34 lines
827 B
Python
34 lines
827 B
Python
"""
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Input: Start and finish time of n jobs
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Output: Schedule with maximum number of non overlapping jobs
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The Strategy: At each step choose the job with earliest finish time
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Algorithm Type: Greedy
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Time Complexity: O(n*log(n))
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"""
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def get_opt_schedule(jobs): # Returns the job_id's in the optimal schedule
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opt_schedule = []
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sorted_jobs = sorted(jobs, key=lambda j: j[2])
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n = len(sorted_jobs)
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opt_schedule.append(sorted_jobs[0][0])
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for i in range(1, n):
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if sorted_jobs[i][1] >= jobs[opt_schedule[-1]][2]:
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opt_schedule.append(sorted_jobs[i][0])
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return opt_schedule
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jobs = [
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[0, 2, 8], # [job_id, start_time, finish_time]
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[1, 6, 10],
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[2, 1, 3],
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[3, 4, 7],
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[4, 3, 6],
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[5, 1, 2],
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[6, 8, 10],
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[7, 10, 15],
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[8, 12, 16],
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[9, 14, 16]
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]
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opt_schedule = get_opt_schedule(jobs)
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print(opt_schedule)
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