HDU-1158dp

Employment Planning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6073    Accepted Submission(s): 2627

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

题意：给n个项目，做一个项目一个月。雇佣一个人需要hire，工作一个月需要salary的钱，解雇一个人需要fire的钱。问做完n项目最少需要多少钱

思路：用dp[i][j]表示第i个项目j个人最少需要多少钱（我们可以从前面项目的人数推过来）

坑点：没说数据多大，尽量开到超时边缘

AC代码：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int INF=0x7f7f7f7f;
int dp[20][10000],num[20];
int main()
{
    int n,i,j,k;
    int hire,fire,salary;
    int Max,Min;
    while (cin>>n&&n)
        {
            cin>>hire>>salary>>fire;
            Max=-INF;
            memset(dp,0,sizeof(dp));
            memset(num,0,sizeof(num));
            for (i=1;i<=n;i++)
                {
                    cin>>num[i];
                    Max=max(Max,num[i]);            //找到最多需要雇佣多少人
                }
            for (i=1;i<=Max;i++)
                dp[1][i]=i*hire+i*salary;        //第一个项目雇佣人的费用 ，初始化 
            for (i=2;i<=n;i++)
                for (j=num[i];j<=Max;j++)
                    {
                        Min=INF;                    //找到完成从i-1项目到i项目最少的钱 
                        for (k=num[i-1];k<=Max;k++)
                            if (k<j)
                                Min=min(Min,dp[i-1][k]+(j-k)*hire+j*salary);
                            else Min=min(Min,dp[i-1][k]+(k-j)*fire+j*salary);
                        dp[i][j]=Min;
                    }
            Min=INF;
            for (i=num[n];i<=Max;i++)
                Min=min(Min,dp[n][i]);
            cout<<Min<<endl;
        }
    return 0;
}