hdu 5423 统计树每一层的节点树 dfs

roblem 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 1 and vertice i.(The
length of each edge is 1).

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 A and
tree B when
vertice 1 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 100 testcases.

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

Then n−1 lines
follow. Each line contains two numbers u,v(1≤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

3
1 2
2 3
4
1 2
2 3
1 4

Sample Output

YES
NO

HintFor the second testcase, this tree is similiar with the given tree:
4
1 2
1 4
3 4    如果一个节点有兄弟节点还有儿子，那么把儿子节点挪到兄弟节点上则他就是有相似的树，所以，我们只有当每一个节点只有一个儿子，叶子节点除外，最后一层可以有多个节点则无这棵树相似的树，也就是 一条串 。-。-。-。-。-。最后一层可有多个叶子节点，
#include<bits/stdc++.h>
using namespace std;
int maxdepth = 0;
int m[1005][1005];
int d[1005];
int cnt[1005];
int n;
void dfs(int u ,int depth)
{
maxdepth = max(depth,maxdepth);
d[u] = depth;
for(int i = 1; i<= n; i++)
{
if(d[i]==-1&&m[u][i])
{
dfs(i,depth+1);
}
}
}
void Init()
{
memset(d,-1,sizeof(d));
memset(m,0,sizeof(m));
memset(cnt,0,sizeof(cnt));
maxdepth = 0;
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
Init();
int a,b;
for(int i = 1 ; i < n; i++)
{
scanf("%d%d",&a,&b);
m[a][b]=1;
m[b][a]=1;
}
dfs(1,1);
for(int i = 1; i<= n; i++)
{
cnt[d[i]]++;
}
int flag = 0;
for(int i = 1 ; i <maxdepth; i++)
{
if(cnt[i]>1)
{
flag = 1;
break;
}
}
if(!flag)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}