电梯调度算法

在高峰时间，实习生小飞常常会被电梯每层楼都停弄得很不耐烦，于是他想出了这样一个办法：由于楼层并不高，那么在繁忙的时间，每次电梯从一层往上走时，我们只允许电梯停在其中的某一层。所有乘客都从一楼上电梯，到达某层楼后，电梯听下来，所有乘客再从这里爬楼梯到自己的目的层。在一楼时，每个乘客选择自己的目的层，电梯则自动计算出应停的楼层。
问：电梯停在哪一层楼，能够保证这次乘坐电梯的所有乘客爬楼梯的层数之和最少？

#! /usr/bin/python
# coding=utf-8
import random,math
from itertools import groupby

floor = 5
def 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 nTargetFloor


if __name__ == '__main__':
    main()