hdu 5326（floyd递推人际关系过）

最新推荐文章于 2020-01-11 09:51:30 发布

mengxiang000000

最新推荐文章于 2020-01-11 09:51:30 发布

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分类专栏：最短路及其拓展文章标签： hdu 5326 杭电5326

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本博客探讨如何在给定公司组织关系的情况下，计算特定级别管理的员工人数，通过应用图论中的Floyd算法解决实际问题。

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Work

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1303 Accepted Submission(s): 791

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

Sample Output

题目大意：给你n-1个关系，表示a比b官大，就是说a管辖b。问，有多少个官管着k个员工。

第一次发现自己意淫出来点肥猪流解题算法好开森~~~~~~~~

题干明确给出全序关系，一方管另一方，是绝对不可能矛盾的东西，刷图论刷的多了，突然想起来这个题还能用floyd的递推关系思想来做~

思路很简单：如果j管i并且i管k，则能退出来j管着k。这是floyd的思想。

复杂度：N^3.一共100个点，最坏的打算100*100*100 ，1000MS，足够了，是不会超时的。

这里直接上AC代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int a[105][105];
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                a[i][j]=0x1f1f1f1f;
            }
        }
        for(int i=0;i<n-1;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            if(a[x][y]==0x1f1f1f1f)
            {
                a[x][y]=1;

            }
        }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                for(int k=1;k<=n;k++)
                {
                    if(a[j][i]==1&&a[i][k]==1)
                     a[j][k]=1;
                }//递推人际关系，如果j管i，i管k，那么j就能管k
        int output=0;
        for(int i=1;i<=n;i++)
        {
            int cont=0;
            for(int j=1;j<=n;j++)//遍历每一个人，如果a【i】【j】==1，说明i管j
            {
                if(a[i][j]==1)
                cont++;
            }
            if(cont==k)
            output++;
        }
        printf("%d\n",output);
    }
}