解析

我的实现不够简洁，导致bug越调越谜…
关键就是**找被割点分开的分量的方法：dfs！！！
一下子就豁然开朗了

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=2e3+100;
const double eps=1e-6;
inline ll read(){ll x=0,f=1;char c=getchar();while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}while(isdigit(c)){x=(x<<1)+(x<<3)+c-'0';c=getchar();}return f*x;
}
int n,m;
int ans;
struct node{int to,nxt,from;
}p[N<<1];
int fi[N],cnt;
inline void addline(int x,int y){//printf("%d->%d w=%d\n",x,y,w);p[++cnt]=(node){y,fi[x],x};fi[x]=cnt;
}
int dfn[N],low[N],zhan[N],tim,top,cut[N],r,tot;
void tarjan(int x){dfn[x]=low[x]=++tim;zhan[++top]=x;int nm=0;for(int i=fi[x];~i;i=p[i].nxt){int to=p[i].to;//printf("x=%d to=%d\n",x,to);if(!dfn[to]){++nm;tarjan(to);if(low[to]>=dfn[x]){cut[x]=1;}low[x]=min(low[x],low[to]);}else low[x]=min(low[x],dfn[to]);}if(x==r) cut[x]=nm>1;
}
int num,Cut,tag[N],t;
bool vis[N];
void dfs(int x){if(vis[x]||tag[x]==t) return;num++;//printf("x=%d\n",x);if(cut[x]){tag[x]=t;Cut++;return;}vis[x]=1;for(int i=fi[x];~i;i=p[i].nxt){int to=p[i].to;dfs(to);}
}
int main(){#ifndef ONLINE_JUDGEfreopen("a.in","r",stdin);freopen("a.out","w",stdout);#endifint o=0;while(1){memset(fi,-1,sizeof(fi));cnt=-1;memset(vis,0,sizeof(vis));memset(dfn,0,sizeof(dfn));tim=0;top=0;memset(cut,0,sizeof(cut));memset(tag,0,sizeof(tag));m=read();n=0;if(!m) return 0;for(int i=1;i<=m;i++){int x=read(),y=read();addline(x,y);addline(y,x);n=max(n,max(x,y));}tot=n;for(int i=1;i<=n;i++){r=i;if(!dfn[i]) tarjan(i);}ll nm=0,ans=1;for(int i=1;i<=n;i++){//printf("i=%d cut=%d\n",i,cut[i]);if(!vis[i]&&!cut[i]){num=0;Cut=0;t=i;dfs(i);printf("i=%d num=%d Cut=%d\n",i,num,Cut);if(Cut==0){nm+=2;ans*=num*(num-1)/2;}else if(Cut==1){nm++;ans*=(num-1);}}}printf("Case %d: %lld %lld\n",++o,nm,ans);}
}
/*
4
1 2
2 3
1 3
3 4
*/