Doves and bombs UVA

题意:给定一个n个点的连通的无向图，一个点的“鸽子值”定义为将它从图中删去后连通块的个数。

求对应的点 和 每个点的“鸽子值”

用一个数组在判断割顶的那个地方 累加标记一下所连接的连通块的数量即可

初始化为1。。从1开始累加

饿的发懒。。。看别人的代码吧。原代码地址：https://blog.csdn.net/u014664226/article/details/46622001

#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<cstdlib>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<queue>  
#include<stack> 
#include<string>
#include<map> 
#include<set>
#define eps 1e-6 
#define LL long long  
using namespace std;  
 
const int maxn = 10000 + 100;
const int INF = 0x3f3f3f3f;
int n, m;
vector<int> G[maxn];
int val[maxn], node[maxn]; //node数组记录结点id间接排序 
bool cmp(int x, int y) {
    return val[x] == val[y] ? x < y : val[x] > val[y];
}
 
int pre[maxn], dfs_clock;
int dfs(int u, int fa) { //u在dfs树中的父节点为fa 
    int lowu = pre[u] = ++dfs_clock;
    int child = 0;        //子节点个数 
    for(int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if(!pre[v]) {   //没有访问过v 
            child++;
            int lowv = dfs(v, u);
            lowu = min(lowu, lowv);    //用后代的low函数更新u的low函数 
            if(lowv >= pre[u]) {
                val[u]++;
            }
                
        }
        else if(pre[v] < pre[u] && v != fa)  lowu = min(lowu, pre[v]);  //用反向边更新u的low函数 
    } 
    if(fa < 0 && child == 1) val[u] = 1;
    return lowu; 
} 
 
 
void init() {
    dfs_clock = 0;
    memset(pre, 0, sizeof(pre));
    for(int i = 0; i < n; i++) {
        G[i].clear();
        node[i] = i;
        val[i] = 1;
    }
    int x, y;
    while(scanf("%d%d", &x, &y) == 2 && x >= 0) {
        G[x].push_back(y);
        G[y].push_back(x);
    }
}
 
void solve() {
    dfs(0, -1);
    sort(node, node+n, cmp);
    for(int i = 0; i < m; i++) cout << node[i] << " " << val[node[i]] << endl; 
    cout << endl;
}
 
int main() {
    //freopen("input.txt", "r", stdin);
    while(scanf("%d%d", &n, &m) == 2 && n) {
        init();
        solve();
    }
    return 0;
}