hdu 3738 The Sweat Shop 单调栈加贪心思想-优快云博客

本文介绍了一种解决工人调度问题的算法，旨在通过合理的人员增减来最小化招聘、解雇及工资支出的成本。该算法利用单调不增栈优化求解过程，确保了在满足每日用工需求的同时，实现了总成本的最小化。

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The Sweat Shop

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 90 Accepted Submission(s): 28

Problem Description

Many people spend thier lives in sweat shop in China. The boss of one of the famous sweat shop, GXX, is considering about the employment plan of her company.
He already knows that in the next N days, how many worker are needed for this day's work, a[i]. In additional, he will spend x yuan on recruit a worker, y yuan on fire a worker. If a worker spend a day on work in his company, GXX must pay z yuan for the wages per day.
GXX has no work before the first day and he decide to fire all the workers after these N days. Now comes to the question, how much he will pay on the employment plan in minimum.

Input

The first line contains only one integer T, denoting the number of test cases.
For each test cases, the first line contains only one integer N, denoting the number of days. (1 <= N <= 100000)
The next line contains three integers, x, y and z, denoting the amount of spend on recruit, fire a worker, and the daily salary of a worker. (1 <= x,y,z <= 100000)
The third line contains N integers, a[i], denoting the minimum number of workers needed for the i-th day. (1 <= a[i] <= 100000)

Output

For each test cases, output a single number denoting the minimum amount GXX should spend on employment plan.

Sample Input

Sample Output

题意：需要一些工人来做n天工作，每一天需要的工人数量告诉了，招聘一个工人需要x元，开除一个工人需要y元，工人每天的工资是z元，n天工作做完后，需要开除所有的工人，问怎么安排才能使得所需的总费用最少。

思路: 单调不增栈中存的是满足i-1天的可能人数，相当于一个优化。第i天时，只用遍历栈中比a[i]小的和第一个比a[i]大的即可，不需要雇佣更多的人。

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
typedef __int64 ll;

using namespace std;

ll a[100009];
ll q[100009];
ll dp[100009];

int main()
{
    int T,n;
    ll x,y,z;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        scanf("%I64d%I64d%I64d",&x,&y,&z);

        for(int i=1;i<=n;i++)
            scanf("%I64d",&a[i]);

        int t=0;
        q[t]=0;
        a[0]=0;
        dp[0]=0;

        for(int i=1;i<=n;i++)
        {
            dp[i]=1ll<<60;
            while(t>=0 && a[i]>a[q[t]])
            {
                ll tmp=dp[q[t]]+a[q[t]]*z*(i-q[t]);
                tmp+=(a[i]-a[q[t]])*(x+z);
                dp[i]=min(dp[i],tmp);
                t--;
            }

            if(t>=0)//处理第一个比当前大的情况
            {
                ll tmp=dp[q[t]]+(a[q[t]]-a[i])*y+a[i]*(i-q[t])*z;
                dp[i]=min(dp[i],tmp);
            }
            q[++t]=i;
        }

        dp[n]+=a[n]*y;

        printf("%I64d\n",dp[n]);
    }

    return 0;
}