HDU 5326 Work(并查集变种)

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): 1543 Accepted Submission(s): 933

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

解题思路:并查集+暴力枚举。
其实简单点就是判断任意两个点是不是有领导与下属的关系。
下面是AC代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int pre[110];int num[110];int fin(int x,int y){    if(x!=pre[x])    {        if(pre[x]==y)        {            return 1;        }        else        {            return fin(pre[x],y);        }    }    if(x==pre[x])    {        return 0;    }}int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        int a,b;        for(int i=1; i<=n; i++)        {            pre[i]=i;        }        for(int i=0;i<n-1;i++)        {            scanf("%d%d",&a,&b);            pre[b]=a;        }        memset(num,0,sizeof(num));        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(i!=j)                if(fin(i,j))                {                    num[j]++;                }            }        }        int sum=0;        for(int i=1;i<=n;i++)        {            if(num[i]==m)            {                sum++;            }        }        printf("%d\n",sum);    }    return 0;}