K - Work 分类： 比赛 2015-07-29 19:13 3人阅读 评论(0) 收藏

K - Work

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 5326

Appoint description:  System Crawler  (2015-07-28)

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 
题意就是，有n个人，判断有几个人领导的人数是k，
输入两个数，前面的领导后面的；
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <string>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>

using namespace std;

const int MAX = 110;

int Map[MAX][MAX];

int Dp[MAX];

int n,k;

int DFS(int s)//DFS搜索每个人所领导的人数；
{
    if(Dp[s])
    {
        return Dp[s];
    }
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        if(Map[s][i])
        {
            sum++;
            sum+=DFS(i);
        }
    }
    Dp[s]=sum;
    return Dp[s];
}

int main()
{
    int a,b;
    while(~scanf("%d %d",&n,&k))
    {
        memset(Map,0,sizeof(Map));
        memset(Dp,0,sizeof(Dp));
        for(int i=1;i<n;i++)
        {
            scanf("%d %d",&a,&b);
            Map[a][b]=1;
        }
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            if(DFS(i)==k)
            {
                ans++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。