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

题解:

规定一个特殊边,也就是这条边的两个端点的颜色是不一样的,找出所有的特殊边,同时记录相关颜色
出现的次数,如果有一种颜色出现的次数和边的数目是相同的,那么就存在这么一个点。


 #include<set>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1e5+;
struct Node
{
int u,v;
};
int n;
int d[maxn];
int col[maxn];
Node e[maxn];
int main()
{
scanf("%d",&n);
for(int i=; i<n; i++)
scanf("%d%d",&e[i].u,&e[i].v);
for(int i=; i<=n; i++)
scanf("%d",&col[i]);
int tot=;
for(int i=; i<n; i++)
{
if(col[e[i].u]!=col[e[i].v])
{
tot++;
d[e[i].u]++,d[e[i].v]++;
}
}
for(int i=;i<=n;i++)
{
if(d[i]==tot)
{
printf("YES\n%d\n",i);
return ;
}
}
printf("NO\n");
return ;
}

Codeforces 764C Timofey and a tree的更多相关文章

  1. Codeforces 763A. Timofey and a tree

    A. Timofey and a tree 题意:给一棵树,要求判断是否存在一个点,删除这个点后,所有连通块内颜色一样.$N,C \le 10^5$ 想法:这个叫换根吧.先求出一个点合法即其儿子的子树 ...

  2. 【codeforces 764C】Timofey and a tree

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  3. Codeforces Round #395 (Div. 2) C. Timofey and a tree

    地址:http://codeforces.com/contest/764/problem/C 题目: C. Timofey and a tree time limit per test 2 secon ...

  4. 763A - Timofey and a tree

    A. Timofey and a tree time limit per test 2 seconds memory limit per test 256 megabytes input standa ...

  5. codeforces 741D Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths(启发式合并)

    codeforces 741D Arpa's letter-marked tree and Mehrdad's Dokhtar-kosh paths 题意 给出一棵树,每条边上有一个字符,字符集大小只 ...

  6. codeforces 812E Sagheer and Apple Tree(思维、nim博弈)

    codeforces 812E Sagheer and Apple Tree 题意 一棵带点权有根树,保证所有叶子节点到根的距离同奇偶. 每次可以选择一个点,把它的点权删除x,它的某个儿子的点权增加x ...

  7. codeforces 220 C. Game on Tree

    题目链接 codeforces 220 C. Game on Tree 题解 对于 1节点一定要选的 发现对于每个节点,被覆盖切选中其节点的概率为祖先个数分之一,也就是深度分之一 代码 #includ ...

  8. Codeforces E. Alyona and a tree(二分树上差分)

    题目描述: Alyona and a tree time limit per test 2 seconds memory limit per test 256 megabytes input stan ...

  9. Codeforces Round #395 C. Timofey and a tree

    package codeforces; import java.util.*; public class CodeForces_764C_Timofey_and_a_tree { static fin ...

随机推荐

  1. Linux监控-历史细项数据回溯

    Linux监控数据回溯 网络服务监控 应用场景: lvs 后端内网端机器网络波动监控: nginx 80.443端口连接监控: mysql 连接监控 以上为抛砖引玉,根据环境安装到监控工具(open ...

  2. Sublime text3的安装及python开发环境的搭建

    作者:struct_mooc 博客地址:https:////www.cnblogs.com/structmooc/p/12376592.html 一. Sublime text3的安装 1.subli ...

  3. redis教程-redis环境搭建安装(qq:1324981084)

    需要整套redis缓存高可用集群教学视频的加qq:1324981084,本套视频从安装到集群的搭建和源码的解析,从零基础讲解. 1.利用命令将redis下载到/usr/local/文件夹下: wget ...

  4. C#制作Wincc组件进行配方管理

    1,安装WinccV7.4并破解: 安装WinccV7.4SP1. 安装授权文件---根据提示 安装免狗驱动,根据提示 安装SImatic.net v13. 2,连接PLC, 首先在同一个局域网里面, ...

  5. Linux btrfs文件系统

    btrfs,它名字挺多:B-tree fs;Butter fs;Better fs 开源协议是GPL,2007年由Oracle研发 核心特性: 多物理卷支持,btrfs可由多个物理卷组成:支持RAID ...

  6. 关于css背景的一点总结

    background默认背景区域覆盖内容和内边距及边框,分别有以下属性: 1.background-clip(定义背景绘制区域) border-box 背景覆盖边框最外面 padding-box 背景 ...

  7. 【Java】实验代码整理(多线程、自定义异常、界面)

    1.界面+文件输入输出流 package finalExam; import java.awt.BorderLayout; import java.awt.Container; import java ...

  8. Unity比较常用的数据类型

    几种常见数据结构的使用情景 Array需要处理的元素数量确定并且需要使用下标时可以考虑,不过建议使用List<T> ArrayList不推荐使用,建议用List<T> List ...

  9. 修改 div 的滚动条的样式

    修改 div 的滚动条的样式 需要用到浏览器专属的伪元素,没有万能的办法,支持的浏览器不是很多. 假设有一个(你已经)设好宽高.定好位的 div, <div class="group- ...

  10. Bash脚本编程学习笔记04:测试命令test、状态返回值、位置参数和特殊变量

    我自己接触Linux主要是大学学习的Turbolinux --> 根据<鸟哥的Linux私房菜:基础篇>(第三版) --> 马哥的就业班课程.给我的感觉是这些课程对于bash的 ...