[ABC318G] Typical Path Problem

Problem Statement

You are given a simple connected undirected graph $G$ with $N$ vertices and $M$ edges.

The vertices and edges of $G$ are numbered as vertex $1$, vertex $2$, $\ldots$, vertex $N$, and edge $1$, edge $2$, $\ldots$, edge $M$, respectively, and edge $i$ $(1\leq i\leq M)$ connects vertices $U_i$ and $V_i$.

You are also given distinct vertices $A,B,C$ on $G$.
Determine if there is a simple path connecting vertices $A$ and $C$ via vertex $B$.

What is a simple connected undirected graph?

A graph $G$ is said to be a simple connected undirected graph when $G$ is an undirected graph that is simple and connected.

A graph $G$ is said to be an undirected graph when the edges of $G$ have no direction.

A graph $G$ is simple when $G$ does not contain self-loops or multi-edges.

A graph $G$ is connected when one can travel between all vertices of $G$ via edges.

Constraints

• $3 \leq N \leq 2\times 10^5$
• $N-1\leq M\leq\min\left(\frac{N(N-1)}{2},2\times 10^5\right)$
• $1\leq A,B,C\leq N$
• $A$, $B$, and $C$ are all distinct.
• $1\leq U_i<V_i\leq N$
• The pairs $(U_i,V_i)$ are all distinct.
• All input values are integers.

点不能重复，考虑圆方树。

如果 \(A,B,C\) 在一个点双内，那么一定没问题。因为删掉B 这个点一定还有路径可以到 \(C\)

那么在圆方树上的体现就是 A,B,C 连着同一个方点。

以此类推，如果 \(A,C\) 在圆方树上的路径通过了一个 B 所连的方点，那么就可以，否则割点不能重复走，所以不行。

#include<bits/stdc++.h>
using namespace std;
const int N=4e5+5;
int n,m,a,b,c,vs[N],st[N],tp,idx,dfn[N],low[N];
struct graph{
	int hd[N],e_num;
	struct edge{
		int v,nxt;
	}e[N<<1];
	void add_edge(int u,int v)
	{
		e[++e_num]=(edge){v,hd[u]};
		hd[u]=e_num;
		e[++e_num]=(edge){u,hd[v]};
		hd[v]=e_num;
	}
}g,h;
int read()
{
	int s=0;
	char ch=getchar();
	while(ch<'0'||ch>'9')
		ch=getchar();
	while(ch>='0'&&ch<='9')
		s=s*10+ch-48,ch=getchar();
	return s;
}
void tarjan(int x)
{
	dfn[x]=low[x]=++idx;
	st[++tp]=x;
	for(int i=g.hd[x];i;i=g.e[i].nxt)
	{
		int v=g.e[i].v;
		if(!dfn[v])
		{
			tarjan(v);
			low[x]=min(low[x],low[v]);
			if(low[v]==dfn[x])
			{
				++n;
				while(st[tp]^v)
					h.add_edge(n,st[tp--]);
				h.add_edge(n,st[tp--]);
				h.add_edge(x,n);
			}
		}
		else
			low[x]=min(low[x],dfn[v]);
	}
}
void dfs(int x,int y)
{
	st[++tp]=x;
	if(x==c)
	{
		for(int i=1;i<=tp;i++)
			vs[st[i]]=1;
		return;
	}
	for(int i=h.hd[x];i;i=h.e[i].nxt)
		if(h.e[i].v^y)
			dfs(h.e[i].v,x);
	--tp;
}
int main()
{
	n=read(),m=read(),a=read(),b=read(),c=read();
	for(int i=1,u,v;i<=m;i++)
		g.add_edge(read(),read());
	tarjan(a);
	dfs(a,0);
	for(int i=h.hd[b];i;i=h.e[i].nxt)
		if(vs[h.e[i].v])
			return puts("Yes"),0;
	puts("No");
}