Work
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 618 Accepted Submission(s): 412
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
//这题是要求管理了m个人的有多少个
//利用pre[]数组 记录每个b的pre[b]为a; 接着从b到根的中的人掌控的个数+1 例如
3 2 1 2 2 3 pre[2]=1; c[1]++ pre[3]=2; c[2]++-> pre[2]=1 c[1]++; 所以掌握个数为2的有一个c[1]
#include <stdio.h>
#include <string.h>
int pre[110];
int c[110];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
memset(c,0,sizeof(c));
for(int i=0;i<=n;i++)
pre[i]=i;
int a,b;
for(int i=0;i<n-1;i++)
{
scanf("%d%d",&a,&b);
pre[b]=a;
while(pre[b]!=b) //从b到根 每个上级掌控人数加1；
{
b=pre[b];
c[b]++;
}
}
int cnt=0;
for(int i=1;i<=n;i++)
{
if(c[i]==m)
cnt++;
}
printf("%d\n",cnt);
}
return 0;
}
