HDU

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 TT, let F(T,i)F(T,i) be the distance between vertice 1 and vertice ii.(The length of each edge is 1). 

Two trees AA and BB are similiar if and only if the have same number of vertices and for each ii meet F(A,i)=F(B,i)F(A,i)=F(B,i). 

Two trees AA and BB are different if and only if they have different numbers of vertices or there exist an number ii which vertice ii have different fathers in tree AAand tree BB when vertice 1 is root. 

Tree AA is special if and only if there doesn't exist an tree BB which AA and BB are different and AA and BB 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 100 testcases. 

For each testcase, the first line contains a number n(1≤n≤1000)n(1≤n≤1000). 

Then n−1n−1 lines follow. Each line contains two numbers u,v(1≤u,v≤n)u,v(1≤u,v≤n) , which means there is an edge between uu and vv.

Output

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

Sample Input

31 22 341 22 31 4

Sample Output

YESNO          

Hint

For the second testcase, this tree is similiar with the given tree:41 21 43 4        
 
中文题意：
问题描述
众所周知，萌萌哒六花不擅长数学，所以勇太给了她一些数学问题做练习，其中有一道是这样的：对于一棵树$T$,令$F(T,i)$为点1到点$i$的最短距离（边长是1）. 两棵树$A$和$B$是相似的当且仅当他们顶点数相同且对于任意的$i$都有$F(A,i)=F(B,i)$.两棵树$A$和$B$是不同的当且仅当他们定点数不同或者存在一个$i$使得以1号点为根的时候$i$在两棵树中的父亲不同。一棵树$A$是特殊的当且仅当不存在一棵和它不同的树和它相似。现在勇太想知道一棵树到底是不是特殊的。当然，这个问题对于萌萌哒六花来说实在是太难了，你可以帮帮她吗？
输入描述
数据组数不超过100组。每组数据的第一行一个整数$n(2\le n\le 1000)$。接下来$n-1$行。每行两个整数$u,v(1\le u,v\le n)$，代表给定树上的一条边。
输出描述
对于每一组数据，如果给定树是特殊的输出"YES"否则输出"NO"。
输入样例
31 22 341 22 31 4
输出样例
YESNO
Hint
对于第二组数据下面这棵树和它相似。41 21 43 4
思路：根据题意总共只有三种情况：

所我们只需要用深搜一层一层进行搜索即可。然后记录每个节点的深度即可。
代码：
#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;int n;vector<int>g[1005];vector<int>dis[1005];bool vis[1005];void dfs(int u, int d){    vis[u] = true;    dis[d].push_back(u);//判断此点属于的层数    for (int i = 0; i < g[u].size(); i++)    {        int v = g[u][i];        if (!vis[v])        {            //printf("##%d %d\n",v,d+1);            dfs(v, d + 1);//向与之相连的点进行搜索        }    }}int main(){    while (~scanf("%d", &n))    {        if (n == 1)        {            puts("YES");            continue;        }        for (int i = 0; i <= n; i++)g[i].clear(), dis[i].clear();//之前进行清空        memset(vis, false, sizeof vis);        for (int i = 1; i < n; i++)        {            int u, v;            scanf("%d %d", &u, &v);            g[u].push_back(v);//表示：v属于u，目的用于统计属于u的点的集合            g[v].push_back(u);//u属于v        }        bool ans = true;        dfs(1, 0);        int t = 0;        while ((int)dis[t].size() == 1)t++;//如果属于层数等于一的话的点数；        int num = dis[t].size();//不等一的点数        if (num + t != n)ans = false;//相加等于n的话证明是特殊树，否则不是。        if (ans)puts("YES");        else puts("NO");    }    return 0;}

用queue数组写：
#include<stdio.h>#include <iostream>#include<cstdio>#include<iostream>#include<algorithm>#include<math.h>#include<string.h>#include<map>#include<queue>#include<vector>#include<deque>#define ll long long#define inf 0x3f3f3f3f#define mem(a,b) memset(a,b,sizeof(a))using namespace std;int vis[1001],n;vector<int> p[1001];int dfs(){    queue<int>q;    q.push(1);    vis[1]=1;    int sum=1;    while(!q.empty())    {        int x=q.front();        q.pop();int temp=0;        for(int i=0; i<p[x].size(); i++)        {            int y=p[x][i];            if(vis[y]==0)            {                vis[y]=1;                q.push(y);                temp++;            }        }        sum+=temp;        if(temp>1)        {            break;        }    }    if(sum==n)        return 1;    else        return 0;}int main(){    //int n;    while(~scanf("%d",&n))    {        //mem(vis,0);        int u,v;mem(vis,0);        for(int i=0; i<=n; i++)        {            p[i].clear();        }        for(int i=0; i<n-1; i++)        {            scanf("%d%d",&u,&v);            p[u].push_back(v);            p[v].push_back(u);        }        int ans=dfs();        if(ans==1)            printf("YES\n");        else            printf("NO\n");    }}