给定一个表示作业数量的整数N和一个矩阵range[] ，该矩阵由需要完成的每个作业的范围[开始日，结束日] 组成，任务是找到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：上述问题可以使用优先队列来解决。请按照以下步骤解决问题：
- 找出作业范围内的最小和最大天数。
- 按开始日的升序对所有作业进行排序。
- 从最小天到最大天迭代，每第i 天，选择当天可以完成的结束日期最少的作业。
- 为了执行上述步骤，维护一个Min Heap ，并在每个第i 天，将当天可以完成的作业插入到按结束日期排序的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。
- 对所有天重复此过程，最后打印已完成的作业计数。
下面是上述方法的一个实现：
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 是最大天数和最小天数之间的差值，N 是作业数。
辅助空间： O(N 2 )