P5058 [ZJOI2004]嗅探器

题面:https://www.luogu.org/problem/P5058

本题为我们提供了一个很好的思路:若存在u,v,满足dfn[v]<=dfn[u]则v一定在以u为根的子树内,然后这题就变成了找割点的题目.

Code:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<ctime>
#include<queue>
using namespace std;
const int N=200005;
int n,c1,c2,head[N],cnt,dfn[N],low[N],tim;
bool cur[N];
struct Node{
    int u,v,nxt;
}edge[N*2];
void add(int u,int v){
    ++cnt;
    edge[cnt].u=u;
    edge[cnt].v=v;
    edge[cnt].nxt=head[u];
    head[u]=cnt;
}
void tarjan(int u,int fa){
    dfn[u]=low[u]=++tim;
    for(int i=head[u];i;i=edge[i].nxt){
        int v=edge[i].v;
        if(v!=fa){
            if(dfn[v]==-1){
                tarjan(v,u);
                low[u]=min(low[u],low[v]);
                if(low[v]>=dfn[u]&&u!=c1&&dfn[v]<=dfn[c2]){
                    cur[u]=1;
                }
            }
            low[u]=min(low[u],dfn[v]);
        }
    }
}
int main(){
    int u,v;
    memset(dfn,-1,sizeof(dfn));
    scanf("%d",&n);
    while(~scanf("%d%d",&u,&v)){
        if(u==0&&v==0){
            break;
        }
        add(u,v);
        add(v,u);
    }
    scanf("%d%d",&c1,&c2);
    tarjan(c1,0);
    for(int i=1;i<=n;i++){
        if(cur[i]){
            printf("%d\n",i);
            return 0;
        }
    }
    puts("No solution");
    return 0;
}

来源：https://www.cnblogs.com/ukcxrtjr/p/11643866.html