国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁(yè) > 學(xué)院 > 開(kāi)發(fā)設(shè)計(jì) > 正文

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

2019-11-14 13:03:59
字體:
來(lái)源:轉(zhuǎn)載
供稿:網(wǎng)友

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 題意:給你一個(gè)n個(gè)節(jié)點(diǎn)n-1條邊的樹(shù)。每個(gè)節(jié)點(diǎn)有顏色ci。問(wèn)是否存在一個(gè)點(diǎn),將它看為根后,所有子樹(shù)的顏色相同(子樹(shù)和子樹(shù)之間可以不同) 題解:我們將顏色相同的并相連的點(diǎn)進(jìn)行縮點(diǎn)。最后就會(huì)形成這里寫(xiě)圖片描述類(lèi)似這樣的圖,如果有解的話就像圖上紅色圈內(nèi)樣,與其他都相連(則它的度為染色數(shù)-1.) 代碼:

#include <bits/stdc++.h>#define ll long longusing namespace std;const int N=1e6;int n,cor,u,v;vector<int>edge[N];struct node{int f,l;}p[N];int color[N];int vis[N];int dp[N];int next_color[N];void dfs(int po,int fa){ vis[po]=1; next_color[po]=cor; for(int i=0;i<edge[po].size();i++) { int v=edge[po][i]; if(v!=fa&&color[v]==color[po]) { dfs(v,po); } }}int main(){ cin>>n; cor=0; for(int i=1;i<n;i++) { cin>>u>>v; edge[u].push_back(v); edge[v].push_back(u); p[i].f=u; p[i].l=v; } for(int i=1;i<=n;i++) cin>>color[i]; for(int i=1;i<=n;i++) { if(!vis[i]) { cor++; dfs(i,-1); } } for(int i=1;i<n;i++) { u=p[i].f; v=p[i].l; if(next_color[u]!=next_color[v]) { dp[u]++; dp[v]++; } } for(int i=1;i<=n;i++) { if(dp[i]==cor-1) { cout<<"YES"<<endl; cout<<i<<endl; return 0; } } cout<<"NO"<<endl; return 0;}
發(fā)表評(píng)論 共有條評(píng)論
用戶名: 密碼:
驗(yàn)證碼: 匿名發(fā)表
主站蜘蛛池模板: 烟台市| 庐江县| 雷山县| 饶河县| 新龙县| 原平市| 罗源县| 荣成市| 东方市| 灵台县| 碌曲县| 河池市| 宝丰县| 禄劝| 盘山县| 峨边| 兴安县| 江源县| 马关县| 霸州市| 青铜峡市| 开原市| 怀化市| 乐亭县| 韶山市| 潼关县| 行唐县| 饶河县| 怀柔区| 河津市| 鄂伦春自治旗| 丹巴县| 健康| 连平县| 金川县| 渑池县| 黄平县| 巫山县| 长阳| 沈阳市| 万载县|