hdu5326 Work 并查集

最新推荐文章于 2017-11-02 15:41:50 发布

kyoma

最新推荐文章于 2017-11-02 15:41:50 发布

阅读量378

点赞数

CC 4.0 BY-SA版权

分类专栏： ------------解题报告------------- 并查集

本文链接：https://blog.youkuaiyun.com/kyoma/article/details/52875025

------------解题报告------------- 同时被 2 个专栏收录

208 篇文章

订阅专栏

并查集

11 篇文章

订阅专栏

本文介绍了一道关于公司内部管理关系的问题，利用并查集数据结构进行求解。问题要求找出公司中直接或间接管理着特定数量员工的管理者数目。通过去除状态压缩，并新增一个用于记录每位管理者下属员工数目的数组，最终实现了有效的解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Work

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1781 Accepted Submission(s): 1065

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

Author

ZSTU

Source

2015 Multi-University Training Contest 3

题意：给你公司几个人的关系，表示a是b的直接上司，问你手下管理着k个人的人有多少个

这题想到并查集还是比较自然的。。。把并查集模版的状态压缩去掉，新开一个数组表示每个人手下有多少人。然后每次回溯找父节点的时候对相应的管理人数＋＋

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
#include <map>
#include <set>
using namespace std;
const int maxn = 150;
int save[maxn];
int manager[maxn];
int pre[maxn];
int find(int x){
    int r=x;
    while(pre[r]!=r){
        manager[pre[r]]++;
        r=pre[r];
        //cout<<"aaa";
    }
    /*if(pre[x]!=r){
        pre[x]=r;
    }*/
    return r;
}
void join(int a,int b){
    pre[b]=a;
}
int main(){
    int n,k;
    while(~scanf("%d%d",&n,&k)){
        int a,b,i;
        memset(manager, 0, sizeof(manager));
        for(i=1;i<=n;i++){
            pre[i]=i;
        }
        for(i=1;i<=n-1;i++){
            scanf("%d%d",&a,&b);
            join(a, b);
        }
        for(i=1;i<=n;i++){
            find(i);
        }
        int ans=0;
        for(i=1;i<=n;i++){
            if(manager[i]==k){
                ans++;
            }
        }
        printf("%d\n",ans);
    }
    
}