Light OJ 1026 - Critical Links （图论-双向图tarjan求割边,桥）

题目大意：双向联通图， 现在求减少任意一边使图的联通性改变，按照起点从小到大列出所有这样的边

解题思路：双向边模版题 tarjan算法

代码如下：

#include<bits/stdc++.h>
using namespace std;
 
const int N = ;
vector<int>vec[N];
pair<int, int>edge[N];
int dfn[N], low[N];
int res, ans;
 
void tarjan(int u, int f)
{
    dfn[u] = low[u] = res ++;
    for(int i = ; i < vec[u].size(); ++ i)
    {
        int v = vec[u][i];
        if(dfn[v] == -)
        {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else if(f != v)
            low[u] = min(low[u], dfn[v]);
 
        if(dfn[u] < low[v])
        {
            if(u > v)
                edge[ans ++] = make_pair(v, u);
            else
                edge[ans ++] = make_pair(u, v);
        }
    }
}
 
void solve(int cases)
{
    for(int i = ; i < N; ++ i)
        vec[i].clear();
 
    int n;
    scanf("%d", &n);
 
    for(int i = ; i < n; ++ i)
    {
        int a, b, c;
        scanf("%d (%d)", &a, &b);
        for(int j = ; j <= b; ++ j)
        {
            scanf("%d", &c);
            vec[a].push_back(c);
        }
    }
 
    memset(dfn, -, sizeof(dfn));
    memset(low, -, sizeof(low));
 
    ans = res = ;
    for(int i = ; i < n; ++ i)
    {
        if(dfn[i] == -)
            tarjan(i, -);
    }
    sort(edge, edge+ans);
    printf("Case %d:\n", cases);
    printf("%d critical links\n", ans);
    for(int i=; i<ans; ++ i)
        printf("%d - %d\n", edge[i].first, edge[i].second);
}
 
int main()
{
    int T;
    scanf("%d", &T);
    for(int i = ; i <= T; ++ i)
        solve(i);
    return ;
}