Manthan, Codefest 18 (rated, Div. 1 + Div. 2) D.Valid BFS? (模拟）

D. Valid BFS?

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The BFS algorithm is defined as follows.

1. Consider an undirected graph with vertices numbered from $\mathrm{11 to nn.\; Initialize qq as\; a\; new queue containing\; only\; vertex 11,\; mark\; the\; vertex 11 as\; used.}$
2. Extract a vertex $\mathrm{vv from\; the\; head\; of\; the\; queue qq.}$
3. Print the index of vertex $\mathrm{vv.}$
4. Iterate in arbitrary order through all such vertices $\mathrm{uu that uu is\; a\; neighbor\; of vv and\; is\; not\; marked\; yet\; as\; used.\; Mark\; the\; vertex uu as\; used\; and\; insert\; it\; into\; the\; tail\; of\; the\; queue qq.}$
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 $\mathrm{11.\; The tree is\; an\; undirected\; graph,\; such\; that\; there\; is\; exactly\; one\; simple\; path\; between\; any\; two\; vertices.}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 2\cdot 1051\le n\le 2\cdot 105)\; which\; denotes\; the\; number\; of\; nodes\; in\; the\; tree.}$

The following $\mathrm{n-1n-1 lines\; describe\; the\; edges\; of\; the\; tree.\; Each\; of\; them\; contains\; two\; integers xx and yy (1\le x,y\le n1\le x,y\le n) —\; the\; endpoints\; of\; the\; corresponding\; edge\; of\; the\; tree.\; It\; is\; guaranteed\; that\; the\; given\; graph\; is\; a\; tree.}$

The last line contains $\mathrm{nn distinct\; integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le n1\le ai\le n) —\; the\; sequence\; to\; check.}$

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

Copy

4
1 2
1 3
2 4
1 2 3 4

output

Copy

Yes

input

Copy

4
1 2
1 3
2 4
1 2 4 3

output

Copy

No

Note

Both sample tests have the same tree in them.

In this tree, there are two valid BFS orderings:

• $\mathrm{1,2,3,41,2,3,4,}$
• $\mathrm{1,3,2,41,3,2,4.}$

The ordering $\mathrm{1,2,4,31,2,4,3 doesn\text{'}t\; correspond\; to\; any\; valid BFS order.}$

分析：根据题意，进行第一个样例的模拟，我们可以发现，既然1一定是第一个数，就从1开始扩展，它连接2,3,那么后面一定是2,3，所以需要扫描2,3位置，然后2连接4，扫描4位置

我们可以发现即使有很多数，扫描过程也是从前往后线性进行的，所以可以直接根据题意进行模拟，先标记该位置连接的数，再扫描具体的区间看是否相符。

代码如下:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
#define INF 0x3f3f3f3f
const int MAXN=2e5+100;
int n,x,y;
vector<int>V[MAXN];
int vis[MAXN];
int a[MAXN];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n-1;i++)
    {
      scanf("%d%d",&x,&y);
      V[x].push_back(y);
      V[y].push_back(x);
    }
     for(int i=1;i<=n;i++)
     scanf("%d",&a[i]);
    vis[1]=1;
    int p=2;
    bool flag=true;
    if(a[1]!=1)
    flag=false;
     int len=V[a[1]].size();
    for(int i=1;i<=n;i++)
    {
       for(int j=0;j<V[a[i]].size();j++)
        {
            int x=V[a[i]][j];
            vis[x]=1;
        }
       for(int j=p;j<p+len&&j<=n;j++)
       {
           if(!vis[a[j]])
           {
               flag=false;
               break;
           }
       }
        p=p+len;
        len=V[a[i+1]].size()-1;
       if(flag==false)
       break;
    }
    if(flag==true)printf("Yes
");
    else printf("No
");
    return 0;
}