树形结构 Codeforces Round #629 (Div. 3) E题

Tree Queries

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.

---

题目大意：给你一颗n个结点的树，m个询问，每个询问包含 k 个结点，问是否存在一条路径（由根结点 1 开始的路径）包含这k个结点，或者这些结点距离这条路径的长度为 1；

非常思维的一道树上题目，mx[] 数组表示按照dfs序表示结点顺序，mi[] 数组表示该结点可以往下到达的最大结点（结点按dfs序表示）；可以发现，如果两个点x，y不在一条路径上，那么深度更小的点 mi[x] 一定小于深度更大的点的 mx[y]；

所以只要记录 k 个点的mi的最小值，mx的最大值，比较大小就可以；

代码：

#include
#define LL long long
#define pa pair
#define ls k<<1
#define rs k<<1|1
#define inf 0x3f3f3f3f
using namespace std;
const int N=200100;
const int M=50100;
const int mod=1e9;
int n,m;
struct Node{
	int to,nex;
}edge[N*2];
int head[N],cnt;
int fa[N],mi[N],mx[N],num;
void add(int p,int q){
	edge[cnt].to=q,edge[cnt].nex=head[p],head[p]=cnt++;
}
void dfs(int p,int ft){
	fa[p]=ft;
	mx[p]=++num;
	for(int i=head[p];~i;i=edge[i].nex){
		int q=edge[i].to;
		if(q!=ft) dfs(q,p);
	}
	mi[p]=num;
}
int main(){
	memset(head,-1,sizeof(head));
	scanf("%d%d",&n,&m);
	for(int i=1;i<n;i++){
		int u,v;scanf("%d%d",&u,&v);
		add(u,v),add(v,u);
	}
	dfs(1,0);
	while(m--){
		int k;scanf("%d",&k);
		int m_i=n,m_x=1;
		for(int i=1;i<=k;i++){
			int x;scanf("%d",&x);
			if(x!=1) x=fa[x];
			m_i=min(m_i,mi[x]);
			m_x=max(m_x,mx[x]);
		}
		if(m_x<=m_i) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}