hdoj1158

Employment Planning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2356    Accepted Submission(s): 914

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 5 610 9 110

 

Sample Output

199

 

Source

Asia 1997, Shanghai (Mainland China)

 

Recommend

Ignatius

 用dp[i][j]表示前i个月末人数为j时的最小开支.则dp[i][j]=min{dp[i-1][k]+cost(i,j)},其中k表示第i-1个月的人数,cost(i,j)为第i个月的开支.当k<=j时,cost(i,j)=hire*(j-k)+j*salary;当k>j时,cost(i,j)=fire*(k-j)+j*salary

#include<iostream>using namespace std;int mon[13];int dp[13][10000];int main() {    int n;    while (cin >> n, n) {        int hire, sal, fire;        cin >> hire >> sal >> fire;        int max = -1;        for (int i = 0; i < n; i++) {            cin >> mon[i];            if (mon[i] > max)                max = mon[i];        }        for (int i = mon[0]; i <= max; i++)            dp[0][i] = hire * i + sal * i;        for (int i = 1; i < n; i++)            for (int j = mon[i]; j <= max; j++) {                int min = INT_MAX;                for (int k = mon[i - 1]; k <= max; k++) {                    if (j >= k && dp[i - 1][k] + hire * (j - k) + j * sal < min)                        min = dp[i - 1][k] + hire * (j - k) + j * sal;                    else if (j < k && dp[i - 1][k] + fire * (k - j) + j * sal < min)                        min = dp[i - 1][k] + fire * (k - j) + j * sal;                }                dp[i][j] = min;            }        int ans = INT_MAX;        for (int i = mon[n - 1]; i <= max; i++)            if (dp[n - 1][i] < ans)                ans = dp[n - 1][i];        cout << ans << endl;    }    return 0;}