hdu 1158 Employment Planning【dp】

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本文探讨了一个项目管理问题，即如何在不同月份内动态调整员工数量以达到项目成本最小化。通过使用动态规划算法，文章详细介绍了如何构建状态转移方程来解决这一问题，并提供了一个AC代码实例，帮助读者理解并实现解决方案。

Employment Planning

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4582 Accepted Submission(s): 1922

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

最开始觉得是一道贪心题，但是找了半天找不到贪心的点，所以转战其他算法，我们既然不能贪心，那么我们只能遍历所有情况，所以我首先想到了动态规划的方向；

设dp【i】【j】表示从第一个月一直到第i个月现在有j个员工的最少花费。

推导出动态规划转移方程：dp【i】【j】=min（dp【i】【j】，dp【i-1】【k】+fire/hire abs（j-k）+slary*j）

AC代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int dp[12][1000];//month number of compony now pay
int a[15];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(dp,0x3f3f3f3f,sizeof(dp));
        if(n==0)break;
        int hire,slary,fire;
        int maxn=0;
        scanf("%d%d%d",&hire,&slary,&fire);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            maxn=max(maxn,a[i]);
        }
        int output=0x3f3f3f3f;
        int j,k=0;
        for(int i=0;i<n;i++)
        {
            for(j=a[i];j<=maxn;j++)
            {
                if(i==0)
                dp[i][j]=j*hire+slary*j;
                else
                {
                    for(k=a[i-1];k<=maxn;k++)
                    {
                        int cha;
                        if(j>=k)//如果j大于k，说明这个月的员工要比上个月员工多，即hire
                        {
                            cha=j-k;
                            dp[i][j]=min(j*slary+cha*hire+dp[i-1][k],dp[i][j]);
                        }
                        if(j<k)//相反即fire
                        {
                            cha=k-j;
                            dp[i][j]=min(j*slary+cha*fire+dp[i-1][k],dp[i][j]);
                        }
                    }

                }
                if(i==n-1)
                {
                    output=min(output,dp[i][j]);
                }
            }
        }
        printf("%d\n",output);
    }
}