hdu 5423 Rikka with Tree（dfs）

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:
 
For a tree T, let F(T,i) be the distance between vertice  and vertice i.(The length of each edge is ). 
 
Two trees A and B are similiar if and only if the have same number of vertices and for each i meet F(A,i)=F(B,i). 
 
Two trees A and B are different if and only if they have different numbers of vertices or there exist an number i which vertice i have different fathers in tree Aand tree B when vertice  is root.
 
Tree A is special if and only if there doesn't exist an tree B which A and B are different and A and B are similiar.
 
Now he wants to know if a tree is special.
 
It is too difficult for Rikka. Can you help her?

 

Input

There are no more than  testcases. 
 
For each testcase, the first line contains a number n(≤n≤).
 
Then n− lines follow. Each line contains two numbers u,v(≤u,v≤n) , which means there is an edge between u and v.

 

Output

For each testcase, if the tree is special print "YES" , otherwise print "NO".

 

Sample Input

 

Sample Output

YES
NO

Hint

For the second testcase, this tree is similiar with the given tree:

                                                                                       4

                                                                                       1 2

                                                                                       1 4

                                                                                       3 4

 

Source

BestCoder Round #53 (div.2)

 

根据题意可以构造出来的特殊树有三种情况（全是以结点1为根节点），，第一种是一条直线，第二种是一条直线末尾开花，第三种是直接开花。如图所示。然后就可以知道对于每一个深度的点数为1,1,1，...，x，0，0，...。然后dfs一遍找一下就可以了。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<vector>
 #include<set>
 #include<map>
 #include<queue>
 #include<algorithm>
 using namespace std;
 #define N 1006
 vector<int>g[N];
 vector<int>deep[N];
 int vis[N];
 void dfs(int st,int d){
     vis[st]=;
     deep[d].push_back(st);
     for(int i=;i<g[st].size();i++){
         int u=g[st][i];
         if(!vis[u]){
             dfs(u,d+);
         }
     }
 }
 int main()
 {
     int n;
     while(scanf("%d",&n)==){
 
         for(int i=;i<=n;i++) {
             g[i].clear();
             deep[i].clear();
         }
         memset(vis,,sizeof(vis));
 
         for(int i=;i<n;i++){
             int u,v;
             scanf("%d%d",&u,&v);
             g[u].push_back(v);
             g[v].push_back(u);
         }
         dfs(,);
 
         int t=;
         while(deep[t].size()==) t++;
 
         int num=deep[t].size();
         if(num+t!=n) {
             printf("NO\n");
         }else{
             printf("YES\n");
         }
 
     }
     return ;
 }