codeforces 1328E LCA

没想到LCA，最初的思路是想把查询的每个点的字典序搞出来然后判断

题意：给你一个有n个顶点的无向简单图，顶点1为根节点。然后再给m个询问，每个询问给k个数，问这 k 个点能否在其中某个点到根节点 1 的路径上或者与路径的距离为 1。

思路：问这 k 个点能否在其中某个点到根节点 1 的路径上或者与路径的距离为 1，可以转化为这个点的父节点是否在这个路径上，然后直接裸LCA。

　　　　n*logn + m*logn

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '
'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e5+10;
int t, m, n, k;
int tot;
int ver[2*maxn],edge[2*maxn],nexts[2*maxn];
int head[2*maxn];
int d[maxn],f[maxn][20],dist[maxn];    
int q[maxn];
void add(int x,int y){
    ver[++tot] = y;
    nexts[tot] = head[x];
    head[x] = tot;
    //edge[tot] = z;
}
void bfs(){
    queue<int> que;
    que.push(1);
    d[1] = 1;
    while(!que.empty()){
        int x = que.front();
        que.pop();
        for(int i = head[x]; i; i = nexts[i]){
            int y = ver[i];
            if(d[y]) continue;
            //dist[y] = dist[x] + edge[i];
            d[y] = d[x] + 1;
            f[y][0] = x;
            for(int j = 1; j <= k;j++){
                f[y][j] = f[f[y][j-1]][j-1];
            }
            que.push(y);
        }
    }
}
int lca(int x, int y){
    if(d[x] > d[y]) swap(x,y);
    for(int i = k;i >= 0;i--){
        if(d[f[y][i]] >= d[x]) y = f[y][i];
    }
    if(x == y) return x;
    for(int i = k;i >= 0;i--){
        if(f[x][i] != f[y][i]){
            x = f[x][i];
            y = f[y][i];
        }
    }
    return f[x][0];
}
int main()
{
    cin>>n>>m;
    rep(i,1,n-1){
        int x,y;
        cin>>x>>y;
        add(x,y);
        add(y,x);
    }
    k = (int)(log(n) / log(2)) + 1;
    bfs();    
    while(m--){
        int nub;
        cin>>nub;
        int pos = 0;
        int maxdeep = -1;
        rep(i,1,nub){
            int x;
            cin>>x;
            q[i] = f[x][0];
            if(maxdeep < d[q[i]]){
                maxdeep = d[q[i]];
                pos = q[i];
            }
        }
        int flag = 0;
        rep(i,1,nub){
            int LCA = lca(pos,q[i]);
            if(LCA != q[i]){
                flag = 1;
                break;
            }
        }
        cout<<(flag==1?"NO":"YES")<<endl;


    }
}