HDU 1158 Employment Planning （DP)
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)
雇傭人：有三項費用：雇傭費（h），工資（w），解雇費（f）
給你n個月需要的人數，求最小的支出費用。。。。。
dp[i][j]:第i天雇傭j個人的費用，確定最大的天數maxn
#include
#include
#include
#include
#include
using namespace std;
int h,w,f;
int a[100000];
;int dp[15][100000];
int main()
{
int n,maxn,minn;
while(~scanf("%d",&n)&&n)
{
scanf("%d%d%d",&h,&w,&f);
int maxn=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(a[i]>maxn)
maxn=a[i];
}
for(int j=a[1];j<=maxn;j++)//第一天的狀態可以確定
dp[1][j]=(h+w)*j;
for(int i=2;i<=n;i++)
{
for(int j=a[i];j<=maxn;j++)//j天雇傭的人數
{
minn=INT_MAX;
for(int k=a[i-1];k<=maxn;k++)//j的前一天雇傭人數
{
if(j>=k)
dp[i][j]=(j-k)*h+j*w+dp[i-1][k];
else
dp[i][j]=(k-j)*f+j*w+dp[i-1][k];
if(dp[i][j]
