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Work
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.
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
7 2 1 2 1 3 2 4 2 5 3 6 3 7
2
下面是我的代码：
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAX=1e3+10;
int par[MAX],num[MAX];void find(int m)
{if(par[m]!=m){num[par[m]]++;//记录 每个上级所带领的下属个数 find(par[m]);return ;}}void un(int x,int y)
{if(x!=y)par[y]=x;
}int main()
{int n,m,ans;while(scanf("%d%d",&n,&m)!=EOF){int a,b;memset(num,0,sizeof(num));for(int i=1;i<=n;i++)par[i]=i;for(int j=1;j<n;j++){scanf("%d%d",&a,&b);un(a,b);}for(int i=1;i<=n;i++)find(i);//寻找每个点的下属个数 ans=0;for(int i=1;i<=n;i++){if(num[i]==m)ans++;}printf("%d\n",ans); }return 0;
}