UVALive 6044(双连通分量的应用)

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=34902

思路：首先是双连通缩点，然后就是搜索一下，搜索时要跳过连通分量的点的个数>=2的点，最后的答案是n*(n-1)/2.

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stack>
 #include<vector>
 using namespace std;
 #define MAXN 11111
 #define MAXM 444444

 struct Edge{
     int v,next;
 }edge[MAXM];

 int n,m,NE,cnt,_count;
 int head[MAXN];

 void Insert(int u,int v)
 {
     edge[NE].v=v;
     edge[NE].next=head[u];
     head[u]=NE++;
 }

 int low[MAXN],dfn[MAXN];
 int color[MAXN];
 bool mark[MAXN];
 stack<int>S;
 void Tarjan(int u,int father)
 {
     int flag=;
     low[u]=dfn[u]=++cnt;
     mark[u]=true;
     S.push(u);
     for(int i=head[u];i!=-;i=edge[i].next){
         int v=edge[i].v;
         if(v==father&&!flag){ flag=;continue; }
         if(dfn[v]==){
             Tarjan(v,u);
             low[u]=min(low[u],low[v]);
         }else if(mark[v]){
             low[u]=min(low[u],dfn[v]);
         }
     }
     if(low[u]==dfn[u]){
         _count++;
         int x=S.top();
         if(x==u)S.pop();
         else {
             do{
                 x=S.top();
                 S.pop();
                 mark[x]=false;
                 color[x]=_count;
             }while(x!=u);
         }
     }
 }

 int ans;
 void dfs(int u,int father)
 {
     color[u]=;
     _count++;
     for(int i=head[u];i!=-;i=edge[i].next){
         int v=edge[i].v;
         if(v==father)continue;
         if(color[v])continue;
         dfs(v,u);
     }
 }

 int main()
 {
     int _case,u,v,t=;
     scanf("%d",&_case);
     while(_case--){
         scanf("%d%d",&n,&m);
         NE=;
         memset(head,-,sizeof(head));
         while(m--){
             scanf("%d%d",&u,&v);
             Insert(u,v);
             Insert(v,u);
         }
         cnt=_count=;
         memset(dfn,,sizeof(dfn));
         memset(color,,sizeof(color));
         for(int i=;i<=n;i++){
             if(dfn[i]==)Tarjan(i,-);
         }
         ans=;
         for(int i=;i<=n;i++){
             if(color[i]==){
                 _count=;
                 dfs(i,-);
                 ans+=_count*(_count-)/;
             }
         }
         printf("Case #%d: %d\n",t++,ans);
     }
     return ;
 }