【lightoj-1026】Critical Links(桥)

题意：

给出无向图，求桥的模板题。

#include <bits/stdc++.h>
using namespace std;
const int N = 10004;
int dfn[N], low[N];//时间戳;low[i]以i为根子树的最小祖先的时间戳
bool vis[N];
vector<int>V[N];
int n, num, c;
pair<int, int>P[N];
void dfs(int s, int f)
{
    low[s] = dfn[s] = ++num;
    for(unsigned int i = 0; i < V[s].size(); i++)
    {
        int v = V[s][i];
        if(!dfn[v])//未访问过（s-v为树边）
        {
            dfs(v, s);//dfs完更新出low[v]
            low[s] = min(low[s], low[v]);
            if(low[v] > dfn[s])//割边不加=
                P[++c].first = min(v, s), P[c].second = max(v, s);
        }
        else if(v != f) low[s] = min(low[s], dfn[v]);//回边且不是父子
    }
}
int main()
{
    int t, n, cas = 0, m, a, b;
    cin>>t;
    while(t--)
    {
        memset(dfn, 0, sizeof dfn);
        memset(low, 0, sizeof low);
        memset(vis, 0, sizeof vis);
        scanf("%d", &n);
        for(int i = 0 ; i < n; i++) V[i].clear();
        num = 0, c = 0;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d (%d)", &a, &m);
            while(m--)
            {
                scanf("%d", &b);
                V[a].push_back(b);
                V[b].push_back(a);
            }
        }
        for(int i = 0; i < n; i++)
        {
            if(!vis[i])
                dfs(i, i);
        }
        sort(P+1, P+1+c);
        printf("Case %d:\n%d critical links\n", ++cas, c);
        for(int i = 1; i <= c; i++)
            printf("%d - %d\n", P[i].first, P[i].second);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/lesroad/p/8747237.html