杭电1158 DP

Employment Planning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5128    Accepted Submission(s): 2191

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 5 610 9 110

 

Sample Output

199

 

题意：老板知道雇佣一个工人的花费，工人月薪，辞去一个工人的花费。老板要做一项工程，每个月有最低工人人数要求，多余工人不打工但也拿月薪，让你帮老板求出完成这个工程的最低花费。

思路：用一个二维数组表示状态，dp[ j ][ i ]表示第 i 个月雇佣 j 个工人的最低花费。

三重循环遍历：

i :1——>m(最多需要月数)

j : a[i]（第 i 个月最低人数） ——>maxn（总的最多人数）

k:a[i-1] ——> maxn

当k>j时花费为dp[k][i-1]+j*s+(j-k)*h ；当k<j时花费为dp[k][i-1]+j*s+(k-j)*f。通过k循环遍历，求出每次dp[ j ][ i ]的最小值。

AC代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define INF 0xFFFFFFusing namespace std;int a[13],dp[10050][13];int main(){    //freopen("D://in.txt", "r", stdin);    int min1,m,h,s,f;    while(~scanf("%d",&m)&&m)    {        scanf("%d %d %d",&h,&s,&f);        int maxn=-1;        for(int i=1;i<=m;i++)        {            scanf("%d",&a[i]);            maxn=max(maxn,a[i]);        }        for(int j=a[1];j<=maxn;j++)            dp[j][1]=j*(h+s);        for(int i=2;i<=m;i++)        {            for(int j=a[i];j<=maxn;j++)            {                min1=INF;                for(int k=a[i-1];k<=maxn;k++)                    if(j>=k)                        min1=min(min1,dp[k][i-1]+j*s+(j-k)*h);                    else                        min1=min(min1,dp[k][i-1]+j*s+(k-j)*f);                dp[j][i]=min1;            }        }        min1=INF;        for(int i=a[m];i<=maxn;i++)            min1=min(min1,dp[i][m]);        printf("%d\n",min1);    }    return 0;}