Employment Planning[HDU1158]

最新推荐文章于 2021-05-22 13:04:13 发布

转载最新推荐文章于 2021-05-22 13:04:13 发布 · 121 阅读

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原文链接：http://www.cnblogs.com/dramstadt/p/6207777.html

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#java

本文介绍了一个关于项目管理中就业计划的问题求解方法，通过动态规划算法确定每个月的员工雇佣和解雇数量以达到最低成本的目标。输入包含项目的月份、成本参数及每月所需最少员工数。

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Employment Planning

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5292 Accepted Submission(s): 2262

Problem Description
A project manager wants to determine the number of the workers needed in every month. He does know the minimal number of the workers needed in each month. When he hires or fires a worker, there will be some extra cost. Once a worker is hired, he will get the salary even if he is not working. The manager knows the costs of hiring a worker, firing a worker, and the salary of a worker. Then the manager will confront such a problem: how many workers he will hire or fire each month in order to keep the lowest total cost of the project.

Input
The input may contain several data sets. Each data set contains three lines. First line contains the months of the project planed to use which is no more than 12. The second line contains the cost of hiring a worker, the amount of the salary, the cost of firing a worker. The third line contains several numbers, which represent the minimal number of the workers needed each month. The input is terminated by line containing a single '0'.

Output
The output contains one line. The minimal total cost of the project.

Sample Input
3
4 5 6
10 9 11
0

Sample Output
199

#include<stdio.h>
#include<string.h>

const int INF=99999999;

int dp[15][10010];
int people[15];

int main(){

    //freopen("input.txt","r",stdin);

    int n;
    int hire,salary,fire;
    while(scanf("%d",&n) && n){
        scanf("%d%d%d",&hire,&salary,&fire);
        int max_people=0;
        int i,j,k;
        for(i=1;i<=n;i++){
            scanf("%d",&people[i]);
            if(max_people<people[i])
                max_people=people[i];
        }
        for(i=people[1];i<=max_people;i++)      //初始化第一个月
            dp[1][i]=i*salary+i*hire;
        int min;
        for(i=2;i<=n;i++){
            for(j=people[i];j<=max_people;j++){
                min=INF;        //有了这个前面就不需要用O（n^2）初始化dp了。
                for(k=people[i-1];k<=max_people;k++)
                    if(min>dp[i-1][k]+(j>=k?(j*salary+(j-k)*hire):(j*salary+(k-j)*fire)))
                        min=dp[i-1][k]+(j>=k?(j*salary+(j-k)*hire):(j*salary+(k-j)*fire));
                dp[i][j]=min;
            }
        }
        min=INF;
        for(i=people[n];i<=max_people;i++)
            if(min>dp[n][i])
                min=dp[n][i];
        printf("%d\n",min);
    }
    return 0;
}