HDU 1158 Employment Planning

最新推荐文章于 2020-07-11 20:50:23 发布

GentleH

最新推荐文章于 2020-07-11 20:50:23 发布

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本文链接：https://blog.youkuaiyun.com/acbron/article/details/8785225

项目经理面临确定项目每月所需员工数量的任务，已知最低需求并考虑到雇佣、解雇和工资成本，需优化成本，通过制定合理的雇佣和解雇策略来最小化整个项目的总成本。

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Employment Planning

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2754 Accepted Submission(s): 1084

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

4 5 6

10 9 11

Sample Output

199

View Code

 1 /*
 2   dp[i][j]表示第i个月雇佣了j个人花费的最少费用
 3   if(j>=k) dp[i][j] = min(dp[i][j],dp[i-1][k]+hire*(j-k)+j*sal);
 4   else dp[i][j] = min(dp[i][j],dp[i-1][k]+fire*(k-j)+j*sal);
 5   其实就是遍历一遍第i个月可以雇佣的人数，再遍历一遍第i-1个月可以雇佣的人数，
 6   若i-1个月时人多，则炒掉，否则则雇佣，因为第i组仅参考第i-1组的话，没必要
 7   留下不干活的人。这样遍历下去，所有情况就都考虑到了。
 8   
 9   15MS    280K
10   */
11 #include <iostream>
12 #include <cstdio>
13 #include <cstring>
14 #include <algorithm>
15 
16 using namespace std;
17 
18 int dp[15][1005];
19 int hire,sal,fire;
20 int month[15];
21 int n;
22 
23 int main()
24 {
25     while(~scanf("%d",&n) && n)
26     {
27         scanf("%d%d%d",&hire,&sal,&fire);
28         int Max = 0;
29         for(int i=1; i<=n; i++)
30         {
31             scanf("%d",&month[i]);
32             if(month[i] > Max)
33                 Max = month[i];
34         }
35         for(int i=month[1]; i<=Max; i++)
36             dp[1][i] = (hire + sal)*i;
37         for(int i=2; i<=n; i++)
38         {
39             for(int j=month[i]; j<=Max; j++)
40             {
41                 dp[i][j] = 0xfffffff;
42                 for(int k=month[i-1]; k<=Max; k++)
43                 {
44                     if(j >= k)
45                         dp[i][j] = min(dp[i][j],dp[i-1][k]+hire*(j-k)+j*sal);
46                     else
47                         dp[i][j] = min(dp[i][j],dp[i-1][k]+fire*(k-j)+j*sal);
48                 }
49             }
50         }
51         int ans = 0xfffffff;
52         for(int i=month[n]; i<=Max; i++)
53             if(ans > dp[n][i])
54                 ans = dp[n][i];
55         printf("%d\n",ans);
56     }
57     return 0;
58 }