hdu 5296 Annoying problem(LCA)

题目链接：hdu 5296 Annoying problem

先求出dfs序，然后每次修改一个节点u，找到dfs最接近u的两个点，a，b的dfs序分别大于和小于u的。修改值即为dp[u] - dp[lca(a,u)] - dp[lca(b,u)] + dp[lca(a, b)]，dp[i]表示i到根节点的权值和。但集合中的节点的dfs序都大于或小于u时，a，b即用最大和最小的即可。

#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <algorithm>using namespace std;
const int maxn = 100005;
typedef set<int>::iterator iter;set<int> S;
int N, Q, dp[maxn], vis[maxn];
int E, first[maxn], jump[maxn * 2], link[maxn * 2], val[maxn * 2];
int id, idx[maxn], ridx[maxn], dep[maxn], top[maxn], far[maxn], son[maxn], cnt[maxn];inline void addEdge(int u, int v, int w) {link[E] = v;val[E] = w;jump[E] = first[u];first[u] = E++;
}void dfs (int u, int pre, int d, int w) {far[u] = pre;dep[u] = d;cnt[u] = 1;son[u] = 0;dp[u] = w;for (int i = first[u]; i + 1; i = jump[i]) {int v = link[i];if (v == pre)continue;dfs(v, u, d + 1, w + val[i]);cnt[u] += cnt[v];if (cnt[son[u]] < cnt[v])son[u] = v;}
}void dfs (int u, int rot) {top[u] = rot;if (son[u])dfs(son[u], rot);for (int i = first[u]; i + 1; i = jump[i]) {int v = link[i];if (v == far[u] || v == son[u])continue;dfs(v, v);}idx[u] = ++id;ridx[id] = u;
}int LCA (int u, int v) {int p = top[u], q = top[v];while (p != q) {if (dep[p] < dep[q]) {swap(p, q);swap(u, v);}u = far[p];p = top[u];}if (dep[u] > dep[v])swap(u, v);return u;
}void init () {S.clear();scanf("%d%d", &N, &Q);id = E = 0;memset(first, -1, sizeof(first));memset(vis, 0, sizeof(vis));int u, v, w;for (int i = 1; i < N; i++) {scanf("%d%d%d", &u, &v, &w);addEdge(u, v, w);addEdge(v, u, w);}dfs(1, 0, 0, 0);dfs(1, 1);
}int solve (int u) {if (S.size() == 0)return 0;int a, b;iter mv = S.upper_bound(idx[u]);if (mv == S.begin() || mv == S.end()) {a = ridx[*S.begin()];b = ridx[*S.rbegin()];} else {a = ridx[*mv];mv--;b = ridx[*mv];}return dp[u] - dp[LCA(u, a)] - dp[LCA(u, b)] + dp[LCA(a, b)];
}int main () {int cas;scanf("%d", &cas);for (int kcas = 1; kcas <= cas; kcas++) {init();printf("Case #%d:\n", kcas);int k, u, ans = 0;for (int i = 0; i < Q; i++) {scanf("%d%d", &k, &u);if (k == 1 && vis[idx[u]] == 0) {ans += solve(u);S.insert(idx[u]);vis[idx[u]] = 1;}if (k == 2 && vis[idx[u]]) {S.erase(idx[u]);vis[idx[u]] = 0;ans -= solve(u);}printf("%d\n", ans);}}return 0;
}

hdu 5296 Annoying problem(LCA)相关推荐

hdu A + B Problem II(大数相加，数组实现)
hdu A + B Problem II(大数相加,数组实现) 题目走起注意最后一个case 不需要换行下面代码 #include<stdio.h> #include<strin ...
HDU.1001 Sum Problem
原题 HDU.1001 Sum Problem 分类杂题题意计算从1到正整数n的累加和. 输入/输出要求与格式输入内容每行输入一个正整数输出结果结果为累加和输出格式每个输出结果独占 ...
[HDU 5445]Food Problem[多重背包]
题目链接:[HDU 5445]Food Problem[多重背包] 题意分析: 有n种类型的点心,每种提供t的能量,占据u的空间,有v个: 有m种类型的卡车,每种容量x,雇佣花费y,能提供z辆: 点心 ...
hdu 5266(线段树+LCA)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5266 解题思路: 考虑dfs序,通过在简单的证明可知L~R的LCA为L ~R 中dfs序较小的那个位置 ...
hdu 3549 Flow Problem（最大流模板题）
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549 Problem Description Network flow is a well-known ...
hdu 1023 Train Problem II
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1212 Train Problem II Description As we all know the ...
HDU 5427 A problem of sorting 水题
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5427 A problem of sorting Time Limit: 2000/1000 MS ( ...
HDU - 3078 Network 倍增LCA
倍增lca,把路径上点全部拉出来拍个序输出第k大就过了. 不知道怎么过的,咱也不敢问. #include<stdio.h> #include<string.h> #includ ...
HDU 1022[Train Problem I] 栈的应用
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1022 题目大意:一列序列为s1的车厢是否能通过车站的中转以序列s2出站.白皮上有. 关键思想:栈的应用 ...
HDU 3549 Flow Problem (dinic模版 isap模版)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549 题意: 给你一个有向图,问你1到n的最大流. dinic模版 (n*n*m) 1 #includ ...

hdu 5296 Annoying problem(LCA)

hdu 5296 Annoying problem(LCA)相关推荐

最新文章

热门文章