Codeforces 1037D Valid BFS? (BFS)

The BFS algorithm is defined as follows.

1. Consider an undirected graph with vertices numbered from 1 to n. Initialize q as a new queue containing only vertex 1, mark the vertex 1 as used.
2. Extract a vertex v from the head of the queue q.
3. Print the index of vertex v.
4. Iterate in arbitrary order through all such vertices uthat u is a neighbor of v and is not marked yet as used. Mark the vertex u as used and insert it into the tail of the queue q.
5. If the queue is not empty, continue from step 2.
6. Otherwise finish.

 

Since the order of choosing neighbors of each vertex can vary, it turns out that there may be multiple sequences which BFS can print.

In this problem you need to check whether a given sequence corresponds to some valid BFS traversal of the given tree starting from vertex 1. The tree is an undirected graph, such that there is exactly one simple path between any two vertices.

Input

The first line contains a single integer n(1≤n≤2⋅10^5) which denotes the number of nodes in the tree.

The following n−1 lines describe the edges of the tree. Each of them contains two integers x and y (1≤x,y≤n) — the endpoints of the corresponding edge of the tree. It is guaranteed that the given graph is a tree.

The last line contains n distinct integers a1,a2,…,an (1≤ai≤n) — the sequence to check.

Output

Print "Yes" (quotes for clarity) if the sequence corresponds to some valid BFS traversal of the given tree and "No" (quotes for clarity) otherwise.

You can print each letter in any case (upper or lower).

Examples

Input

4
1 2
1 3
2 4
1 2 3 4

Output

Yes

Input

4
1 2
1 3
2 4
1 2 4 3

Output

No

Note

Both sample tests have the same tree in them. In this tree, there are two valid BFS orderings:

• 1,2,3,4,

• 1,3,2,4.

The ordering 1,2,4,3 doesn't correspond to any valid BFS order.

 

題意：給你一棵樹和一個序列，問你是不是bfs序？

 

解題思路： 正常的bfs,當我們出一個父親節點的時候，我們用set來維護它的兒子節點是不是都出現。如果都出現了，那麼我們就按題目給的bfs序來把兒子節點壓入到隊列中。

 

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
using namespace std;
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
const int maxn=2e5+10;
int n,fa[maxn],flag,du[maxn];
int a[maxn];
set<int> st1,st2;
vector<int> ve[maxn];
void bfs(){
	if(a[1]!=1){
		flag=0;return ;
	}
	queue<int> qu;
	while(!qu.empty()) qu.pop();
	qu.push(1);
	int st=2;
	while(!qu.empty()){
		int u=qu.front();qu.pop();
		int i,sum=0,te=du[u],k=st;
		st1.clear();st2.clear();
		for(i=0;i<te;i++){
			if(ve[u][i]!=fa[u]){
				st1.insert(a[k++]);
				st2.insert(ve[u][i]);
			}
		}
		if(st1!=st2){
			flag=0;return ;
		}
		for(i=0;i<te;i++) if(ve[u][i]!=fa[u]) qu.push(a[st++]),fa[a[st-1]]=u; 
		if(st==n){
			flag=1;
			return ;
		}
	}
}
int main(){
	int i,j;
	while(scanf("%d",&n)!=EOF){
		for(i=1;i<=n;i++) ve[i].clear();
		int u,v;mem(du,0);
		for(i=1;i<n;i++){
			scanf("%d%d",&u,&v);
			ve[u].push_back(v);ve[v].push_back(u);
			du[u]++;du[v]++;
		}
		for(i=1;i<=n;i++) scanf("%d",&a[i]);
		flag=1;
		bfs();
		if(flag) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}