在给定的约束下最大化可以完成的工作

给定一个表示作业数量的整数N ，以及一个矩阵range []，其中包含要完成的每个作业的范围[开始日期，结束日期] ，任务是查找可以完成的最大可能作业。
例子：

Input: N = 5, Ranges = {{1, 5}, {1, 5}, {1, 5}, {2, 3}, {2, 3}} 
Output: 5 
Explanation: Job 1 on day 1, Job 4 on day 2, Job 5 on day 3, Job 2 on day 4, Job 3 on day 5

Input: N=6, Ranges = {{1, 3}, {1, 3}, {2, 3}, {2, 3}, {1, 4}, {2, 5}} 
Output: 5 
 

方法：可以使用优先级队列来解决上述问题。请按照以下步骤解决问题：

• 在工作范围内找到最短和最长的一天。
• 按开始日期的升序对所有作业进行排序。
• 从最小日期到最大日期进行迭代，并在^第i天中选择可以在该天完成的具有最少结束日期的作业。
• 为了执行上述步骤，请维护一个Min Heap ，然后在^第i天，将当天可以完成的作业插入按结束日期排序的Min Heap中。如果有任何的工作可以在^第i^个一天内完成，考虑一个最低的结束日期，并增加完成作业的数量。
• 全天重复此过程，最后打印完成的作业数。

以下是上述方法的实现：

C++

// C++ Program to implement the
// above approach
 
#include 
using namespace std;
 
// Function to find maxiumum
// number of jobs
int find_maximum_jobs(
    int N,
    vector > ranges)
{
    // Min Heap
    priority_queue,
                   greater >
        queue;
 
    // Sort ranges by start day
    sort(ranges.begin(), ranges.end());
 
    // Stores the minimum and maximum
    // day in the ranges
    int min_day = ranges[0].first;
    int max_day = 0;
 
    for (int i = 0; i < N; i++)
        max_day
            = max(max_day,
                  ranges[i].second);
 
    int index = 0, count_jobs = 0;
 
    // Iterating from min_day to max_day
    for (int i = min_day; i <= max_day; i++) {
 
        // Insert the end day of the jobs
        // which can be completed on
        // i-th day in a priority queue
        while (index < ranges.size()
               && ranges[index].first <= i) {
 
            queue.push(ranges[index].second);
            index++;
        }
 
        // Pop all jobs whose end day
        // is less than current day
        while (!queue.empty()
               && queue.top() < i)
            queue.pop();
 
        // If queue is empty, no job
        // can be completed on
        // the i-th day
        if (queue.empty())
            continue;
 
        // Increment the count of
        // jobs completed
        count_jobs++;
 
        // Pop the job with
        // least end day
        queue.pop();
    }
 
    // Return the jobs
    // on the last day
    return count_jobs;
}
 
// Driver Code
int main()
{
 
    int N = 5;
    vector > ranges;
    ranges.push_back({ 1, 5 });
    ranges.push_back({ 1, 5 });
    ranges.push_back({ 1, 5 });
    ranges.push_back({ 2, 3 });
    ranges.push_back({ 2, 3 });
 
    cout << find_maximum_jobs(N, ranges);
}

Java

// Java Program to implement the
// above approach
import java.io.*;
import java.util.*;
 
class GFG {
 
  // Function to find maxiumum
  // number of jobs
  static int find_maximum_jobs(int N, ArrayList> ranges)
  {
    // Min Heap
    ArrayList queue = new ArrayList();
    // Sort ranges by start day
    Collections.sort(ranges, new Comparator>() {   
      @Override
      public int compare(ArrayList o1, ArrayList o2) {
        return o1.get(0).compareTo(o2.get(0));
      }              
    });
 
    // Stores the minimum and maximum
    // day in the ranges
    int min_day = ranges.get(0).get(0);
    int max_day = 0;
    for (int i = 0; i < N; i++)
      max_day = Math.max(max_day, ranges.get(i).get(1));
    int index = 0, count_jobs = 0;
 
    // Iterating from min_day to max_day
    for (int i = min_day; i <= max_day; i++)
    {
      // Insert the end day of the jobs
      // which can be completed on
      // i-th day in a priority queue
      while (index < ranges.size() && ranges.get(index).get(0) <= i)
      {
        queue.add(ranges.get(index).get(1));
        index++;
      }
      Collections.sort(queue);
 
      // Pop all jobs whose end day
      // is less than current day
      while (queue.size() > 0 && queue.get(0) < i)
        queue.remove(0);
 
      // If queue is empty, no job
      // can be completed on
      // the i-th day
      if (queue.size() == 0)
        continue;
      // Increment the count of
      // jobs completed
      count_jobs++;
 
      // Pop the job with
      // least end day
      queue.remove(0);
    }
    // Return the jobs
    // on the last day
    return count_jobs;
 
  }
 
  // Driver code
  public static void main (String[] args) {
 
    int N = 5;
    ArrayList> ranges = new ArrayList>();
    ranges.add(new ArrayList(Arrays.asList(1, 5)));
    ranges.add(new ArrayList(Arrays.asList(1, 5)));
    ranges.add(new ArrayList(Arrays.asList(1, 5)));
    ranges.add(new ArrayList(Arrays.asList(2, 3)));
    ranges.add(new ArrayList(Arrays.asList(2, 3)));
    System.out.println(find_maximum_jobs(N, ranges));
  }
}
// This code is contributed by avanitrachhadiya2155

Python3

# Python3 Program to implement the
# above approach
 
# Function to find maxiumum
# number of jobs
def find_maximum_jobs(N, ranges) :
 
    # Min Heap
    queue = []
     
    # Sort ranges by start day
    ranges.sort()
     
    # Stores the minimum and maximum
    # day in the ranges
    min_day = ranges[0][0]
    max_day = 0
    for i in range(N) :
      max_day = max(max_day, ranges[i][1])
    index, count_jobs = 0, 0
     
    # Iterating from min_day to max_day
    for i in range(min_day, max_day + 1) :
     
      # Insert the end day of the jobs
      # which can be completed on
      # i-th day in a priority queue
      while (index < len(ranges) and ranges[index][0] <= i) :
       
        queue.append(ranges[index][1])
        index += 1
       
      queue.sort()
     
      # Pop all jobs whose end day
      # is less than current day
      while (len(queue) > 0 and queue[0] < i) :
        queue.pop(0)
     
      # If queue is empty, no job
      # can be completed on
      # the i-th day
      if (len(queue) == 0) :
        continue
     
      # Increment the count of
      # jobs completed
      count_jobs += 1
     
      # Pop the job with
      # least end day
      queue.pop(0)
     
    # Return the jobs
    # on the last day
    return count_jobs
     
    # Driver code
N = 5
ranges = []
ranges.append((1, 5))
ranges.append((1, 5))
ranges.append((1, 5))
ranges.append((2, 3))
ranges.append((2, 3))
 
print(find_maximum_jobs(N, ranges))
 
# This code is contributed by divyeshrabadiya07.

C#

// C# Program to implement the
// above approach
using System;
using System.Collections.Generic;
class GFG {
 
  // Function to find maxiumum
  // number of jobs
  static int find_maximum_jobs(int N, List> ranges)
  {
    // Min Heap
    List queue = new List();
 
    // Sort ranges by start day
    ranges.Sort();
 
    // Stores the minimum and maximum
    // day in the ranges
    int min_day = ranges[0].Item1;
    int max_day = 0;
    for (int i = 0; i < N; i++)
      max_day = Math.Max(max_day, ranges[i].Item2);
    int index = 0, count_jobs = 0;
 
    // Iterating from min_day to max_day
    for (int i = min_day; i <= max_day; i++)
    {
 
      // Insert the end day of the jobs
      // which can be completed on
      // i-th day in a priority queue
      while (index < ranges.Count && ranges[index].Item1 <= i)
      {
        queue.Add(ranges[index].Item2);
        index++;
      }
      queue.Sort();
 
      // Pop all jobs whose end day
      // is less than current day
      while (queue.Count > 0 && queue[0] < i)
        queue.RemoveAt(0);
 
      // If queue is empty, no job
      // can be completed on
      // the i-th day
      if (queue.Count == 0)
        continue;
 
      // Increment the count of
      // jobs completed
      count_jobs++;
 
      // Pop the job with
      // least end day
      queue.RemoveAt(0);
    }
 
    // Return the jobs
    // on the last day
    return count_jobs;
  }
 
  // Driver code
  static void Main()
  {
    int N = 5;
    List> ranges = new List>();
    ranges.Add(new Tuple(1, 5));
    ranges.Add(new Tuple(1, 5));
    ranges.Add(new Tuple(1, 5));
    ranges.Add(new Tuple(2, 3));
    ranges.Add(new Tuple(2, 3));
 
    Console.Write(find_maximum_jobs(N, ranges));
  }
}
 
// This code is contributed by divyesh072019.

输出：

5

时间复杂度： O(Xlog(N))，其中X是最大和最小天数之差，N是作业数。
辅助空间： O(N ² )