模拟算法:打印任务 Queue来实现
队列(queue)是一种有次序的数据集合,其特征是新数据项的添加总发生在一端(通常称为“尾rear”端)而现存数据项的移除总发生在另一端(通常称为“首front”端)
问题:多人共享一台打印机,采取“先到先服务”的队列策略来执行打印任务
在这种设定下,一个首要的问题就是:
1、这种打印作业系统的容量有多大?
2、在能够接受的等待时间内,系统能容纳多少用户
3、以多高频率提交多少打印任务?
如何对问题建模?
1、首先对问题进行抽象,确定相关的对象和过程抛弃那些对问题实质没有关系的学生性别、
年龄、打印机型号、打印内容、纸张大小等等众多细节
class Queue:
def __init__(self):
self.items = []
def isEmpty(self):
return self.items == []
def enqueue(self, item):
# 队列首段加选项
self.items.insert(0, item)
def dequeue(self):
# 队列尾端出
return self.items.pop()
def size(self):
return len(self.items)
import random
class Printer:
def __init__(self, ppm):
self.pagerate = ppm
self.currentTask = None
self.timeRemaining = 0
def tick(self):
if self.currentTask is not None:
self.timeRemaining = self.timeRemaining - 1
if self.timeRemaining <= 0:
self.currentTask = None
def busy(self):
if self.currentTask != None:
return True
else:
return False
def startNext(self, newtask):
self.currentTask = newtask
self.timeRemaining = newtask.getPages() * 60 / self.pagerate
class Task:
def __init__(self, time):
self.timestamp = time
self.pages = random.randrange(1, 21)
def getStamp(self):
return self.timestamp
def getPages(self):
return self.pages
def waitTime(self, currenttime):
return currenttime - self.timestamp
def newPrintTask():
num = random.randrange(1, 181)
if num == 180:
return True
else:
return False
def simulation(numSeconds, pagesPerMinute):
labprinter = Printer(pagesPerMinute)
printQueue = Queue()
waitingtimes = []
for currentSecond in range(numSeconds):
if newPrintTask():
task = Task(currentSecond)
printQueue.enqueue(task)
if (not labprinter.busy()) and (not printQueue.isEmpty()):
nexttask = printQueue.dequeue()
waitingtimes.append(nexttask.waitTime(currentSecond))
labprinter.startNext(nexttask)
labprinter.tick()
averageWait = sum(waitingtimes) / len(waitingtimes)
print("average wait %6.2f secs %3d task remaining" % (averageWait, printQueue.size()))
调用
for i in range(10):
simulation(3600,5)
作业问题:饭馆的餐桌设置,使得顾客排队时间变短?
心有猛虎,细嗅蔷薇