谷歌Codejam奶昔例子：找不到錯誤

Question

我在練這些問題： http://code.google.com/codejam/contest/32016/dashboard#s=p1&a=1

我相信我有正確的解決方案，（即使在輸入和輸出的人工檢查）; 我仍然收到Incorrect Output

任何人都可以指出我哪裏可能是錯的？

def process_file(file): 
    fsock = open(file) 
    text = fsock.read() 
    fsock.close() 
    lines = text.split('\n') 
    return lines 


def process_lines(lines): 
    cur = 1 
    ans = [] 
    while cur < len(lines) - 1: 
     N = int(lines[cur]) 
     cur += 1 
     M = int(lines[cur]) 
     cur += 1 
     struct = [] 
     for i in range(0, M): 
      cust_pref = [int(n) for n in lines[cur].split(' ')] 
      cust_drinks = [a-1 for a in cust_pref[1::2]] 
      cust_drinks_malt_pref = cust_pref[2::2] 
      cust_choice = [(cust_drinks[i], cust_drinks_malt_pref[i]) for i in range(0, len(cust_drinks))] 
      cur += 1 
      struct.append(cust_choice) 
     ans.append((N, struct)) 
    return ans 


def process_case(case): 
    milkshake_menu = [0] * case[0] # our default menu 
    i = 0 
    impossible = False 
    while i < len(case[1]): # i represents the customer number, case[1] represents customers 
     acceptable = False 
     customer = case[1][i] 
     for drink_preferred in customer: 
      if milkshake_menu[drink_preferred[0]] == drink_preferred[1]: 
       acceptable = True 
       i += 1 # ok, next customer 
       break 
     if not acceptable: 
      for drink_preferred in customer: 
       # find a malted preference 
       if drink_preferred[1] == 1 and milkshake_menu[drink_preferred[0]] != 1: 
        # he needs a malted one 
        milkshake_menu[drink_preferred[0]] = 1 
        # but then we have to test previous customers, reset i 
        i = 0 
        break 
       #if you have come here, the customer does not have a malted preference, or has a unmalted preference that conflicts with other customer 
       impossible = True 
      #impossible is True, break outer loop 
      if impossible: 
       break 
    if impossible: 
     return "IMPOSSIBLE" 
    else: 
     return " ".join([str(n) for n in milkshake_menu]) 



if __name__ == "__main__": 
    import sys 
    filename = sys.argv[1] 
    lines = process_file(filename) 
    inp = process_lines(lines) 
    for k, v in enumerate(inp): 
     a = process_case(v) 
     print "Case #%d: %s" % (k + 1, a) 

引擎收錄輸出：http://pastebin.com/uXJQKbSr

Answer 1

一兩件事，看起來很奇怪是在循環

 for drink_preferred in customer: 
      # find a malted preference 
      if drink_preferred[1] == 1 and milkshake_menu[drink_preferred[0]] != 1: 
       # he needs a malted one 
       milkshake_menu[drink_preferred[0]] = 1 
       # but then we have to test previous customers, reset i 
       i = 0 
       break 
      impossible = True 

我希望這是

 for drink_preferred in customer: 
      # find a malted preference 
      if drink_preferred[1] == 1 and milkshake_menu[drink_preferred[0]] != 1: 
       # he needs a malted one 
       milkshake_menu[drink_preferred[0]] = 1 
       # but then we have to test previous customers, reset i 
       i = 0 
       break 
     else: 
      impossible = True 

我想此刻的你在檢查所有可能性之前將判斷任務是不可能的。