HLD 加区间加线段树维护有向路径上的最大买卖差值。
OJ: luogu
题目 ID: P3976
难度:提高
标签:重链剖分线段树最大子段差python
日期: 2026-07-17 02:00
题意
沿 a -> b 的路径选择先买后卖的两个城市,输出最大利润,然后让路径上所有价格增加 v。
思路
线段树节点维护最小值、最大值、正向最大差和反向最大差。正向合并的跨段候选是 right.max-left.min,反向则是 left.max-right.min。HLD 拆段时,a 侧区间反向、b 侧区间正向,按旅行顺序合并;查询后再做路径区间加。
Python 知识
- 元组
(min, max, forward, backward)让方向信息随区间一起返回。 reverse_info只交换两个方向的差值,最值不变。None作为空合并结果,避免构造无效哨兵。
代码
python
import sys
sys.setrecursionlimit(1_000_000)
input = sys.stdin.buffer.readline
n = int(input())
prices = [0] + list(map(int, input().split()))
graph = [[] for _ in range(n + 1)]
for _ in range(n - 1):
u, v = map(int, input().split())
graph[u].append(v)
graph[v].append(u)
parent = [0] * (n + 1)
depth = [0] * (n + 1)
order = [1]
for node in order:
for neighbor in graph[node]:
if neighbor != parent[node]:
parent[neighbor] = node
depth[neighbor] = depth[node] + 1
order.append(neighbor)
subtree = [1] * (n + 1)
heavy = [0] * (n + 1)
for node in reversed(order[1:]):
subtree[parent[node]] += subtree[node]
if subtree[node] > subtree[heavy[parent[node]]]:
heavy[parent[node]] = node
top = [0] * (n + 1)
dfn = [0] * (n + 1)
timer = 0
chains = [(1, 1)]
while chains:
node, chain_top = chains.pop()
while node:
top[node] = chain_top
timer += 1
dfn[node] = timer
for neighbor in graph[node]:
if neighbor != parent[node] and neighbor != heavy[node]:
chains.append((neighbor, neighbor))
node = heavy[node]
minimum = [0] * (4 * n)
maximum = [0] * (4 * n)
forward = [0] * (4 * n)
backward = [0] * (4 * n)
lazy = [0] * (4 * n)
base = [0] * (n + 1)
for node in range(1, n + 1):
base[dfn[node]] = prices[node]
def pull(node):
left, right = node * 2, node * 2 + 1
minimum[node] = min(minimum[left], minimum[right])
maximum[node] = max(maximum[left], maximum[right])
forward[node] = max(forward[left], forward[right], maximum[right] - minimum[left])
backward[node] = max(backward[left], backward[right], maximum[left] - minimum[right])
def build(node, left, right):
if left == right:
minimum[node] = maximum[node] = base[left]
return
middle = (left + right) // 2
build(node * 2, left, middle)
build(node * 2 + 1, middle + 1, right)
pull(node)
def apply(node, value):
minimum[node] += value
maximum[node] += value
lazy[node] += value
def push(node):
if lazy[node]:
apply(node * 2, lazy[node])
apply(node * 2 + 1, lazy[node])
lazy[node] = 0
def update(node, left, right, query_left, query_right, value):
if query_left <= left and right <= query_right:
apply(node, value)
return
push(node)
middle = (left + right) // 2
if query_left <= middle:
update(node * 2, left, middle, query_left, query_right, value)
if middle < query_right:
update(node * 2 + 1, middle + 1, right, query_left, query_right, value)
pull(node)
def merge(first, second):
if first is None:
return second
if second is None:
return first
first_min, first_max, first_best, first_reverse = first
second_min, second_max, second_best, second_reverse = second
return (min(first_min, second_min), max(first_max, second_max),
max(first_best, second_best, second_max - first_min),
max(first_reverse, second_reverse, first_max - second_min))
def query(node, left, right, query_left, query_right):
if query_left <= left and right <= query_right:
return minimum[node], maximum[node], forward[node], backward[node]
push(node)
middle = (left + right) // 2
result = None
if query_left <= middle:
result = query(node * 2, left, middle, query_left, query_right)
if middle < query_right:
result = merge(result, query(node * 2 + 1, middle + 1, right, query_left, query_right))
return result
def reverse_info(info):
return info[0], info[1], info[3], info[2]
build(1, 1, n)
def path_info(x, y):
left_parts = []
right_parts = []
while top[x] != top[y]:
if depth[top[x]] >= depth[top[y]]:
left_parts.append(reverse_info(query(1, 1, n, dfn[top[x]], dfn[x])))
x = parent[top[x]]
else:
right_parts.append(query(1, 1, n, dfn[top[y]], dfn[y]))
y = parent[top[y]]
if depth[x] >= depth[y]:
left_parts.append(reverse_info(query(1, 1, n, dfn[y], dfn[x])))
else:
right_parts.append(query(1, 1, n, dfn[x], dfn[y]))
result = None
for info in left_parts:
result = merge(result, info)
for info in reversed(right_parts):
result = merge(result, info)
return result
query_count = int(input())
answers = []
for _ in range(query_count):
x, y, increase = map(int, input().split())
info = path_info(x, y)
answers.append(str(max(0, info[2])))
while top[x] != top[y]:
if depth[top[x]] < depth[top[y]]:
x, y = y, x
update(1, 1, n, dfn[top[x]], dfn[x], increase)
x = parent[top[x]]
if depth[x] > depth[y]:
x, y = y, x
update(1, 1, n, dfn[x], dfn[y], increase)
print("\n".join(answers))原有 C++ 版本仍保留:
cpp
/**
* Author by Rainboy blog: https://rainboylv.com github: https://github.com/rainboylvx
* rbook: -> https://rbook.roj.ac.cn https://rbook2.roj.ac.cn
* rainboy的学习导航网站: https://idx.roj.ac.cn
* create_at: 2026-07-17 01:04
* update_at: 2026-07-17 01:04
*/
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
return 0;
}复杂度
每次查询和更新 O(log^2 n),空间 O(n)。
总结
路径有方向时,区间统计量必须同时保存正向和反向两套信息。