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 大意： 给你n-1个关系，a直接领导b。问这些人中有多少人可以领导K个人。领导包括直接领导和间接领导。思路： 用数组num[]记录每个结点领导的人数，令i从1~n表示结点，若i不是父结点，让num[fa[i]]+1,表示i上面的结点领导的人数加1，一直加到父结点，到i=n时sun[1~n]表示1~n个结点领导的人数，再找出人数等于k的。
1 #include<cstdio>
2 #include<string.h>
3 #include<algorithm>
4 using namespace std;
5 int n,k,i,ans,fa[110],a,b,num[110];
6 void f1(int a) //若a不是父结点，让a上面所有结点都加1；表示领导的人数
7 {
8 while(a != fa[a])
9 {
10 num[fa[a]]++;
11 a=fa[a];
12 }
13 }
14 int main()
15 {
16 while(scanf("%d %d",&n,&k)!=EOF)
17 {
18 memset(num,0,sizeof(num));
19 ans=0;
20 for(i = 1 ; i <= n ; i++)
21 {
22 fa[i]=i;
23 }
24 for(i = 1 ; i < n ; i++)
25 {
26 scanf("%d %d",&a,&b);
27 fa[b]=a;
28 }
29 for(i = 1 ; i <= n ; i++)
30 {
31 f1(i);
32 }
33 for(i = 1; i<=n;i++)
34 {
35 if(num[i] == k) //找出领导的人数为k的结点
36 {
37 ans++;
38 }
39 }
40 printf("%d\n",ans);
41 }
42 }