HUD5423 Rikka with Tree（DFS）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5423

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 321    Accepted Submission(s): 162

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
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
.

 

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

此题满足条件的只有一种情况，即： 树从上到下的结点数依次为： 1， 1， 1， ，，，，x（x为任意）. 也即是说，只有最后一层的结点数才能大于 1 .

dfs 求出每层的结点数， 就能判断出答案。

#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
using namespace std;

bool ok;
vector<int> a[1010];
int num[1010];

void dfs(int u, int fa, int d)
{
    num[d]++;
    int len = a[u].size();
    for(int i=0; i<len; i++)
    {
        if(a[u][i]==fa) continue;
        dfs(a[u][i], u, d+1);
    }
}

int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        for(int i=0; i<1010; i++)
        a[i].clear();
        memset(num, 0, sizeof(num));
        for(int i=1; i<n; i++)
        {
            int x, y;
            scanf("%d%d", &x, &y);
            a[x].push_back(y);
            a[y].push_back(x);
        }
        ok = false;
        dfs(1, -1, 1);
        for(int i=1; i<1010; i++)
        {
            if(num[i-1]>1&&num[i])
            ok = true;
        }
        if(ok) printf("NO\n");
        else printf("YES\n");
    }
    return 0;
}