查找加权作业计划中涉及的作业

给定N个工作，其中每个工作都由以下三个元素表示。

1.开始时间
2.完成时间
3.关联的利润或价值

找到与最大利润相关的工作子集，以使子集中没有两个工作重叠。

例子：

Input:  
Number of Jobs n = 4
Job Details {Start Time, Finish Time, Profit}
Job 1:  {1, 2, 50} 
Job 2:  {3, 5, 20}
Job 3:  {6, 19, 100}
Job 4:  {2, 100, 200}

Output: 
Jobs involved in maximum profit are
Job 1:  {1, 2, 50}
Job 4:  {2, 100, 200}

在上一篇文章中，我们讨论了加权作业调度问题。但是，该帖子仅涵盖与获取最大利润有关的代码。在这篇文章中，我们还将打印出获利最大的职位。

令arr [0..n-1]为Jobs的输入数组。我们定义一个数组DP []，以便DP [i]存储要获得最大利润的数组arr [0..i]所涉及的Jobs。即DP [i]存储子问题arr [0..i]的解。其余算法与以前的文章中讨论的相同。

以下是其C++实现–

// C++ program for weighted job scheduling using Dynamic
// Programming and Binary Search
#include 
using namespace std;
  
// A job has start time, finish time and profit.
struct Job
{
    int start, finish, profit;
};
  
// to store subset of jobs
struct weightedJob
{
    // vector of Jobs
    vector job;
  
    // find profit associated with included Jobs
    int value;
};
  
// A utility function that is used for sorting events
// according to finish time
bool jobComparator(Job s1, Job s2)
{
    return (s1.finish < s2.finish);
}
  
// A Binary Search based function to find the latest job
// (before current job) that doesn't conflict with current
// job. "index" is index of the current job. This function
// returns -1 if all jobs before index conflict with it. The
// array jobs[] is sorted in increasing order of finish time
int latestNonConflict(Job jobs[], int index)
{
    // Initialize 'lo' and 'hi' for Binary Search
    int lo = 0, hi = index - 1;
  
    // Perform binary Search iteratively
    while (lo <= hi)
    {
        int mid = (lo + hi) / 2;
        if (jobs[mid].finish <= jobs[index].start)
        {
            if (jobs[mid + 1].finish <= jobs[index].start)
                lo = mid + 1;
            else
                return mid;
        }
        else
            hi = mid - 1;
    }
  
    return -1;
}
  
// The main function that finds the subset of jobs
// associated with maximum profit such that no two
// jobs in the subset overlap.
int findMaxProfit(Job arr[], int n)
{
    // Sort jobs according to finish time
    sort(arr, arr + n, jobComparator);
  
    // Create an array to store solutions of subproblems.
    // DP[i] stores the Jobs involved and their total profit
    // till arr[i] (including arr[i])
    weightedJob DP[n];
  
    // initialize DP[0] to arr[0]
    DP[0].value = arr[0].profit;
    DP[0].job.push_back(arr[0]);
  
    // Fill entries in DP[] using recursive property
    for (int i = 1; i < n; i++)
    {
        // Find profit including the current job
        int inclProf = arr[i].profit;
        int l = latestNonConflict(arr, i);
        if (l != - 1)
            inclProf += DP[l].value;
  
        // Store maximum of including and excluding
        if (inclProf > DP[i - 1].value)
        {
            DP[i].value = inclProf;
  
            // including previous jobs and current job
            DP[i].job = DP[l].job;
            DP[i].job.push_back(arr[i]);
  
        }
        else
            // excluding the current job
            DP[i] = DP[i - 1];
    }
  
    // DP[n - 1] stores the result
    cout << "Optimal Jobs for maximum profits are\n" ;
    for (int i=0; i

输出：

Optimal Jobs for maximum profits are
(1, 2, 50)
(2, 100, 200)

Total Optimal profit is 250