Codeforces 763 A. Timofey and a tree （dfs）

Codeforces 763 A. Timofey and a tree （dfs）

Description

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 ≤ 10^5) — 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 ≤ 10^5), 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

41 22 33 41 2 1 1

Examples output

YES2

题意

给定一棵树，树中的每个节点都有其固定的颜色，问能否找到一个节点，使得与该节点相连的每个子树只包含一种颜色。

思路

仔细想想便可以知道，我们只需要找出与当前节点不同颜色的邻接点数目最大的那个点，然后对其相连的子树一一判断即可。

若其他子树每个只包含一种颜色则为 ​​YES​​​ ，否则为 ​​NO​​ 。

AC 代码

#include#define IO ios::sync_with_stdio(false);\ cin.tie(0);\ cout.tie(0);using namespace std;const int maxn = 1e5+10;typedef __int64 LL;struct node{ int to; int next;} edge[maxn<<1];int head[maxn],tot;int color[maxn];int n,now;bool flag = true;void init(){ memset(head,-1,sizeof(head)); tot = 0;}void addedge(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++;}void dfs(int x,int fa){ if(color[x]!=now) { flag = false; return; } for(int i=head[x]; i!=-1; i=edge[i].next) { int to = edge[i].to; if(to==fa) continue; dfs(to,x); if(!flag) return; }}int main(){ IO; init(); cin>>n; for(int i=1; i>u>>v; addedge(u,v); addedge(v,u); } for(int i=1; i<=n; i++) { cin>>color[i]; } int ansv = -1,maxv = -1; for(int i=1; i<=n; i++) { int tmp = 0; for(int j=head[i]; j!=-1; j=edge[j].next) { if(color[edge[j].to]!=color[i]) tmp++; } if(tmp>maxv) { maxv = tmp; ansv = i; } } for(int i=head[ansv]; i!=-1; i=edge[i].next) { now = color[edge[i].to]; dfs(edge[i].to,ansv); if(!flag) { cout<<"NO"<

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