Description
You are given a rooted tree consisting of n vertices numbered from 1 to n. The root of the tree is a vertex number 1.
A tree is a connected undirected graph with n−1 edges.
You are given m queries. The i-th query consists of the set of ki distinct vertices vi[1],vi[2],…,vi[ki]. Your task is to say if there is a path from the root to some vertex u such that each of the given k vertices is either belongs to this path or has the distance 1 to some vertex of this path.
Input
The first line of the input contains two integers n and m (2≤n≤2⋅105, 1≤m≤2⋅105) — the number of vertices in the tree and the number of queries.
Each of the next n−1 lines describes an edge of the tree. Edge i is denoted by two integers ui and vi, the labels of vertices it connects (1≤ui,vi≤n,ui≠vi).
It is guaranteed that the given edges form a tree.
The next m lines describe queries. The i-th line describes the i-th query and starts with the integer ki (1≤ki≤n) — the number of vertices in the current query. Then ki integers follow: vi[1],vi[2],…,vi[ki] (1≤vi[j]≤n), where vi[j] is the j-th vertex of the i-th query.
It is guaranteed that all vertices in a single query are distinct.
It is guaranteed that the sum of ki does not exceed 2⋅105 (∑i=1mki≤2⋅105).
Output
For each query, print the answer — “YES”, if there is a path from the root to some vertex u such that each of the given k vertices is either belongs to this path or has the distance 1 to some vertex of this path and “NO” otherwise.
Note
The picture corresponding to the example:
Consider the queries.
The first query is [3,8,9,10]. The answer is “YES” as you can choose the path from the root 1 to the vertex u=10. Then vertices [3,9,10] belong to the path from 1 to 10 and the vertex 8 has distance 1 to the vertex 7 which also belongs to this path.
The second query is [2,4,6]. The answer is “YES” as you can choose the path to the vertex u=2. Then the vertex 4 has distance 1 to the vertex 1 which belongs to this path and the vertex 6 has distance 1 to the vertex 2 which belongs to this path.
The third query is [2,1,5]. The answer is “YES” as you can choose the path to the vertex u=5 and all vertices of the query belong to this path.
The fourth query is [4,8,2]. The answer is “YES” as you can choose the path to the vertex u=9 so vertices 2 and 4 both have distance 1 to the vertex 1 which belongs to this path and the vertex 8 has distance 1 to the vertex 7 which belongs to this path.
The fifth and the sixth queries both have answer “NO” because you cannot choose suitable vertex u.
题目大意
给出一个树，和m个查询，每个查询有一些点，判断是否有一条路径使这些点在该路径上或与这条路径的距离为1
思路
找到最深的点，该点便是路径的终点。
判断其余点x和该点的最近公共祖先是否为x或者是x的父节点
#include <bits/stdc++.h>
using namespace std;
const int maxn=2e5+5;
const int inf=0x3f3f3f3f;
struct node{
int to,next;
}edge[maxn<<1];
int head[maxn],cnt,w[maxn],dep[maxn],fa[maxn][20];
void add(int a,int b){
edge[++cnt].to=b;
edge[cnt].next=head[a];
head[a]=cnt;
}
void dfs(int now,int f){
fa[now][0]=f;
dep[now]=dep[f]+1;
for(int i=head[now];i!=-1;i=edge[i].next){
int to=edge[i].to;
if(to==f)continue;
dfs(to,now);
}
}
int lca(int a,int b){
if(dep[a]<dep[b])swap(a,b);
for(int i=18;i>=0;--i){
if(dep[fa[a][i]]>=dep[b])a=fa[a][i];
}
if(a==b)return a;
for(int i=18;i>=0;--i){
if(fa[a][i]!=fa[b][i]){
a=fa[a][i],b=fa[b][i];
}
}
return fa[a][0];
}
int main()
{
int n,m,q;
scanf("%d %d",&n,&m);
memset(head,-1,sizeof(head));
for(int i=1;i<n;++i){
int a,b;
scanf("%d %d",&a,&b);
add(a,b),add(b,a);
}
dfs(1,0);
for(int j=1;j<=18;++j){
for(int i=1;i<=n;++i){
fa[i][j]=fa[fa[i][j-1]][j-1];
}
}
while(m--){
int ans=1,now=0;
scanf("%d",&q);
for(int i=1;i<=q;++i){
scanf("%d",&w[i]);
if(dep[w[i]]>dep[now])now=w[i];
}
for(int i=1;i<=q;++i){
int x=lca(w[i],now);
if(x!=w[i]&&x!=fa[w[i]][0]){
ans=0;
break;
}
}
printf("%s\n",ans?"YES":"NO");
}
return 0;
}