HLD 加线段树最小值,按根到节点顺序寻找路径上的第一个黑点。
OJ: luogu
题目 ID: P4116
难度:提高
标签:重链剖分线段树路径查询python
日期: 2026-07-17 02:00
题意
切换节点黑白颜色,查询根到给定节点路径上的第一个黑点。
思路
黑点叶子保存 dfn,白点保存无穷大。把路径拆成若干重链段后反转段列表,从根侧开始取区间最小 dfn;第一个非无穷结果就是答案。
Python 知识
bytearray保存黑白状态,切换用异或。- 迭代线段树区间最小值支持点更新和区间查询。
reversed(segments)保持查询方向与根到节点一致。
代码
python
import sys
input = sys.stdin.buffer.readline
n, operations = 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)
inverse = [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
inverse[timer] = node
for neighbor in graph[node]:
if neighbor != parent[node] and neighbor != heavy[node]:
chains.append((neighbor, neighbor))
node = heavy[node]
size = 1
while size < n:
size <<= 1
infinity = n + 1
segment = [infinity] * (2 * size)
black = bytearray(n + 1)
def toggle(node):
black[node] ^= 1
position = size + dfn[node] - 1
segment[position] = dfn[node] if black[node] else infinity
position //= 2
while position:
segment[position] = min(segment[position * 2], segment[position * 2 + 1])
position //= 2
def range_min(left, right):
left, right = left - 1 + size, right + size
answer = infinity
while left < right:
if left & 1:
answer = min(answer, segment[left])
left += 1
if right & 1:
right -= 1
answer = min(answer, segment[right])
left //= 2
right //= 2
return answer
answers = []
for _ in range(operations):
operation, node = map(int, input().split())
if operation == 0:
toggle(node)
continue
segments = []
while top[node] != top[1]:
segments.append((dfn[top[node]], dfn[node]))
node = parent[top[node]]
segments.append((dfn[1], dfn[node]))
answer = infinity
for left, right in reversed(segments):
answer = range_min(left, right)
if answer != infinity:
break
answers.append(str(-1 if answer == infinity else inverse[answer]))
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
每次操作 O(log^2 n),空间 O(n)。
总结
路径上“第一个”元素要同时考虑拆段顺序和段内最小位置。