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("");
    }
}