電梯調度演算法

在高峰時間，實習生小飛常常會被電梯每層樓都停弄得很不耐煩，於是他想出了這樣一個辦法：由於樓層並不高，那麼在繁忙的時間，每次電梯從一層往上走時，我們只允許電梯停在其中的某一層。所有乘客都從一樓上電梯，到達某層樓後，電梯聽下來，所有乘客再從這裡爬樓梯到自己的目的層。在一樓時，每個乘客選擇自己的目的層，電梯則自動計算出應停的樓層。
問：電梯停在哪一層樓，能夠保證這次乘坐電梯的所有乘客爬樓梯的層數之和最少？

#! /usr/bin/python# coding=utf-8import random,mathfrom itertools import groupbyfloor = 5def main():    arr = [random.randint(1,floor) for i in range(random.randint(3,5))]    print arr    do1(arr)    do2(arr)def do1(arr):    print "-" * 20    data = [(i,sum([abs(item - i) for item in arr])) for i in range(1,floor+1)]    maxdata = min([item for index, item in data])    for index, item in data:        print "%s,%s %s" % (index,item,"*" if item == maxdata else "")def do2(arr):    print "-" * 20    arr = sorted(arr)    nPerson = [0] * (floor + 1)    for k,v in groupby(arr):        nPerson[k] = len(list(v))    #print nPerson    floors = sum([x - 1 for x in arr if x > 1])    n1,n2,n3 = 0, nPerson[1], sum(nPerson) - nPerson[1]    nTargetFloor = format_print(1,n1,n2,n3,floors,None)    for x in range(2,floor+1):        floors += n1 + n2 - n3        n1 += n2        n2 = nPerson[x]        n3 -= n2        nTargetFloor = format_print(x,n1,n2,n3,floors,nTargetFloor)        def format_print(x,n1,n2,n3,floors, nTargetFloor):    print x,n1,n2,n3,floors,    if n1+n2 >= n3 and (nTargetFloor == None or nTargetFloor == floors):        print "*"        return floors    print    return nTargetFloorif __name__ == '__main__':    main()