google code jam exercise——MilkShakes(1)

MilkShakes，这道题有些难度，自己水平太差，还没有找到好的方法来解决。对于small case，得到了正确的答案。先来看看题。

题目的大概意思是，给一些不同flavor(风味)牛奶(1,..N)，可以是malted(1)，也可以是unmalted(0)，但是每一种牛奶只能是malted或者unmalted；有很多顾客，每个顾客喜欢的不一样，也可能喜欢多种。要求给出各种牛奶是否malted，能够满足所有顾客都有自己喜欢的flavor，并且malted的数量尽可能少。具体内容参考http://code.google.com/codejam/contest/32016/dashboard#s=p1

最直接的方法就是产生0，1的序列，检查是否满足。为了使malted的数量尽可能少，所以产生0，1序列的顺序是里面的1从少到多。基于这个想法，有了下面的程序实现。

#!/usr/bin/python
#encoding:UTF-8
#Filename:MilkShakes.py

import sys

def checkSatisfy(customer,choice):
    for i in range(customer[0]):
        j = 2 * i + 1
        if choice[customer[j]-1]==customer[j+1]:
            return 1
    return 0

def binary2dec(binaryArray):
    var = 0
    for i in range(len(binaryArray)):
        if binaryArray[i]==1:
            var = var + (binaryArray[i]<<i)
    return var

def generateFlavorArray(binaryNum):
    #    binaryNum = 4
    initial = [];
    for i in range(binaryNum):
        initial.append(0)

    layer0 = [initial]
    layer1 = []
    record = {} 
    cnt = 0
    record[binary2dec(initial)] = cnt
    flavorArray = [initial]

    while(len(layer0)>0):
        for i in range(len(layer0)):
            tmp = layer0[i]
            for j in range(binaryNum):
                tmp1 = tmp[0:]
                if tmp1[j]==0:
                    tmp1[j] = 1
                    if binary2dec(tmp1) not in record:
                        layer1.append(tmp1)
                        cnt = cnt + 1
                        record[binary2dec(tmp1)] = cnt
                    #    print tmp1
                        flavorArray.append(tmp1)
        layer0 = layer1
        layer1 = []

    return flavorArray



inname = "input.txt"
outname = "output.txt"
if len(sys.argv)>1:
    inname = sys.argv[1]
    outname = inname.rstrip(".in")
    outname = outname + ".out"
fin = open(inname,"r")
fout = open(outname,"w")

line = fin.readline()

testCaseNum = int(line)

caseNum = 0

for caseNum in range(testCaseNum):
    line = fin.readline()
    flavorNum = int(line)
    line = fin.readline()
    customerNum = int(line)
#    lines = fin.readlines(customerNum)
    customers = []
#    for line in lines:
    for i in range(customerNum):
        line = fin.readline()
        line = line.rstrip("\n")
        customer = [int(val) for val in line.split()]
#        print customer
        customers.append(customer)
    customers.sort()

    
    answer = "Case #%d:" %(caseNum+1)

    flavorArray = generateFlavorArray(flavorNum)
    flag = 0
    for choice in flavorArray:
        customerCnt = 0
        for customer in customers:
            if checkSatisfy(customer,choice)!=1:
                break
            customerCnt = customerCnt + 1
        if customerCnt == customerNum:
            an = [" "+str(b) for b in choice]
            for s in an:
                answer = answer + s
            answer = answer + "\n"
            flag = 1
            break

    if flag==0:
        answer = answer + " IMPOSSIBLE\n"

    fout.write(answer)

fin.close()
fout.close()

这个程序测试small case的时候还可以通过，测试large case的时候占用内存太大。这种穷据搜索的方法不太合适，需要找一个更好的算法。

接下来会考虑贪心算法，或者其他的，希望能找到有效的方法。