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Employment Planning
阅读量:192 次
发布时间:2019-02-28

本文共 2357 字,大约阅读时间需要 7 分钟。

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

题目大意:有一个经理想雇佣一些工人,给出计划有的月数n,雇佣、解雇一个工人的花费和一个工人的工资(只要被雇佣,不干活也有工资),以及每个月需要的工人数,求计划的最小花费。

题目分析:

  1. 状态表示:f[i][j] //表示1-i个月,且当前工人数是j时的最小花费
  2. 状态计算:这个题可以划分出两种状态
    当这个月的人数比上个月的人数多时,f[i][j]=min(f[i][j],f[i-1][k]+(j-k)*hire+j*salary),即上个月的最小花费+雇佣了(j-k)人的钱+j个人的工资
    当这个月的人数比上个月的人数少时,f[i][j]=min(f[i][j],f[i-1][k]+(k-j)*fire+j*salary),即上个月的最小花费+解雇了(k-j)人的钱+j个人的工资
  3. 注:第一个月的花费需要单独计算,因为第1个月刚开始雇人,因此雇佣的人就为f[1][i]=(hire+salary)*i

代码如下:

#include 
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long longconst int N=15,M=1e4+5;using namespace std; int main(){ int n; while(cin>>n,n) { int hire,salary,fire,num[N]={ 0},f[N][M]={ 0}; memset(f,0x3f,sizeof f); //因为求最小值,所以f初始化为正无穷 cin>>hire>>salary>>fire; int maxn=0; for(int i=1;i<=n;i++) //输出每个月需要的人,并找出最大值 cin>>num[i],maxn=max(maxn,num[i]); for(int i=num[1];i<=maxn;i++) //预处理第一个月 f[1][i]=(hire+salary)*i; for(int i=2;i<=n;i++) for(int j=num[i];j<=maxn;j++) for(int k=num[i-1];k<=maxn;k++) { //因为雇佣人数要大于等于计划人数,所以从num[i]开始计算 if(j>k) f[i][j]=min(f[i][j],f[i-1][k]+(j-k)*hire+j*salary); else f[i][j]=min(f[i][j],f[i-1][k]+(k-j)*fire+j*salary); } int ans=1e9; for(int i=num[n];i<=maxn;i++) //答案为f[n][i]中的最小值 ans=min(ans,f[n][i]); cout<
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