P5058. [ZJOI2004] 嗅探器

\(\text{Solution}\)

一道简单的\(tarjan\)求割点，只需判断\(B\)点是否在子树内即可。

\(\text{Code}\)

#include<cstdio>
#include<algorithm>
#define RE register
#define IN inline
using namespace std;
const int N = 2e5 + 5;
int n,dfn[N],low[N],siz[N],h[N],tot,dfc,a,b,ans = 2e9;

struct edge{
	int to,nxt;
}e[N << 3];
IN void add(int x,int y) {e[++tot] = edge{y,h[x]},h[x] = tot;}
IN void tarjan(int u,int fa)
{
	dfn[u] = low[u] = ++dfc,siz[u] = 1;
	for (RE int i = h[u]; i; i = e[i].nxt)
	{
		int v = e[i].to;
		if (!dfn[v])
		{
			tarjan(v,u),siz[u] += siz[v],low[u] = min(low[u],low[v]);
			if (low[v] >= dfn[u] && dfn[u] > 1)
				if (dfn[b] >= dfn[v] && dfn[b] < dfn[v] + siz[v]) ans = min(ans,u);
		}
		else 
			if (v != fa) low[u] = min(low[u],dfn[v]);
	}
}
int main()
{
	int q,p;
	scanf("%d%d%d",&n,&q,&p);
	for (; q != 0 && p != 0;) add(q,p),add(p,q),scanf("%d%d",&q,&p);
	scanf("%d%d",&a,&b);
	tarjan(a,a);
	if (ans == 2000000000) printf("No solution\n");
	else printf("%d\n",ans);
}