogu/P2871","target":"luogu/P1734","type":"pre","directed":true,"reason":"B 的主解把 A 教的「内层容量倒序枚举、每件物品至多选一次」原样搬进 main.cpp（for(j=s;j>=i;--j) dp[j]=max(dp[j],dp[j-i]+val[i]) 与 A 的 dp[j]=max(dp[j],dp[j-W[i]]+D[i]) 同形），只把重量价值换成数字与真约数和、并在前面补上真约数和预处理这一额外台阶。"},{"id":"pre:luogu/B4141->luogu/P1510","source":"luogu/B4141","target":"luogu/P1510","type":"pre","directed":true,"reason":"B 把 A 教的「内层容量倒序枚举实现每个物品最多选一次」直接搬进 0/1 背包主体，只是容量从和 n 换成体力 c、价值从项数换成体积，再叠加达到 v 的截断与最小花费扫描"},{"id":"pre:luogu/P2871->luogu/P1510","source":"luogu/P2871","target":"luogu/P1510","type":"pre","directed":true,"reason":"B 的 main.cpp 沿用 A 教的「容量倒序枚举保证每件物品最多选一次」这一步，把容量换成体力做同形转移后，再补上截断与扫描最小花费的新流程"},{"id":"pre:luogu/P1048->luogu/P1510","source":"luogu/P1048","target":"luogu/P1510","type":"pre","directed":true,"reason":"B 直接复用 A 教的「内层容量倒序枚举保证每件物品最多选一次」这一步：main.cpp 的倒序循环 dp[j]=max(dp[j],dp[j-m]+k) 与 A 的 dp[t]=max(dp[t],dp[t-time]+value) 同形，只是把容量换成体力、把答案换成扫描第一个 dp[j]>=v 的最小花费（外加超过 v 的截断）。"},{"id":"pre:luogu/P1060->luogu/P1910","source":"luogu/P1060","target":"luogu/P1910","type":"pre","directed":true,"reason":"B 直接沿用 A 教的 0/1 背包“容量倒序枚举保证每件物品只用一次”这一步（B 的 main.cpp 两层循环都写成倒序，注释即“两维容量都倒序，保证每个间谍最多只选一次”），只是把单个预算容量扩成探查风险与工资两维费用，其余状态设计与转移形状不变。"},{"id":"pre:luogu/P2871->luogu/P1910","source":"luogu/P2871","target":"luogu/P1910","type":"pre","directed":true,"reason":"B 直接复用 A 教的 0/1 背包“每个对象最多选一次 ⇒ 容量必须倒序枚举”这一步，只是把单个重量容量扩成探查风险与工资两维费用（main.cpp:26-30 两层倒序循环），模板级台阶成立。"},{"id":"pre:luogu/P1048->luogu/P1910","source":"luogu/P1048","target":"luogu/P1910","type":"pre","directed":true,"reason":"B 沿用 A 教的 0/1 背包“每个对象最多选一次”的容量倒序枚举这一步（B 代码注释「两维容量都倒序，保证每个间谍最多只选一次」），只是把单个时间容量扩成探查风险与工资两维费用，成为二维费用 0/1 背包。"},{"id":"pre:luogu/P1060->luogu/P1855","source":"luogu/P1060","target":"luogu/P1855","type":"pre","directed":true,"reason":"B 的 main.cpp 把 A 教的一维 dp 倒序容量循环原样扩成 dp[j][k] 的两重倒序，用同一「倒序保证每件物品只参与一次转移」的机制保证愿望不重复选，新增的只是金钱/时间第二维费用和价值恒为 1 的计数目标。"},{"id":"pre:luogu/P2871->luogu/P1855","source":"luogu/P2871","target":"luogu/P1855","type":"pre","directed":true,"reason":"A 教的「容量倒序枚举保证物品只选一次」正是 B 把 dp[j] 扩成 dp[j][k] 时对两维容量都倒序枚举的那一步，转移仍是 dp=max(dp, dp[减去消耗]+价值)，只多加一维费用、把价值固定为 1。"},{"id":"pre:luogu/P1048->luogu/P1855","source":"luogu/P1048","target":"luogu/P1855","type":"pre","directed":true,"reason":"B 把 A 教的“容量倒序枚举保证物品只选一次”的 01 背包模板直接搬到两维费用上：A 的一维 dp[t] 内层倒序扩成 dp[j][k] 的两重倒序，转移形式仍为 dp=max(dp, dp[减去消耗]+价值)，只是新增金钱维度且价值恒为 1。"},{"id":"pre:luogu/P2871->luogu/P2663","source":"luogu/P2871","target":"luogu/P2663","type":"pre","directed":true,"reason":"B 的 main.cpp 把 A 教的「容量倒序枚举保证每件物品只选一次」逐字搬进 dp[j][s] 二维转移（j 与 s 双重倒序、由 dp[j-1][s-a[i]] 推 dp[j][s]），额外台阶只是多了「恰好选 n/2 人」的人数维与最后从 sum/2 倒扫可达值。"},{"id":"pre:luogu/P1164->luogu/P2663","source":"luogu/P1164","target":"luogu/P2663","type":"pre","directed":true,"reason":"B 的 main.cpp 主解把 A 教的「恰好凑和：只给 dp[0]=1 / dp[0][0]=true，且因为每件只能选一次而内层容量倒序枚举」整步搬到二维可达性上：人数 j 与总分 s 两维都倒序枚举（main.cpp:33-34，注释亦写「人数和分数都要倒序」，只在末尾 scan 时用），额外台阶仅是加一维「恰好选 n/2 人」并从 sum/2 倒扫最大可达值。"},{"id":"pre:luogu/U663295->luogu/P2663","source":"luogu/U663295","target":"luogu/P2663","type":"pre","directed":true,"reason":"A 教的正是「布尔可达性 + 容量倒序枚举保证每件只用一次」这一步，B 的 main.cpp（j 与 s 双重倒序、dp[j-1][s-a[i]] 推 dp[j][s]）把同一倒序枚举原样搬进二维可达性转移，只额外叠加「恰好 n/2 人」的人数维与从 sum/2 倒扫可达值；难度由 普及- 升到 普及/提高-，台阶性成立。"},{"id":"pre:luogu/P1060->luogu/P1507","source":"luogu/P1060","target":"luogu/P1507","type":"pre","directed":true,"reason":"B 的正式解就是把 A 教的 0/1 背包容量倒序枚举这一具体步骤照搬过来：main.cpp:22-23 外层枚举物品、内层对体积与质量两重倒序循环做 dp[j][l]=max(dp[j][l],dp[j-h][l-t]+k)，只是把单个容量换成体积、质量两维并改求最大卡路里，难度增量为 +2 的普及/提高-，方向正确。"},{"id":"pre:luogu/P2639->luogu/P1507","source":"luogu/P2639","target":"luogu/P1507","type":"pre","directed":true,"reason":"A 讲透的「每件物品只能选一次 ⇒ 容量必须倒序枚举」正是 B 的二维费用 0/1 背包实际使用的步骤：B 把 A 的一维容量 dp[j] 扩成体积、质量两维 dp[j][k]，并在两维上继续倒序枚举（main.cpp:22-23 的 for(j=H;j>=h;--j) / for(l=T;l>=t;--l)），从而保证每件食品最多选一次；A 是纯 0/1 背包基础题，B 叠加的是体积+质量双重限制这一台阶，难度 普及- → 普及/提高-，方向正确。"},{"id":"pre:luogu/P2871->luogu/P1507","source":"luogu/P2871","target":"luogu/P1507","type":"pre","directed":true,"reason":"B 的二维费用 01 背包复用了 A 的“倒序枚举容量保证每件物品只选一次”这一具体步骤，只是在一维容量基础上叠加上体积、质量两维循环"},{"id":"pre:leetcodecn/pascals-triangle->luogu/P1057","source":"leetcodecn/pascals-triangle","target":"luogu/P1057","type":"pre","directed":true,"reason":"A 教的 `ans[i][j]=ans[i-1][j-1]+ans[i-1][j]` 这种「上一层相邻两项相加」的填表递推，被 B 直接用作 `dp[i][j]=dp[i-1][left(j)]+dp[i-1][right(j)]` 的状态转移，只在相邻位置上加了环形边界处理。"},{"id":"pre:leetcodecn/minimum-path-sum->luogu/P1796","source":"leetcodecn/minimum-path-sum","target":"luogu/P1796","type":"pre","directed":true,"reason":"B 的 main.cpp:36（min(cur[planet], prev[from]+cost)）与 dp[i][j]=min(dp[i-1][from]+cost) 复用了 A 的「min 前驱 + 当前权值 + 滚动数组」转移，只因前驱由固定的上/左换成输入给出的上一层星球，才多出建分层 DAG 与按层读入的流程。"},{"id":"pre:luogu/P1359->luogu/P1796","source":"luogu/P1359","target":"luogu/P1796","type":"pre","directed":true,"reason":"B 直接复用 A 教的『边单向即 DAG，于是按序用 dp[前驱]+边权 更新后继』这步（main.cpp:36 的 min(cur[planet], prev[from]+cost)），只是把 A 的编号序前驱换成输入给出的上一层前驱，并额外处理负权、变长输入与滚动数组。"},{"id":"pre:acwing/1015->luogu/P1796","source":"acwing/1015","target":"luogu/P1796","type":"pre","directed":true,"reason":"B 的状态转移与 A 同一步：枚举前驱取极值再加当前权值，只是把 A 的“上方/左方两前驱取 max 加花生数”换成输入给出的上一层前驱取 min 加航线费用（main.cpp 的 min(cur[planet], prev[from]+cost)），额外难度在建分层 DAG 与滚动数组"},{"id":"common:luogu/P1824--noi_openjudge/ch0111-10","source":"luogu/P1824","target":"noi_openjudge/ch0111-10","type":"common","directed":false,"reason":"同为二分最小距离+从左到右贪心扫描判定：P1824 放置奶牛，ch0111-10 统计需移走的石头数，判定同型。"},{"id":"common:luogu/P1824--luogu/P2678","source":"luogu/P1824","target":"luogu/P2678","type":"common","directed":false,"reason":"同为最大化最小值+贪心扫描判定：P1824 按最小间距尽早放牛，P2678 按最短跳跃距离统计最少移石数，判定过程可互相套用。"},{"id":"pre:luogu/P4014->luogu/P4013","source":"luogu/P4014","target":"luogu/P4013","type":"pre","directed":true,"reason":"先掌握基本的费用流（MCMF）建模，再处理数字梯形中三级容量分层控制的拆点技巧。"},{"id":"pre:luogu/P4014->luogu/P2053","source":"luogu/P4014","target":"luogu/P2053","type":"pre","directed":true,"reason":"先掌握基本费用流建模，再理解修车问题中将调度顺序展开为维度拆点的思维。"},{"id":"pre:luogu/P4014->luogu/P4016","source":"luogu/P4014","target":"luogu/P4016","type":"pre","directed":true,"reason":"先掌握基本费用流建模，再理解环形均分纸牌的费用流建模或中位数解法。"},{"id":"pre:luogu/P2774->luogu/P3749","source":"luogu/P2774","target":"luogu/P3749","type":"pre","directed":true,"reason":"先理解最小割中 S-正权-T-负权的建图思想，再处理最大权闭合子图结合区间 DP 依赖的多层建模。"},{"id":"pre:luogu/P2947->luogu/P1198","source":"luogu/P2947","target":"luogu/P1198","type":"pre","directed":true,"reason":"先掌握单调栈求下一个更大元素的基本模板，再理解在线末尾追加 + 区间最大值的多解法对比。"},{"id":"pre:HDU/3549->luogu/P4014","source":"HDU/3549","target":"luogu/P4014","type":"pre","directed":true,"reason":"先掌握最大流 Dinic 模板，再学习在残余网络上加入费用维度的 MCMF 算法。"},{"id":"pre:HDU/3549->luogu/P2763","source":"HDU/3549","target":"luogu/P2763","type":"pre","directed":true,"reason":"先掌握最大流 Dinic 模板，再学习将二分图多重匹配转化为最大流建图。"},{"id":"pre:POJ/1523->HDU/3394","source":"POJ/1523","target":"HDU/3394","type":"pre","directed":true,"reason":"需要先掌握 v-BCC 分解和边入栈的 Tarjan 写法，才能对每个 BCC 统计 V_cnt 和 E_cnt。"},{"id":"common:HDU/3394--luogu/P3225","source":"HDU/3394","target":"luogu/P3225","type":"common","directed":false,"reason":"同为对 v-BCC 内节点进行计数和分类的题目，对比理解不同应用场景。"},{"id":"pre:HDU/1269->HDU/1827","source":"HDU/1269","target":"HDU/1827","type":"pre","directed":true,"reason":"需要先掌握 Tarjan 求 SCC，再在源 SCC 中取最小通知代价。"},{"id":"common:HDU/1878--OpenJ_Bailian/1300","source":"HDU/1878","target":"OpenJ_Bailian/1300","type":"common","directed":false,"reason":"同为无向图欧拉回路/路径的度数判定题，适合对比基础条件和含指定起终点的条件。"},{"id":"pre:POJ/1144->OpenJ_Bailian/3091","source":"POJ/1144","target":"OpenJ_Bailian/3091","type":"pre","directed":true,"reason":"需要先掌握 Tarjan 算法和 DFS 遍历框架，再理解桥的判断条件 low[v] > dfn[u] 与割点的区别。"},{"id":"pre:POJ/1523->luogu/P3225","source":"POJ/1523","target":"luogu/P3225","type":"pre","directed":true,"reason":"先通过 SPF 理解 v-BCC 分解和割点与点双的关系，再处理矿场搭建中的分类计数。"},{"id":"pre:POJ/1144->luogu/P5058","source":"POJ/1144","target":"luogu/P5058","type":"pre","directed":true,"reason":"需要先掌握 Tarjan 求割点的基本框架，再理解如何在 DFS 中判断割点是否分隔特定两点。"},{"id":"pre:HDU/1269->luogu/P1407","source":"HDU/1269","target":"luogu/P1407","type":"pre","directed":true,"reason":"需要先掌握 Tarjan 求 SCC，再把交替环判定转化为同一强连通分量。"},{"id":"common:HDU/1269--luogu/P2863","source":"HDU/1269","target":"luogu/P2863","type":"common","directed":false,"reason":"同为 Tarjan 求 SCC 的基础模板题，都是直接验证强连通分量性质。"},{"id":"common:POJ/1144--POJ/1523","source":"POJ/1144","target":"POJ/1523","type":"common","directed":false,"reason":"同为割点判定模板题，POJ 1523 额外引入 v-BCC 分解视角，适合对比学习。"},{"id":"pre:HDU/3549->luogu/P2764","source":"HDU/3549","target":"luogu/P2764","type":"pre","directed":true,"reason":"先掌握最大流 Dinic 模板，再理解 DAG 最小路径覆盖的拆点法建模。"},{"id":"common:luogu/P2764--luogu/P2765","source":"luogu/P2764","target":"luogu/P2765","type":"common","directed":false,"reason":"同为 DAG 最小路径覆盖模型，P2764 为静态建图，P2765 为动态加点增广。"},{"id":"common:POJ/3122--noi_openjudge/ch0111-04","source":"POJ/3122","target":"noi_openjudge/ch0111-04","type":"common","directed":false,"reason":"同为二分长度/体积+floor 切分计数判定：网线主管数可切段数，3122 数可切块数，判定同型。"},{"id":"common:POJ/3122--luogu/P2440","source":"POJ/3122","target":"luogu/P2440","type":"common","directed":false,"reason":"同为二分长度/体积+切分计数判定：P2440 数木材段数，3122 数派的块数，check 都是 floor 聚合计数。"},{"id":"common:POJ/3122--luogu/P1873","source":"POJ/3122","target":"luogu/P1873","type":"common","directed":false,"reason":"同为基础二分答案模板题，POJ 3122 为实数二分，P1873 为整数二分，对比例题。"},{"id":"common:roj/1373--roj/1431","source":"roj/1373","target":"roj/1431","type":"common","directed":false,"reason":"同一模型（递减边际收益 + 预算分配）的按分钟版本，衰减量恒 ≥ 1，可对照两种「取前 k 大」的实现"},{"id":"common:luogu/P1106--roj/1321","source":"luogu/P1106","target":"roj/1321","type":"common","directed":false,"reason":"同一题模型：同样是删 k 位使剩余数字最小，单调栈 + 去前导零，可直接互相对照。"},{"id":"common:roj/1276--roj/1298","source":"roj/1276","target":"roj/1298","type":"common","directed":false,"reason":"同为一本通编辑距离模板题，操作集完全相同，可用来对照同一份状态定义和转移。"},{"id":"common:leetcodecn/edit-distance--roj/1298","source":"leetcodecn/edit-distance","target":"roj/1298","type":"common","directed":false,"reason":"同模型（Levenshtein 距离）的经典题，题面只差输入输出格式，适合一起对比训练。"},{"id":"common:noi_openjudge/ch0107-18--roj/1140","source":"noi_openjudge/ch0107-18","target":"roj/1140","type":"common","directed":false,"reason":"同一本通题目在 NOI OpenJudge 的对应版，输入输出与判定完全一致。"},{"id":"pre:luogu/U661988->luogu/P2834","source":"luogu/U661988","target":"luogu/P2834","type":"pre","directed":true,"reason":"A 讲透的完全背包正序枚举理由（dp[c-v] 已含当前物品贡献）正是 B 内层金额 j 正序的那一步，B 只是把 max 求和改成计数取模，并新增先枚举纸币以合并支付顺序"},{"id":"pre:luogu/U661986->luogu/P2834","source":"luogu/U661986","target":"luogu/P2834","type":"pre","directed":true,"reason":"B 的 main.cpp:25-27 把 A:62 讲透的『容量枚举方向决定物品能否重复使用』这一步反向为正序枚举（j 从 a[i] 到 w），用同一维 dp 数组实现纸币无限张；台阶是 A 未教的组合计数语义与先纸币后金额的循环顺序辨析。"},{"id":"pre:luogu/P1255->luogu/P2840","source":"luogu/P1255","target":"luogu/P2840","type":"pre","directed":true,"reason":"B 的 dp[j] 直接沿用 A 教的「方案数记在 dp[状态]、按最后一步分类求和」这一状态设计与转移：A 的最后一步只有走 1 阶或 2 阶两种，B 把可选前驱扩成所有面额并允许无限次重复，得 dp[j]=Σdp[j-a_i]，再叠加先金额后纸币的循环顺序与取模完成排列计数。"},{"id":"pre:luogu/U661986->luogu/P2840","source":"luogu/U661986","target":"luogu/P2840","type":"pre","directed":true,"reason":"B 的 main.cpp:25-31 沿用 A 教的一维容量 dp 骨架与「容量枚举方向决定物品能否重复使用」这一判定，只是把 A 的倒序枚举反向改成正序（A:62 说正序会让同一物品被选两次，B:59 说先枚举纸币正序会把顺序合并成组合），从而换成完全背包排列计数 dp[j]=(dp[j]+dp[j-a_i])%MOD，并额外叠加 A 未讲的取模与排列/组合循环顺序辨析；难度从 入门 升到 普及-，台阶性成立。"},{"id":"pre:luogu/U661988->luogu/P2840","source":"luogu/U661988","target":"luogu/P2840","type":"pre","directed":true,"reason":"B 直接复用 A 教的『容量正序枚举 ⇒ dp 前驱已含当前物品 ⇒ 可无限次使用』这一判定与一维 dp[容量] 状态设计：main.cpp 内层 dp[j]=(dp[j]+dp[j-a[i]])%MOD 且 j 正序，把 A 的物品换成一维被无限张使用的纸币，只把 max 价值换成方案数累加；B 在此完全背包骨架之上再叠加取模与『先金额后纸币』的排列计数顺序辨析，故 A 是它的入门模板。"},{"id":"pre:luogu/P1352->luogu/U661996","source":"luogu/P1352","target":"luogu/U661996","type":"pre","directed":true,"reason":"B 的 dp[u][j] 定义直接沿用 A 教的「父节点是否被选决定子节点能否被选」这一状态刻画（A 用 dp[u][0/1] 表达「父亲来了，孩子不能来」，B 硬编码「u 必须被选，不选则整棵子树都不能选」）并沿用其 DFS 自底向上合并子树的骨架；B 新增的只是容量维与给孩子分配容量的分组背包合并。"},{"id":"pre:luogu/P1910->luogu/U661996","source":"luogu/P1910","target":"luogu/U661996","type":"pre","directed":true,"reason":"B 合并子节点时复用 A 教的『每个对象只选一次，故容量维倒序枚举』：B 的 main.cpp for(j=V;j>=v[u];j--) 与 A 的倒序容量循环同形，B 原文同样以『旧状态隔离』解释该步，只是把 A 的二维费用压回单容量，再叠加依赖树 DFS、容量分配合并与虚拟根 0。"},{"id":"pre:luogu/P1734->luogu/U661996","source":"luogu/P1734","target":"luogu/U661996","type":"pre","directed":true,"reason":"B 在合并子节点时沿用 A 教的『容量维倒序枚举、让 dp 从旧状态转移而不重复选取』这一步（main.cpp 的 for (j=V; j>=v[u]; --j) 配 dp[u][j-k]+dp[c][k]），只是把它从一维 0/1 背包搬到树形依赖背包的容量分配合并上，额外叠加 A 未教的依赖树与虚拟根 0，Δrank=1、方向正确。"},{"id":"pre:luogu/P1855->luogu/U662012","source":"luogu/P1855","target":"luogu/U662012","type":"pre","directed":true,"reason":"B 的 main.cpp（第 38-40 行 for(j=V;j>=0;j--)）直接复用 A 教的『容量倒序枚举保证每个物品只装入一次』这一步，只是把 A 的固定费用 m 换成枚举分配量 x 的插值 val[x] 并去掉金钱维，额外叠加的是 A 完全未涉及的分段线性插值预计算 val[c]；难度从 普及/提高- 升到 普及+/提高-，台阶性成立。"},{"id":"pre:luogu/P1510->luogu/U662012","source":"luogu/P1510","target":"luogu/U662012","type":"pre","directed":true,"reason":"A 教的「内层容量倒序枚举保证每件物品最多用一次」被 B 明确沿用：B 的 main.cpp 用 for(j=V;j>=0;j--) 倒序枚举容量，只是把 A 的单个重量 m_i 扩成对分配量 x 的枚举，并在外层新增分段线性插值预计算 val[c]；即 B 的 01 背包骨架复用 A 的倒序步骤，叠加插值这一新流程，难度从 普及/提高- 升到 普及+/提高-，符合台阶性。"},{"id":"pre:luogu/P1734->luogu/U662012","source":"luogu/P1734","target":"luogu/U662012","type":"pre","directed":true,"reason":"B 的一维滚动背包完全照搬 A 教的『容量倒序枚举 + dp[j]=max(dp[j],dp[j-x]+val)』保证每件物品只用一次，只把固定重量换成枚举分配量 x、价值换成插值 val[x] 这一额外台阶"},{"id":"pre:luogu/U661988->luogu/U663733","source":"luogu/U661988","target":"luogu/U663733","type":"pre","directed":true,"reason":"B 的 main.cpp（for c=1..V 内层 for v in a 的 dp[c]=(dp[c]+dp[c-v])%MOD）直接复用 A 教的关键观察『容量正序推进时 dp[c-v] 可能已包含当前物品的贡献』，从而让同一体积可被无限次追加；B 只是把外层换成容量、把取 max 换成 +1 计数取模，额外叠加的台阶是『枚举顺序决定排列/组合』这一 A 未涉及的区分（以及恰好装满的 dp[0]=1 初值），难度从 入门 升到 普及-，方向正确。"},{"id":"pre:luogu/U661986->luogu/U663733","source":"luogu/U661986","target":"luogu/U663733","type":"pre","directed":true,"reason":"B 的 main.cpp:25-30 逐行复用 A 教的『一维 dp[c] 由 dp[c-v] 转移＝再放入一件物品 v』这一步，只把 A 的 max 取最优值与容量倒序换成计数累加与容量正序，叠加 A 未涉及的排列/组合枚举顺序判别；A 入门、B 普及-，落差 1 级，台阶性成立。"},{"id":"pre:luogu/P1734->luogu/U663791","source":"luogu/P1734","target":"luogu/U663791","type":"pre","directed":true,"reason":"B 的主解把 A 讲透的「一维 dp 内层容量倒序枚举、每件物品至多选一次」原样当作自己的 0/1 背包主体（B 的 main.cpp 为 for (c=V;c>=pack_v;--c) dp[c]=max(dp[c],dp[c-pack_v]+pack_w)，与 A 的 for (j=s;j>=i;--j) 同形），只是先对每种物品做二进制拆分，把 s_i 件压成 O(log s_i) 个包再套这一步；A 是纯 0/1 背包基础题，B 在其上叠加二进制分组台阶，难度 普及/提高- → 普及+/提高 方向正确。"},{"id":"pre:luogu/P2347->luogu/U663791","source":"luogu/P2347","target":"luogu/U663791","type":"pre","directed":true,"reason":"B 的解法骨架正是 A 教的那一步「有件数限制的物品展开成 0/1 物品后跑 01 背包」：A 逐枚展开砝码并把它当 0/1 物品转移，B 用 1,2,4…二进制打包把展开数量从 O(s_i) 压到 O(log s_i)，每个包仍当作独立 01 物品跑同一套一维 01 背包（main.cpp 的 for (c=V;c>=pack_v;--c) dp[c]=max(dp[c],dp[c-pack_v]+pack_w)），B 甚至把 A 的逐枚展开写作「原来的做法」作对照，额外叠加的只是二进制分组流程与价值最大化；难度 普及/提高- → 普及+/提高 台阶方向正确。"},{"id":"pre:acwing/1015->luogu/P2196","source":"acwing/1015","target":"luogu/P2196","type":"pre","directed":true,"reason":"B 直接沿用 A 的关键观察——移动方向单向保证「前驱已算好」，把网格里「上/左两个已算好的前驱取大再加 a[i][j]」推广成按编号拓扑序对多个连通前驱 f[j] 取大再加 a[i]，并加 pre[] 路径还原与 max_i f[i]。"},{"id":"pre:luogu/P1359->luogu/P2196","source":"luogu/P1359","target":"luogu/P2196","type":"pre","directed":true,"reason":"B 直接复用 A 教的『编号即拓扑序、按编号从小到大枚举点对做 DAG 线性转移』这一步（连 O(n^2) 的 i<j 双层循环骨架都相同），只在最短路/自选起点的 min(dp[i]+cost) 上叠加点权最长路、f[i]=a[i] 边界与 pre[] 路径还原，属于加信息、换目标的台阶式叠加。"},{"id":"pre:luogu/P1216->luogu/P2196","source":"luogu/P1216","target":"luogu/P2196","type":"pre","directed":true,"reason":"数字三角形是最基础的线性 DP：从相邻前驱状态取最优再累加，挖地雷的 DAG 最长路转移是同一模式的推广"},{"id":"pre:luogu/U663703->leetcodecn/perfect-squares","source":"luogu/U663703","target":"leetcodecn/perfect-squares","type":"pre","directed":true,"reason":"B 的 dp[i]=min(dp[i-j*j]+1) 就是在 A 教的正序完全背包转移 dp[c]=dp[c]||dp[c-v] 上把物品换成所有平方数、把可达性判定换成取最少件数，无限件这一关键设定被完整复用。"},{"id":"pre:luogu/P1616->leetcodecn/perfect-squares","source":"luogu/P1616","target":"leetcodecn/perfect-squares","type":"pre","directed":true,"reason":"B 的 main.cpp（dp[i]=min(dp[i],dp[i-j*j]+1)）直接复用 A 教的一维完全背包状态设计与「容量正序、让本轮更新后的状态继续参与转移」这一无限次复用步骤，只是把物品集合换成平方数、把取 max 价值换成取 min 项数并把初值改为 INF，Δrank=2 属台阶题而非新范式。"},{"id":"pre:luogu/P6145->luogu/P1983","source":"luogu/P6145","target":"luogu/P1983","type":"pre","directed":true,"reason":"B 在虚拟节点压缩出的 DAG 上沿用 A 教的按拓扑序做 dp[v]=max(dp[v],dp[u]+w) 的带权最长路转移（main.cpp 的 level[v]=max(level[v],level[u]+w)），只把初值换成最低级别 1、边权换成 0/1，新增成本是 A 未教的虚拟节点压缩。"},{"id":"pre:luogu/P3074->luogu/P1983","source":"luogu/P3074","target":"luogu/P1983","type":"pre","directed":true,"reason":"A 教的是“建 DAG + 拓扑序上按边做最长路转移”；B 在虚拟节点压缩出新 DAG 后，代码里仍然用 level[v]=max(level[v],level[u]+w) 这一拓扑最长路转移求出各级别，只是把 A 的点权换成了 0/1 边权。"},{"id":"pre:luogu/P1807->luogu/P1983","source":"luogu/P1807","target":"luogu/P1983","type":"pre","directed":true,"reason":"B 的正式解（main.cpp 拓扑队列 + level[v]=max(level[v],level[u]+w)）直接复用 A 教的 DAG 最长路转移 max(dist[u]+w) 与拓扑序松弛次序，只是把 A 的『编号天然拓扑序』换成自己用入度队列排序，新增成本全在虚拟节点建图与 0/1 边权压缩上。"},{"id":"pre:luogu/U663797->luogu/P6893","source":"luogu/U663797","target":"luogu/P6893","type":"pre","directed":true,"reason":"A 教的按体积余数分组、把组内转移看成等差数列下标上的滑动窗口，正是 B 离散菜凹费用背包（main.cpp 中 for (int r = 0; r < cost; r++) 的逐余数分组单调队列）实际复用的一步；B 的主要新增难度在连续菜的分段二次最优分配与两类菜合并。"},{"id":"pre:leetcodecn/partition-equal-subset-sum->luogu/P7097","source":"leetcodecn/partition-equal-subset-sum","target":"luogu/P7097","type":"pre","directed":true,"reason":"B 的正式解第一步就是 A 教的那类 0/1 背包判定/最值转移：main.cpp:41 用 for (s=200; s>=p[i]; s--) 容量倒序枚举保证每种道具本回合最多选一次，把 A 的“dp[和] 可达 + 倒序避免重复选”换成 best_k[总 p]=最大总 k，之后才叠加极小极大 DP 这一新台阶。"},{"id":"pre:leetcodecn/coin-change->luogu/P3891","source":"leetcodecn/coin-change","target":"luogu/P3891","type":"pre","directed":true,"reason":"B 的 gain[x] 就是 A 那种完全背包升序转移的复用，只是把 min(最少硬币数) 换成 max(最多新增采集效率)，再叠加按时间推进的状态 DP"},{"id":"pre:luogu/P5662->luogu/P3891","source":"luogu/P5662","target":"luogu/P3891","type":"pre","directed":true,"reason":"B 的 gain[c] 预处理逐字复用了 A 教的「一种物品可无限买 ⇒ 容量维正序完全背包」这一步，把 A 每天重跑一次、容量为金币的完全背包抽成与时间无关的一层预处理，再在其上叠加时间推进 DP，属台阶式叠加。"},{"id":"pre:luogu/P1853->luogu/P3891","source":"luogu/P1853","target":"luogu/P3891","type":"pre","directed":true,"reason":"B 的 gain[x] 预处理正是 A 教的「单期内花预算的最优决策用完全背包 best[j]=max(best[j],best[j-cost]+profit) 压成一张与时间无关的表」这一步（B main.cpp:50 与 A main.cpp:38 同形），A 用它把每年利息滚入下一年本金，B 把它抽成 gain[x] 后再叠加自己新的时间推进 DP（dp[资源]=最大已有采集效率、只保留最大效率的论证），属台阶式叠加而非同难度孪生。"},{"id":"pre:leetcodecn/partition-equal-subset-sum->luogu/P4138","source":"leetcodecn/partition-equal-subset-sum","target":"luogu/P4138","type":"pre","directed":true,"reason":"B 的第 3 步把 A 教的「选子集问题写成 dp[累积量]、每件物品至多选一次的 0/1 背包」原样搬来，只把 A 的和维可达性 dp[i] 换成挂钩数维的最小损失 dp[j]（main.cpp:66-92 逐挂饰转移），A 未教的台阶是 sum(A)>=k-1 可行性观察与必选/必不选分类。"},{"id":"pre:luogu/P2627->luogu/P2254","source":"luogu/P2627","target":"luogu/P2254","type":"pre","directed":true,"reason":"A 教的「枚举断点把转移化成 常数 + 窗口内 dp[j-1]-S[j] 最大值，窗口随枚举量单调右移」正是 B 的 move_right/up 等四个函数在做的事：dp[x][y]=y+max(old_dp[x][k]-k) 的窗口最值交给单调队列取队头（B 代码 old_dp[row][q[head]]-q[head]+col），B 只是把 1D 断点窗口按整段时间搬到网格行/列，并加上遇家具清空队列。"},{"id":"pre:luogu/U663797->luogu/P2254","source":"luogu/U663797","target":"luogu/P2254","type":"pre","directed":true,"reason":"B 的位移量 y-k 与 A 的 (k-j)w 一样把与窗口点相关的线性项提出 max，再对变换值 old_dp[x][k]-k 用单调队列取窗口最大值，只是从 1D 容量下标搬到网格行/列，并叠上整段压缩与遇家具清空队列"},{"id":"pre:luogu/P1725->luogu/P2254","source":"luogu/P1725","target":"luogu/P2254","type":"pre","directed":true,"reason":"A 教的“合法前驱是右移滑动窗口、用单调队列维护窗口内最大值”正是 B main.cpp 里 move_right/up 等四个函数在做的转移（old_dp-下标 的滑动窗口取最大），B 只是把 1D 下标窗口搬到网格行/列并叠加上整段压缩、-k 变换与家具清空队列。"},{"id":"pre:luogu/P8687->luogu/P1433","source":"luogu/P8687","target":"luogu/P1433","type":"pre","directed":true,"reason":"B 的状压 DP 逐字复用了 A 教的「把元素压成二进制 mask、用 mask|新元素 累积已选集合」这一步：A 用它做集合覆盖 dp[mask]，B 沿用同一累积方式并叠加「最后停在哪块」维度与距离代价，形成台阶式加难。"},{"id":"pre:luogu/P1853->luogu/P1450","source":"luogu/P1853","target":"luogu/P1450","type":"pre","directed":true,"reason":"B 的第一步直接复用 A 教的完全背包状态设计（一维金额数组、物品外层、容量正序枚举即可重复取用），把 A 的 best[j]=max(best[j],best[j-cost]+profit) 换成方案数累加 dp[x]+=dp[x-c]（P1450 main.cpp 第 35-39 行与 A main.cpp 第 38 行同模板）预处理无限硬币的表，随后才叠加 A 未教的 16 子集容斥扣掉超上界方案，属台阶式叠加。"},{"id":"pre:luogu/P1025->luogu/P1450","source":"luogu/P1025","target":"luogu/P1450","type":"pre","directed":true,"reason":"B 的第一步直接复用 A 教的『物品为外层循环、物品可无限复用、方案不区分顺序』的完全背包计数（A 的 dp[sum][cnt]+=dp[sum-part][cnt-1] 与 B main.cpp 的 for i: for s: dp[s]+=dp[s-coin[i]] 同形），只把 A 的份数维去掉、语义同为方案数，随后才在其上叠加 16 个子集容斥扣掉超上界方案这一额外流程。"},{"id":"pre:leetcodecn/coin-change->luogu/P1450","source":"leetcodecn/coin-change","target":"luogu/P1450","type":"pre","directed":true,"reason":"B 的第一步直接复用 A 教的“硬币无限使用”完全背包（一维金额 dp、由 dp[x-c] 转移），只把 A 的最少硬币数语义换成方案数累加 dp[x]+=dp[x-c] 做预处理，再在其上用 16 个子集容斥扣掉超上界方案，属 A 之上叠加容斥流程。"},{"id":"pre:luogu/P1725->luogu/P2851","source":"luogu/P1725","target":"luogu/P2851","type":"pre","directed":true,"reason":"B 的多重背包优化直接复用 A 教的滑动窗口单调队列：A 用队列在 [i-R,i-L] 内取 dp 最大值，B 对每个 mod v 余数类维护 old[t*v+r]-t 的长度 c+1 窗口最小值，把 O(k) 枚举降为 O(1) 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