我正在尝试使用链接队列来模拟打印机来管理打印作业。一个打印作业将按照提交的顺序处理，并且有 2 台打印机。每个打印作业都有许多页，并且会在用户选择的特定周期内占用打印机。
例如，如果有 3 个作业，作业 1（5 页）、作业 2（7 页）和作业 3（13 页），在 6 个周期（打印页）之后，一台打印机上就会有作业 2还剩 1 页，另一页上还有作业 3，还剩 12 页。
我能够处理队列中的 2 个作业，但我不知道如何处理第三个作业，因为我希望只有在打印机可用时才处理第三个作业。
【问题讨论】：
这是您所找到的最清晰的解决方案 - 三个类（PrintJob、JobQueue 和 Printer）和一个用于模拟的 main() 函数：
class PrintJob:
def __init__(self, name, pages):
self.name = name
self.pages = pages
self.remaining = pages
def tick(self):
if self.remaining:
self.remaining -= 1
print("Job {}: {} / {} pages remain".format(self.name, self.remaining, self.pages))
def is_done(self):
return self.remaining == 0
class JobQueue:
def __init__(self):
self.jobs = []
def enqueue(self, job):
self.jobs.append(job)
def dequeue(self):
return self.jobs.pop(0)
def is_empty(self):
return not self.jobs
class Printer:
def __init__(self, name, job_queue):
self.name = name
self.job_queue = job_queue
self.current_job = None
def tick(self):
# make sure we have a job to work on
if self.current_job is None:
if self.job_queue.is_empty():
print("Printer {} is idle".format(self.name))
return
else:
self.current_job = self.job_queue.dequeue()
print("Printer {} got job {}".format(self.name, self.current_job.name))
# print!
print("Printer {} is printing...".format(self.name))
self.current_job.tick()
# see if job is finished
if self.current_job.is_done():
print("Printer {} finished job {}".format(self.name, self.current_job.name))
self.current_job = None
def is_idle(self):
return self.current_job is None
def main():
# create job queue and two printers
jobs = JobQueue()
p1 = Printer(1, jobs)
p2 = Printer(2, jobs)
# add jobs
jobs.enqueue(PrintJob(1, 5))
jobs.enqueue(PrintJob(2, 7))
jobs.enqueue(PrintJob(3, 13))
# cycle until queue is empty
tick = 0
while not (jobs.is_empty() and p1.is_idle() and p2.is_idle()):
tick += 1
print("\n=== Tick {} ===".format(tick))
p1.tick()
p2.tick()
print("\nAll done! Time for a pint...")
if __name__ == "__main__":
main()
最后的结果是这样的
=== Tick 1 ===
Printer 1 got job 1
Printer 1 is printing...
Job 1: 4 / 5 pages remain
Printer 2 got job 2
Printer 2 is printing...
Job 2: 6 / 7 pages remain
=== Tick 2 ===
Printer 1 is printing...
Job 1: 3 / 5 pages remain
Printer 2 is printing...
Job 2: 5 / 7 pages remain
=== Tick 3 ===
Printer 1 is printing...
Job 1: 2 / 5 pages remain
Printer 2 is printing...
Job 2: 4 / 7 pages remain
=== Tick 4 ===
Printer 1 is printing...
Job 1: 1 / 5 pages remain
Printer 2 is printing...
Job 2: 3 / 7 pages remain
=== Tick 5 ===
Printer 1 is printing...
Job 1: 0 / 5 pages remain
Printer 1 finished job 1
Printer 2 is printing...
Job 2: 2 / 7 pages remain
=== Tick 6 ===
Printer 1 got job 3
Printer 1 is printing...
Job 3: 12 / 13 pages remain
Printer 2 is printing...
Job 2: 1 / 7 pages remain
=== Tick 7 ===
Printer 1 is printing...
Job 3: 11 / 13 pages remain
Printer 2 is printing...
Job 2: 0 / 7 pages remain
Printer 2 finished job 2
=== Tick 8 ===
Printer 1 is printing...
Job 3: 10 / 13 pages remain
Printer 2 is idle
=== Tick 9 ===
Printer 1 is printing...
Job 3: 9 / 13 pages remain
Printer 2 is idle
=== Tick 10 ===
Printer 1 is printing...
Job 3: 8 / 13 pages remain
Printer 2 is idle
=== Tick 11 ===
Printer 1 is printing...
Job 3: 7 / 13 pages remain
Printer 2 is idle
=== Tick 12 ===
Printer 1 is printing...
Job 3: 6 / 13 pages remain
Printer 2 is idle
=== Tick 13 ===
Printer 1 is printing...
Job 3: 5 / 13 pages remain
Printer 2 is idle
=== Tick 14 ===
Printer 1 is printing...
Job 3: 4 / 13 pages remain
Printer 2 is idle
=== Tick 15 ===
Printer 1 is printing...
Job 3: 3 / 13 pages remain
Printer 2 is idle
=== Tick 16 ===
Printer 1 is printing...
Job 3: 2 / 13 pages remain
Printer 2 is idle
=== Tick 17 ===
Printer 1 is printing...
Job 3: 1 / 13 pages remain
Printer 2 is idle
=== Tick 18 ===
Printer 1 is printing...
Job 3: 0 / 13 pages remain
Printer 1 finished job 3
Printer 2 is idle
All done! Time for a pint...
【讨论】：
这取决于。您将打印作业的优先级放在什么方面？完成打印作业所需的时间？队列中接收作业的顺序？
如果是后者，应该是这样的：
# printer is a class that contains a state 'idle' and a method 'print'
current_job = None
while ! job_queue.empty():
if current_job != job_queue.peak(0)
current_job = job_queue.pop(0)
if printer1.idle():
printer1.print_job(current_job)
elif printer2.idle():
printer2.print_job(current_job)
else:
# do nothing
pass
printer1.print()
printer2.print()
在此基础上，您还可以在打印机类中拥有另一个名为“pages_todo”的状态变量，它是一个整数，告诉您当前打印机对象还剩多少页才能将其空闲状态更改为 true。
class Printer:
def __init__(self):
idle = false
pages_todo = 0
def print_job(pages):
pages_todo += pages
def print():
pages_todo -= 1 if pages_todo > 0
【讨论】：
我的解决方案使用 Queue 实现
class Queue:
def __init__(self):
self.items = []
self.current_task = None
def isEmpty(self):
return self.items == []
def enqueue(self, item):
# ideally we should enqueue item by item
self.items = item[1]
self.current_task = item[0]
def dequeue(self):
return self.items.pop()
def size(self):
return len(self.items)
job1 = (1, list(range(5))) # range serves as our list of pages
job2 = (2, list(range(7)))
job3 = (3, list(range(13)))
list_of_jobs = [job3, job2, job1]
# Two printers
printer1 = Queue()
printer2 = Queue()
for _ in range(6):
if not printer1.isEmpty():
printer1.dequeue()
elif list_of_jobs:
printer1.enqueue(list_of_jobs.pop())
printer1.dequeue()
if not printer2.isEmpty():
printer2.dequeue()
elif list_of_jobs:
printer2.enqueue(list_of_jobs.pop())
printer2.dequeue()
输出
print('printer 1 is working on {} page(s) and is on job {} while printer 2 is working on {} page(s) and is on job {}'.format(printer1.size(), printer1.current_task, printer2.size(), printer2.current_task))
【讨论】：