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传送门：http://acm.hdu.edu.cn/showproblem.php?pid=5326
Work
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 749 Accepted Submission(s): 490
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
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
Author
ZSTU
Source
2015 Multi-University Training Contest 3
题意：
公司有n个人，给出n-1对领导关系，问有多少人领导k个人（包括下属的下属）。
数据规模很小，dfs暴搜。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
using namespace std;
#define eps 1e-8
#define LL long long
#define db double
#define maxn 10010
int n,k;
int vis[110];
vector <int> p[110];
void dfs(int u,int &t){int len=p[u].size();for(int i=0;i<len;i++){int v=p[u][i];if(vis[v]==0){vis[u]=1;++t;dfs(v,t);}}
}
int main(){while(~scanf("%d%d",&n,&k)){for(int i=1;i<=n;i++)p[i].clear();for(int i=1;i<n;i++){int a,b;scanf("%d%d",&a,&b);p[a].push_back(b);}int ans=0;for(int i=1;i<=n;i++){memset(vis,0,sizeof vis);int t=0;vis[i]=1;dfs(i,t);if(t==k)ans++;}printf("%d\n",ans);}return 0;
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