学步园

很典型的求无向图的桥个数，并且输出

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

const int N = 110;
int n, ans, g[N][N];
int dfs_clock, pre[N], low[N];
bool is[N][N];

int dfs( int u, int fa )
{
    int lowu = pre[u] = ++dfs_clock;
    int child = 0;
    for ( int v = 0; v < n; ++v ) if ( g[u][v] ) {
        if ( !pre[v] ) {
            child++;
            int lowv = dfs( v, u );
            lowu = min( lowu, lowv );
            if ( lowv > pre[u] ) {
                if ( !is[u][v] && !is[v][u] ) {
                    ans++;
                    is[u][v] = is[v][u] = 1;
                }
            }
        }
        else if ( pre[v] < pre[u] && v != fa ) lowu = min( pre[v], lowu );
    }
    low[u] = lowu;
    return lowu;
}
int main()
{
    while ( scanf("%d", &n) != EOF ) {
        memset(g, 0, sizeof(g));
        memset(pre, 0, sizeof(pre));
        memset(is, 0, sizeof(is));
        ans = dfs_clock = 0;
        for ( int i = 0; i < n; ++i ) {
            int u, v, num;
            scanf("%d (%d)", &u, &num);
            while ( num-- ) {
                scanf("%d", &v);
                g[u][v] = g[v][u] = 1;
            }
        }
        for ( int i = 0; i < n; ++i ) if ( !pre[i] ) dfs(i, -1);
        printf("%d critical links\n", ans);
        for ( int i = 0; i < n; ++i ) 
            for ( int j = i+1; j < n; ++j ) 
                if ( is[i][j] ) printf("%d - %d\n", i, j);
        printf("\n");
    }
}