hdu-3671-tarjin/割点方案

http://acm.hdu.edu.cn/showproblem.php?pid=3671

给出一幅无向图，询问有多少种移除点对的方案使得剩下的连通分量个数大于1.

和上一题差不多的思路直接做n次tarjin即可。

 #include<iostream>
 #include<cstdio>
 #include<vector>
 #include<cstring>
 using namespace std;
 #define LL long long
 #define pb push_back
 const int maxn=;
 vector<int>g[maxn];
 int dfn[maxn],low[maxn],sub[maxn],root,sum,ban;
 bool vis[maxn];
 void dfs(int u){
     dfn[u]=low[u]=sum++;
     vis[u]=;
     int son=;
     for(int i=;i<g[u].size();++i){
         int v=g[u][i];
         if(v==ban) continue;
         if(vis[v]){
             low[u]=min(low[u],dfn[v]);
         }
         else{
             dfs(v);
             if(u==root)son++;
             else{
                 if(low[v]>=dfn[u])sub[u]++;
             }
             low[u]=min(low[u],low[v]);
         }
     }
     if(u==root)sub[u]=son;
     else sub[u]++;
 }
 int main(){
     int n,m,cas=,i,j,u,v;
     while(scanf("%d%d",&n,&m)==){
         if(n==&&m==)break;
         for(i=;i<=n;++i)g[i].clear();
         while(m--){
             scanf("%d%d",&u,&v);
             g[u].pb(v);
             g[v].pb(u);
         }
         int ans=;
         for(i=;i<=n;++i){
             int p=;
             ban=i,sum=;
             memset(vis,,sizeof(vis));
             memset(sub,,sizeof(sub));
             for(j=;j<=n;++j){
                 if(j==ban || vis[j]) continue;
                 root=j;
                 p++;
                 dfs(j);
             }
             //cout<<"p="<<p<<endl;
             for(j=i+;j<=n;++j){
                 if(sub[j]+p->=)ans++;
             }
         }
         printf("Case %d: %d\n",++cas,ans);
     }
     return ;
 }