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[u]<=dfn[v]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;
}