uva 796（求割边）

题意：在一张图中，让你求割边。并按照顺序数出来，注意图并不是连通的。

思路：对多个连通分支，每个执行一次dfs求割边。最后排序输出就好了。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;

typedef long long ll;
typedef pair<int ,int> pii;
typedef pair<unsigned int, unsigned int> puu;
typedef pair<int ,double> pid;
typedef pair<ll, int> pli;

const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int LEN = 1010;
struct Arc{
    int from, to;
}bri[LEN];
int mp[LEN][LEN], n, nbri, low[LEN], dfn[LEN], dfs_clock, vis[LEN];
vector<int> Map[LEN];
inline bool cmp(Arc a, Arc b)
{
    if(a.from != b.from) return a.from<b.from;
    else return a.to<b.to;
}


//割边模版
//初始化
//for(int i=0; i<LEN; i++)Map[i].clear();
//memset(vis, 0, sizeof vis);
void dfs(int u, int fa)
{
    int v, i, son = 0;
    vis[u] = 1;
    dfn[u] = low[u] = dfs_clock++;
    for(int i = 0; i < Map[u].size(); i++){
        v = Map[u][i];
        if(vis[v] == 1 && v != fa)low[u] = min(low[u], low[v]);
        if(vis[v] == 0){
            dfs(v, u);
            son++;
            low[u] = min(low[u], low[v]);
            if(low[v] > dfn[u])bri[nbri].from = min(u, v), bri[nbri++].to = max(v, u);
        }
    }
}

int main()
{
//    freopen("in.txt", "r", stdin);

    int from, to, tn;
    while(scanf("%d", &n)!=EOF)
    {
        memset(mp, 0 ,sizeof mp);
        for(int i=0; i<LEN; i++)Map[i].clear();
        memset(vis, 0, sizeof vis);
        dfs_clock = 0, nbri = 0;
        for(int i=0; i<n; i++){
            scanf("%d (%d)", &from, &tn);
            for(int j=0; j<tn; j++){
                scanf("%d", &to);
                if(!mp[from][to]){
                    mp[from][to] = mp[to][from] = 1;
                    Map[from].PB(to);
                    Map[to].PB(from);
                }
            }
        }
        for(int i=0; i<n; i++){
            if(vis[i]==0){
                dfs(i, -1);
            }
        }
        sort(bri, bri+nbri, cmp);
        printf("%d critical links
", nbri);
        for(int i=0; i<nbri; i++) printf("%d - %d
", bri[i].from, bri[i].to);
        printf("
");
    }
    return 0;
}