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cf2.c

--ZHIYUAN 2017-02-07 22:00 原文

C. Timofey and a tree
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Each New Year Timofey and his friends cut down a tree of n vertices and bring it home. After that they paint all the n its vertices, so that the i-th vertex gets color ci.

Now it's time for Timofey birthday, and his mother asked him to remove the tree. Timofey removes the tree in the following way: he takes some vertex in hands, while all the other vertices move down so that the tree becomes rooted at the chosen vertex. After that Timofey brings the tree to a trash can.

Timofey doesn't like it when many colors are mixing together. A subtree annoys him if there are vertices of different color in it. Timofey wants to find a vertex which he should take in hands so that there are no subtrees that annoy him. He doesn't consider the whole tree as a subtree since he can't see the color of the root vertex.

A subtree of some vertex is a subgraph containing that vertex and all its descendants.

Your task is to determine if there is a vertex, taking which in hands Timofey wouldn't be annoyed.

Input

The first line contains single integer n (2 ≤ n ≤ 105) — the number of vertices in the tree.

Each of the next n - 1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), denoting there is an edge between vertices u and v. It is guaranteed that the given graph is a tree.

The next line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 105), denoting the colors of the vertices.

Output

Print "NO" in a single line, if Timofey can't take the tree in such a way that it doesn't annoy him.

Otherwise print "YES" in the first line. In the second line print the index of the vertex which Timofey should take in hands. If there are multiple answers, print any of them.

Examples
input
4
1 2
2 3
3 4
1 2 1 1
output
YES
2
input
3
1 2
2 3
1 2 3
output
YES
2
input
4
1 2
2 3
3 4
1 2 1 2
output
NO

 题意：

有n个点，给出n-1条边，有给出这n个点的颜色，问其中有没有一个点可以作为根节点使得每个子树上的点的颜色都相同。

代码：

//这题有一个巧妙的解法，如果不是所有的点颜色都相同的情况那么根节点必定和他儿子的颜色不相同
//并且整棵树中只有根节点和他的儿子的颜色不相同，那么只要统计给出的的每条边中两个端点颜色不相同
//的个数sum并且记这两个点的权值加一，最后如果某一点的权值等于sum，该点就是根。
#include<bits/stdc++.h>
using namespace std;
struct edge{
    int u,v;
}e[200005];
int n,c[200005],sum=0,tmp[200005];
int main()
{
    cin>>n;
    for(int i=1;i<n;i++)
        scanf("%d%d",&e[i].u,&e[i].v);
    for(int i=1;i<=n;i++) scanf("%d",&c[i]);
    memset(tmp,0,sizeof(tmp));
    for(int i=1;i<n;i++){
        if(c[e[i].u]!=c[e[i].v]){
            sum++;tmp[e[i].u]++;tmp[e[i].v]++;
        }
    }
    if(sum==0) cout<<"YES\n"<<"1\n";
    else{
        for(int i=1;i<=n;i++){
            if(tmp[i]==sum){
                cout<<"YES\n"<<i<<endl;
                sum=0;
                break;
            }
        }
        if(sum!=0) cout<<"NO\n";
    }
    
    return 0;
}

 

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