HDU6268 Master of Subgraph

时间限制: 5 Sec 内存限制: 128 MB
提交: 68 解决: 14
[提交] [状态] [命题人:admin]

题目描述

You are given a tree with n nodes. The weight of the i-th node is wi . Given a positive integer m,now you need to judge that for every integer i in [1，m] whether there exists a connected subgraph which the sum of the weights of all nodes is equal to i.

输入

The first line contains an integer T (1≤T≤15) representing the number of test cases.
For each test case, the first line contains two integers n (1≤n≤3000) and m (1≤m≤100000),which are mentioned above.
The following n-1 lines each contains two integers ui and vi (1≤ui,vi≤n). It describes an edge between node ui and node vi .
The following n lines each contains an integer wi (0≤wi≤100000) represents the weight of the i-th node.
It is guaranteed that the input graph is a tree.

输出

For each test case, print a string only contains 0 and 1, and the length of the string is equal to m. If there is a connected subgraph which the sum of the weights of its nodes is equal to i, the i-th letter of string is 1 otherwise 0.

样例输入

复制样例数据

2
4 10
1 2
2 3
3 4
3 2 7 5
6 10
1 2
1 3
2 5
3 4
3 6
1 3 5 7 9 11

样例输出

0110101010
1011111010

树的重心：找到一个点,其所有的子树中最大的子树节点数最少,那么这个点就是这棵树的重心。

#include "bits/stdc++.h"using namespace std;
const int maxn = 3100;
int size_[maxn], sonsize[maxn], mid, val[maxn], vis[maxn];
vector<int> e[maxn];
bitset<100010> ans, s[maxn];void dfs(int now, int fa, int n) {size_[now] = 1;sonsize[now] = 0;for (auto p:e[now]) {if (vis[p] || p == fa) continue;dfs(p, now, n);size_[now] += size_[p];sonsize[now] = max(sonsize[now], size_[p]);}sonsize[now] = max(sonsize[now], n - sonsize[now]);if (sonsize[now] < sonsize[mid]) mid = now;
}void getdp(int now, int fa) {size_[now] = 1;s[now] <<= val[now];for (auto p:e[now]) {if (p == fa || vis[p]) continue;s[p] = s[now];getdp(p, now);s[now] |= s[p];size_[now] += size_[p];}
}void slove(int now) {vis[now] = 1;s[now].reset();s[now][0] = true;getdp(now, -1);ans |= s[now];for (auto p:e[now]) {if (!vis[p]) {mid = 0;dfs(p, -1, size_[now]);slove(mid);}}
}int main() {//freopen("input.txt", "r", stdin);int _;cin >> _;sonsize[0] = 0x3f3f3f3f;while (_--) {int n, m;cin >> n >> m;for (int i = 1; i <= n; i++) {e[i].clear();vis[i] = 0;}ans.reset();int x, y;for (int i = 1; i < n; i++) {cin >> x >> y;e[x].push_back(y);e[y].push_back(x);}for (int i = 1; i <= n; i++) {cin >> val[i];}mid = 0;dfs(1, -1, n);slove(mid);for (int i = 1; i <= m; i++) {printf("%d", (int) ans[i]);}puts("");}
}

转载于:https://www.cnblogs.com/albert-biu/p/10684889.html