hdu1158 Employment Planning

思路：dp[i][j]表示前i个月最后一个月的总人数为j所花的最小费用，状态移动方程:dp[i][j] = min{dp[i-1][k]
+ cost[i][j]},其中cost[i][j]是第i月的花费,

1~当k<=j时，第i个月请了人所以cost[i][j] = j*salary + (j-k)*hire

2~当k>j时，第i个月炒了人，所以cost[i][j] = j*salary + (k-j)*fire

#include<stdio.h>
#include<string.h>

const int INF=99999999;

int dp[15][10010];
int people[15];

int main(){

//freopen("input.txt","r",stdin);

int n;
int hire,salary,fire;
while(scanf("%d",&n) && n){
scanf("%d%d%d",&hire,&salary,&fire);
int max_people=0;
int i,j,k;
for(i=1;i<=n;i++){
scanf("%d",&people[i]);
if(max_people<people[i])
max_people=people[i];
}
for(i=people[1];i<=max_people;i++)      //初始化第一个月
dp[1][i]=i*salary+i*hire;
int min;
for(i=2;i<=n;i++){
for(j=people[i];j<=max_people;j++){
min=INF;        //有了这个前面就不需要用O（n^2）初始化dp了。
for(k=people[i-1];k<=max_people;k++)
if(min>dp[i-1][k]+(j>=k?(j*salary+(j-k)*hire):(j*salary+(k-j)*fire)))
min=dp[i-1][k]+(j>=k?(j*salary+(j-k)*hire):(j*salary+(k-j)*fire));
dp[i][j]=min;
}
}
min=INF;
for(i=people
;i<=max_people;i++)
if(min>dp
[i])
min=dp
[i];
printf("%d\n",min);
}
return 0;
}

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