树剖求LCA

我们可以发现,两条路径ab,cd相交,当且仅当 dep[lca(a,b)]>=dep[lca(c,d)]&(lca(lca(a,b),c)lca(a,b)||lca(lca(a,b),d)lca(a,b))或把abcd交换一下即可

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cstdlib>

using namespace std;

const int MAXN=200005;

int init(){

	int rv=0,fh=1;

	char c=getchar();

	while(c<'0'||c>'9'){

		if(c=='-') fh=-1;

		c=getchar();

	}

	while(c>='0'&&c<='9'){

		rv=(rv<<1)+(rv<<3)+c-'0';

		c=getchar();

	}

	return rv*fh;

}

int head[MAXN],dep[MAXN],fa[MAXN],siz[MAXN],son[MAXN],n,m,nume,top[MAXN],id[MAXN],ind;

struct edge{

	int to,nxt;

}e[MAXN<<1];

void adde(int from,int to){

	e[++nume].to=to;

	e[nume].nxt=head[from];

	head[from]=nume;

}

void dfs1(int u,int rt){

	fa[u]=rt;

	dep[u]=dep[rt]+1;

	siz[u]=1;

	int ma=0;

	for(int i=head[u];i;i=e[i].nxt){

		int v=e[i].to;

		if(v==rt) continue;

		dfs1(v,u);

		siz[u]+=siz[v];

		if(ma<siz[v]){

			ma=siz[v];

			son[u]=v;

		}

	}

}

void dfs2(int u,int topf){

	top[u]=topf;

	id[u]=++ind;

	if(!son[u]) return;

	dfs2(son[u],topf);

	for(int i=head[u];i;i=e[i].nxt){

		int v=e[i].to;

		if(v==fa[u]||v==son[u]) continue;

		dfs2(v,v);

	}

}

int LCA(int a,int b){

	while(top[a]!=top[b]){

		if(dep[top[a]]>dep[top[b]]) a=fa[top[a]];

		else b=fa[top[b]];

	}

	return dep[a]>dep[b]?b:a;

}

int main(){

	n=init();m=init();

	for(int i=1;i<n;i++){

		int u=init(),v=init();

		adde(u,v);adde(v,u);

	}

	dep[1]=1;

	dfs1(1,0);

	dfs2(1,1);

	for(int i=1;i<=m;i++){

		int a=init(),b=init(),c=init(),d=init();

		int r1=LCA(a,b),r2=LCA(c,d);

		if(dep[r1]<dep[r2]){

			swap(r1,r2);

			swap(a,c);

			swap(b,d);

		}

		if(LCA(r1,c)==r1||LCA(r1,d)==r1) cout<<"Y"<<endl;

		else cout<<"N"<<endl;

	}

	return 0;

}