UVA 10765 (割点)

来源：互联网发布：苗疆真的有蛊吗知乎编辑：程序博客网时间：2024/05/21 17:28

题意：给出一个无向图，求出鸽子值，鸽子值即是每删除一个点后有多少个连通块。

思路：求割点，与割点相连的连通分量就是割点的鸽子值，非割点的点的鸽子值均为1。

#include <stdio.h>#include <stdlib.h>#include <iostream>#include <algorithm>#include <cstring>#include <math.h>#include <vector>#include <queue>#include <stack>using namespace std;vector <int> g[10010];struct C {int no, num;};C cut[10010];int dfs_clock;int pre[10010];bool comp(C a, C b){    if(a.num != b.num) return a.num > b.num;    else return a.no < b.no;}int dfs(int u, int fa){    int lowu = pre[u] = ++dfs_clock;    int child = 0;    for(int i = 0; i < g[u].size(); i++)    {        int v = g[u][i];        if(!pre[v])        {            child++;            int lowv = dfs(v, u);            lowu = min(lowu, lowv);            if(lowv >= pre[u])            {                cut[u].num++;            }        }        else if(pre[v] < pre[u] && v != fa)        {            lowu = min(lowu, pre[v]);        }    }    if(fa < 0 && child == 1) cut[u].num = 0;    return lowu;}int main(){    int n, m;    while(scanf("%d%d", &n, &m))    {        if(n == 0 && m == 0) break;        for(int i=0; i<10010; i++)        {            g[i].clear();            cut[i].no = i, cut[i].num = 0;        }        memset(pre, 0, sizeof(pre));        int x, y;        while(scanf("%d%d", &x, &y))        {            if(x == -1 && y == -1) break;            g[x].push_back(y);            g[y].push_back(x);        }        dfs_clock = 0;        dfs(0, -1);        sort(cut, cut+n, comp);        for(int i = 0; i < m; i++)        {            printf("%d %d\n", cut[i].no, cut[i].num + 1);        }        printf("\n");    }    return 0;}

0 0