一次允许两个作业的作业调度
原文:https://www . geesforgeks . org/job-scheduling-two-jobs-允许-time/
给我们 N 个工作，以及它们的开始和结束时间。在特定时刻，我们可以同时做两件事。如果一项工作在同一时刻结束，另一个节目开始，那么我们就做不到。我们需要检查是否有可能完成所有的工作。
示例:
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.
我们首先根据工作的开始时间对它们进行分类。然后我们同时开始两个作业，并检查第三个作业的开始时间是否大于前两个作业的和的结束时间。 以上思路的实现如下。
C++
// CPP program to check if all jobs can be scheduled
// if two jobs are allowed at a time.
#include <bits/stdc++.h>
using namespace std;
bool checkJobs(int startin[], int endin[], int n)
{
// making a pair of starting and ending time of job
vector<pair<int, int> > 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;
}
版权属于：月萌API www.moonapi.com，转载请注明出处
