HDU 5326 Work

Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 731    Accepted Submission(s): 477

Problem Description

It’s an interesting experience to move from ICPC to work, end my college life and start a brand new journey in company.

As is known to all, every stuff in a company has a title, everyone except the boss has a direct leader, and all the relationship forms a tree. If A’s title is higher than B(A is the direct or indirect leader of B), we call it A manages B.

Now, give you the relation of a company, can you calculate how many people manage k people. 

Input

There are multiple test cases.
Each test case begins with two integers n and k, n indicates the number of stuff of the company.
Each of the following n-1 lines has two integers A and B, means A is the direct leader of B.

1 <= n <= 100 , 0 <= k < n
1 <= A, B <= n

Output

For each test case, output the answer as described above.

Sample Input

7 2 1 2 1 3 2 4 2 5 3 6 3 7

Sample Output

2

每次递归查找，查找下属的下属的下属。。。一直到没有下属位置，每次查找统计个数。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
struct node
{
    int num, s[110];
};
struct node res[110];

int ans;
void judge(int i)
{
    if(res[i].num == 0) return;

    ans += res[i].num;
    for(int j = 0; j < res[i].num; j++)
    {
        judge(res[i].s[j]);
    }
}

int main()
{
    int n, k, a, b;

    while(~ scanf("%d%d", &n, &k))
    {
        for(int i = 0; i <= n; i++)
            res[i].num = 0;

        for(int i = 0; i < n - 1; i++)
        {
            scanf("%d%d", &a, &b);
            res[a].s[res[a].num] = b;
            res[a].num++;
        }

        int cnt = 0;
        for(int i = 1; i <= n; i++)
        {
            ans = res[i].num;
            for(int j = 0; j < res[i].num; j++)
                judge(res[i].s[j]);

            if(ans == k) cnt++;
        }
        printf("%d\n", cnt);
    }

    return 0;
}