HDU1158：Employment Planning（暴力DP）

Employment Planning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7097    Accepted Submission(s): 3034

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)

 

 

题意：

有多组数据，每组数据给出一个工程每月最少须要的工人数量，雇佣工人的花费，每月的工资，解雇工人的花费，求出最少花费的钱。

题解：

由题意最多十二个月能够无脑暴力DP（测试数据真的很水……），枚举出每月雇佣最少的人到最多的人的最少花费便可（与上个月的全部状况进行比较）。

状态转移为：

dp[i][f] = min(dp[i][f], dp[i - 1][h] + (f - h) * hire + f * salary);（上个月的人数少于等于need[i]的人数）

dp[i][f] = min(dp[i][f], dp[i - 1][h] + (h - f) * fire + f * salary);（上个月的人数多于need[i]的人数）

具体见代码；

#define _CRT_SECURE_NO_DepRECATE
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <iostream>
#include <cmath>
#include <iomanip>
#include <string>
#include <algorithm>
#include <bitset>
#include <cstdlib>
#include <cctype>
#include <iterator>
#include <vector>
#include <cstring>
#include <cassert>
#include <map>
#include <queue>
#include <set>
#include <stack>
#define ll long long
#define INF 0x3f3f3f3f
#define ld long double
const ld pi = acos(-1.0L), eps = 1e-8;
int qx[4] = { 0,0,1,-1 }, qy[4] = { 1,-1,0,0 }, qxx[2] = { 1,-1 }, qyy[2] = { 1,-1 };
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    int hire, salary, fire, need[20], dp[20][300], max, ans;
    while (cin >> n && n)
    {
        memset(dp, 0, sizeof(dp));
        max = 0;
        ans = INF;
        cin >> hire >> salary >> fire;
        for (int i = 0; i < n; i++)
        {
            cin >> need[i];
            max = ::max(need[i], max);//记录最多须要的人
        }
        for (int i = need[0]; i <= max; i++)//先初始化第一个月须要的钱，枚举从一月最少须要的人到最多须要的人
        {
            dp[0][i] = i * (hire + salary);
        }
        for (int i = 1; i < n; i++)
        {
            for (int f = need[i]; f <= max; f++)//枚举从最少须要的人到最多须要的人的状况花费的钱
            {
                dp[i][f] = INF;
                for (int h = need[i - 1]; h <= max; h++)//与上个月的每种状况进行对比
                {
                    if (h <= f)//人数少于或等于须要的人数
                    {
                        dp[i][f] = min(dp[i][f], dp[i - 1][h] + (f - h) * hire + f * salary);
                    }
                    else//人数多于须要的人数
                    {
                        dp[i][f] = min(dp[i][f], dp[i - 1][h] + (h - f) * fire + f * salary);
                    }
                }
            }
        }
        for (int i = need[n - 1]; i <= max; i++)//找到最小值
        {
            ans = min(ans, dp[n - 1][i]);
        }
        cout << ans << endl;
    }
    return 0;
}