HDU-5326 Work

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暑假集训同时被 2 个专栏收录

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本文探讨了如何通过算法解决公司内部员工管理关系的问题。利用并查集与递归遍历的方法，计算出每位管理者直接或间接管理的员工数量。

Work

这里写图片描述

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1694 Accepted Submission(s): 1012

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

题意：每个人只能由一个直属上司管辖，但一个上司可以有多个下属，给出关系数据和管辖人数，问有·多少人满足条件。
思路：一个关系得出一条指向上司的有向线段，每人都是一个点，用并查数组保存关系，再用一数组保存每个人的管辖人数。枚举每一个人，对于他的上司，管辖人数都加1。将所有人枚举完，所有人管辖人数就得到了。
失误：根据题意就是要找出一个根结点有多少子结点，但用并查集只能由下向上查找来记录一个子结点有多少父节点，没有想想出合适的方法去转化。
代码如下：

#include<stdio.h>

const int MAXN=1e6+10;
int fa[MAXN],num[MAXN];

void find2(__int64 x)//对于一个人，能管辖他的人是多少 
{
    __int64 r=x;
    while(r!=fa[r]) 
    {
        r=fa[r];
        ++num[r];//对于他的每个上司 
    }
}

int main()
{
    __int64 n,k,i,cnt,x,y;
    while(~scanf("%I64d %I64d",&n,&k))
    {
        for(i=1;i<=n;++i)
        {
            fa[i]=i;
            num[i]=0;
        }
        for(i=1;i<=n-1;++i)
        {
            scanf("%I64d %I64d",&x,&y);
            fa[y]=x;//无需合并 
        }
        for(i=1;i<=n;++i)//枚举所有人 
        {
            find2(i);  
        }
        cnt=0;
        for(i=1;i<=n;++i)//寻找满足条件的人数 
        {
            if(num[i]==k)  ++cnt;
        }
        printf("%I64d\n",cnt); 
    }
    return 0;
 }