一次允许同时执行两个作业的作业调度

我们得到N个工作，以及他们的开始和结束时间。我们可以在特定时刻同时完成两项工作。如果一项工作在同一时间结束，则其他节目将开始，那么我们将无法完成它们。我们需要检查是否有可能完成所有作业。

例子：

Input :  Start and End times of Jobs
         1 2 
         2 3
         4 5 
Output : Yes
By the time third job starts, both jobs
are finished. So we can schedule third
job.

Input : Start and End times of Jobs
        1 5
        2 4
        2 6
        1 7
Output : No
All 4 jobs needs to be scheduled at time
3 which is not possible.

我们首先根据作业的开始时间对其进行排序。然后，我们同时启动两个作业，并检查第三个作业的开始时间等是否大于前两个作业的和的结束时间。
下面给出上述想法的实现。

C++

// CPP program to check if all jobs can be scheduled
// if two jobs are allowed at a time.
#include 
using namespace std;
  
bool checkJobs(int startin[], int endin[], int n)
{
    // making a pair of starting and ending time of job
    vector > a;
    for (int i = 0; i < n; i++)
        a.push_back(make_pair(startin[i], endin[i]));
  
    // sorting according to starting time of job
    sort(a.begin(), a.end());
  
    // starting first and second job simultaneously
    long long tv1 = a[0].second, tv2 = a[1].second;
  
    for (int i = 2; i < n; i++) {
          
        // Checking if starting time of next new job
        // is greater than ending time of currently
        // scheduled first job
        if (a[i].first >= tv1)
        {
            tv1 = tv2;
            tv2 = a[i].second;
        }
          
        // Checking if starting time of next new job
        // is greater than ending time of ocurrently
        // scheduled second job
        else if (a[i].first >= tv2)
            tv2 = a[i].second;
  
        else
            return false;
    }
    return true;
}
  
// Driver code
int main()
{
    int startin[] = { 1, 2, 4 }; // starting time of jobs
    int endin[] = { 2, 3, 5 }; // ending times of jobs
    int n = sizeof(startin) / sizeof(startin[0]);
    cout << checkJobs(startin, endin, n);
    return 0;
}

Python3

# Python3 program to check if all 
# jobs can be scheduled if two jobs 
# are allowed at a time. 
  
def checkJobs(startin, endin, n): 
  
    # making a pair of starting and
    # ending time of job 
    a = [] 
    for i in range(0, n):
        a.append([startin[i], endin[i]])
          
    # sorting according to starting
    # time of job 
    a.sort()
  
    # starting first and second job 
    # simultaneously 
    tv1 = a[0][1]
    tv2 = a[1][1]
  
    for i in range(2, n): 
          
        # Checking if starting time of next
        # new job is greater than ending time 
        # of currently scheduled first job 
        if (a[i][0] >= tv1) :
  
            tv1 = tv2 
            tv2 = a[i][1] 
  
        # Checking if starting time of next 
        # new job is greater than ending time 
        # of ocurrently scheduled second job 
        elif (a[i][0] >= tv2) :
            tv2 = a[i][1] 
  
        else:
            return 0
      
    return 1
  
# Driver Code 
if __name__ == '__main__': 
    startin = [1, 2, 4] # starting time of jobs 
    endin = [2, 3, 5] # ending times of jobs 
    n = 3
    print(checkJobs(startin, endin, n))
  
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

输出：

1

另一种解决方案是找到随时需要调度的最大作业数。如果此计数大于2，则返回false。否则返回true。