1523：嗅探器

找割点，且去掉这个点后 S,T 两个点在2个块中

 满足 U!=S && dfn[U] < dfn[T]

#include <bits/stdc++.h>
using namespace std ;
 const int N=2e5+2;
 
 vector<int> g[N];
 int n,m,low[N],dfn[N],ans=1e8,pool,S,T;
 
  void dfs(int x) {
      dfn[x]=low[x]=++pool; 
      int y,i;
      for(i=0;i<g[x].size();i++){
           y=g[x][i];
           if(!dfn[y]){
              dfs(y),low[x]=min(low[x],low[y]);
            if(x!=S&&low[y]>=dfn[x]&&dfn[y]<=dfn[T]){
                ans=min(ans,x);
            }    
         }
           else low[x]=min(low[x],dfn[y]);
      }
 }
 void add(int x,int y){
     g[x].push_back(y);
 }
 int main(){
    cin>>n;
    int x,y;
    while(cin>>x>>y,x&&y){
        add(x,y);add(y,x);
    }
    cin>>S>>T;
    dfs(S);
    if(ans==1e8) cout<<"No solution";else cout<<ans;
}