hdu 5326

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;}