HDU1158Employment Planning题解动态规划DP

Employment Planning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1288    Accepted Submission(s): 481

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
   
   
   
   

 

 

Source

Asia 1997, Shanghai (Mainland China)

 

 

Recommend

Ignatius

状态：

设d[i][j]表示第i个月有j个雇员的最小花费

状态转移方程：

d[i][j]=min{d[i-1][k]+w(j,k)}       (month[i-1]<=k<=maxs)

if(j>k)w(j,k)=(j-k)*a+b*j//需要雇佣j-k个人

if(j<k)w(j,k)=(k-j)*c+b*j//需要裁员k-j个人

边界：

d[1][i]=(a+b)*i

others d[i][j]=INF

参考了 白杨 的代码

http://hi.baidu.com/ynegfly/blog/item/685cf1c89d89618bc91768f0.html

代码：

#include<cstdio> int n,d[15][1005],m[15]; int main() { int a,b,c,i,j,k,t,max,min,ans; while(scanf("%d",&n),n) { max=0; ans=1<<30; scanf("%d%d%d",&a,&b,&c); for(i=1;i<=n;i++) { scanf("%d",m+i); if(m[i]>max) max=m[i]; } for(i=1;i<=n;i++) for(j=m[i];j<=max;j++) d[i][j]=1<<30; for(i=m[1];i<=max;i++) d[1][i]=(a+b)*i; for(i=2;i<=n;i++) for(j=m[i];j<=max;j++) { min=1<<30; for(k=m[i-1];k<=max;k++) { if(k<j) t=(j-k)*a+b*j+d[i-1][k]; else t=(k-j)*c+b*j+d[i-1][k]; if(t<min) min=t; } d[i][j]=min; } for(i=m[n];i<=max;i++) if(d[n][i]<ans) ans=d[n][i]; printf("%d/n",ans); } }