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。
1 #include<iostream> 2 #include<cstdio> 3 #include<vector> 4 #include<cstring> 5 using namespace std; 6 #define LL long long 7 #define pb push_back 8 const int maxn=1010; 9 vector<int>g[maxn]; 10 int dfn[maxn],low[maxn],sub[maxn],root,sum,ban; 11 bool vis[maxn]; 12 void dfs(int u){ 13 dfn[u]=low[u]=sum++; 14 vis[u]=1; 15 int son=0; 16 for(int i=0;i<g[u].size();++i){ 17 int v=g[u][i]; 18 if(v==ban) continue; 19 if(vis[v]){ 20 low[u]=min(low[u],dfn[v]); 21 } 22 else{ 23 dfs(v); 24 if(u==root)son++; 25 else{ 26 if(low[v]>=dfn[u])sub[u]++; 27 } 28 low[u]=min(low[u],low[v]); 29 } 30 } 31 if(u==root)sub[u]=son; 32 else sub[u]++; 33 } 34 int main(){ 35 int n,m,cas=0,i,j,u,v; 36 while(scanf("%d%d",&n,&m)==2){ 37 if(n==0&&m==0)break; 38 for(i=1;i<=n;++i)g[i].clear(); 39 while(m--){ 40 scanf("%d%d",&u,&v); 41 g[u].pb(v); 42 g[v].pb(u); 43 } 44 int ans=0; 45 for(i=1;i<=n;++i){ 46 int p=0; 47 ban=i,sum=0; 48 memset(vis,0,sizeof(vis)); 49 memset(sub,0,sizeof(sub)); 50 for(j=1;j<=n;++j){ 51 if(j==ban || vis[j]) continue; 52 root=j; 53 p++; 54 dfs(j); 55 } 56 //cout<<"p="<<p<<endl; 57 for(j=i+1;j<=n;++j){ 58 if(sub[j]+p-1>=2)ans++; 59 } 60 } 61 printf("Case %d: %d ",++cas,ans); 62 } 63 return 0; 64 }