洛谷 P3398

1. 若两条路径lca相同，则在lca处两者相交
2. 两条路径相交点不可能在非lca处，即两条路径如果相交，一定是某一条路径的lca为另一条路径某个端点的祖先。
3. 需要判断两条路径lca深度大小。再判断两次lca，做完了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<deque>
#include<cstdlib>
#include<ctime>
#define dd double
#define ld long double
#define ll long long
#define ull unsigned long long
#define N 100100
#define M 200200
using namespace std;

const int INF=0x3f3f3f3f;

inline ll read(){
	ll x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

struct edge{
	int to,next;
	inline void intt(int to_,int ne_){
		to=to_;next=ne_;
	}
};
edge li[M];
int head[N],tail;

inline void add(int from,int to){
	li[++tail].intt(to,head[from]);
	head[from]=tail;
}

int size[N],deep[N],top[N],son[N],fa[N];

inline void dfs1(int k,int f){
	size[k]=1;deep[k]=deep[f]+1;fa[k]=f;
	for(int x=head[k];x;x=li[x].next){
		int to=li[x].to;
		if(to==f) continue;
		dfs1(to,k);
		size[k]+=size[to];
		if(size[to]>size[son[k]]) son[k]=to;
	}
}

inline void dfs2(int k,int t){
	top[k]=t;
	if(!son[k]) return;
	dfs2(son[k],t);
	for(int x=head[k];x;x=li[x].next){
		int to=li[x].to;
		if(to==son[k]||to==fa[k]) continue;
		dfs2(to,to);
	}
}

inline int lca(int x,int y){
	while(top[x]!=top[y]){
		if(deep[top[x]]<deep[top[y]]) swap(x,y);
		x=fa[top[x]];
	}
	if(deep[x]>deep[y]) swap(x,y);
	return x;
}

int n,q;

int main(){
	n=read();q=read();
	for(int i=1;i<=n-1;i++){
		int from=read(),to=read();
		add(from,to);add(to,from);
	}
	dfs1(1,0);dfs2(1,1);
	for(int i=1;i<=q;i++){
		int a1=read(),a2=read(),b1=read(),b2=read();
		int lca1=lca(a1,a2),lca2=lca(b1,b2);
		if(deep[lca1]>deep[lca2]){
			int k1=lca(lca1,b1),k2=lca(lca1,b2);
			if(k1==lca1||k2==lca1) printf("Y\n");
			else printf("N\n");
		}
		else{
			int k1=lca(lca2,a1),k2=lca(lca2,a2);
			if(k1==lca2||k2==lca2) printf("Y\n");
			else printf("N\n");
		}
	}
	return 0;
}