uva 610(tarjan的应用)

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=23727

思路：首先是Tarjan找桥，对于桥，只能是双向边，而对于同一个连通分量而言，只要重新定向为同一个方向即可。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stack>
 #include<queue>
 #include<vector>
 using namespace std;
 #define MAXN 1111
 typedef pair<int,int>PP;

 int low[MAXN],dfn[MAXN];
 bool mark[MAXN];
 int _count,cnt;
 vector<int>g[MAXN];
 vector<PP>bridge;
 bool vis[MAXN][MAXN];
 int n,m;

 void Tarjan(int u,int father)
 {
     int flag=;
     low[u]=dfn[u]=++cnt;
     mark[u]=true;
     for(int i=;i<(int)g[u].size();i++){
         int v=g[u][i];
         if(v==father&&!flag){ flag=;continue; }
         if(dfn[v]==){
             Tarjan(v,u);
             low[u]=min(low[u],low[v]);
             if(low[v]>dfn[u]){
                 bridge.push_back(make_pair(u,v));
                 bridge.push_back(make_pair(v,u));
                 vis[u][v]=vis[v][u]=true;
             }else {
                 bridge.push_back(make_pair(u,v));
                 vis[u][v]=vis[v][u]=true;
             }
         }else if(mark[v]){
             low[u]=min(low[u],dfn[v]);
             if(!vis[u][v]){
                 bridge.push_back(make_pair(u,v));
                 vis[u][v]=vis[v][u]=true;
             }
         }
     }
 }

 int main()
 {
     int u,v,t=;
     while(~scanf("%d%d",&n,&m)){
         if(n==&&m==)break;
         for(int i=;i<=n+;i++)g[i].clear();
         bridge.clear();
         while(m--){
             scanf("%d%d",&u,&v);
             g[u].push_back(v);
             g[v].push_back(u);
         }
         memset(mark,false,sizeof(mark));
         memset(dfn,,sizeof(dfn));
         memset(vis,false,sizeof(vis));
         _count=cnt=;
         Tarjan(,-);
         printf("%d\n\n",t++);
         for(int i=;i<(int)bridge.size();i++){
             printf("%d %d\n",bridge[i].first,bridge[i].second);
         }
         puts("#");
     }
     return ;
 }