[code jam 2009试题分析]Qualification Round - Alien Language

题目如下： 
Problem 
After years of study, scientists at Google Labs have discovered an alien language transmitted from a faraway planet. The alien language is very unique in that every word consists of exactly L lowercase letters. Also, there are exactly D words in this language. 
Once the dictionary of all the words in the alien language was built, the next breakthrough was to discover that the aliens have been transmitting messages to Earth for the past decade. Unfortunately, these signals are weakened due to the distance between our two planets and some of the words may be misinterpreted. In order to help them decipher these messages, the scientists have asked you to devise an algorithm that will determine the number of possible interpretations for a given pattern. 
A pattern consists of exactly L tokens. Each token is either a single lowercase letter (the scientists are very sure that this is the letter) or a group of unique lowercase letters surrounded by parenthesis ( and ). For example: (ab)d(dc) means the first letter is either a or b, the second letter is definitely d and the last letter is either d or c. Therefore, the pattern (ab)d(dc) can stand for either one of these 4 possibilities: add, adc, bdd, bdc. 
Input 
The first line of input contains 3 integers, L, D and N separated by a space. D lines follow, each containing one word of length L. These are the words that are known to exist in the alien language. N test cases then follow, each on its own line and each consisting of a pattern as described above. You may assume that all known words provided are unique. 
Output 
For each test case, output 
Case #X: K 
where X is the test case number, starting from 1, and K indicates how many words in the alien language match the pattern. 

Limits 
Small dataset 
1 ≤ L ≤ 10 
1 ≤ D ≤ 25 
1 ≤ N ≤ 10 
Large dataset 
1 ≤ L ≤ 15 
1 ≤ D ≤ 5000 
1 ≤ N ≤ 500 
Sample 

Input 
3 5 4 
abc 
bca 
dac 
dbc 
cba 
(ab)(bc)(ca) 
abc 
(abc)(abc)(abc) 
(zyx)bc

Output 
Case #1: 2 
Case #2: 1 
Case #3: 3 
Case #4: 0 

这道题本身没有什么难度，唯一需要注意的就是速度了。 
我的解法是找到每一个位(token)对应的单词的集合，然后将所有的集合做交集运算，最终结果集合的元素数量就是所求的结果。 
为了快速找到每一个位所对应的单词集合，先便利单词的列表，为每一位建立一个从字母到单词集合的索引。 
Index是一个包含了L个map容器的序列，每一个map容器对应一个位。每个map容器包含了从字母x到该位置的字母为x的映射。 
这样从模式到单词可以简单的变换成集合操作。 
如模式(ab)b(ac)对应的就是 
wordSet = (index[0][a] | index[0][b]) & index[1][b] & (index[2][a] | index[2][c]) 
其中”|”是并集操作，”&”是交集操作。 
Len(wordSet)就是要求的结果。 
具体代码如下： 

import sysf = open(sys.argv[1])L, D, N = [int(i) for i in f.readline().split()]index = [{} for i in range(L)]for i in range(D): word = f.readline().strip() offset = 0 for l in word: if l not in index[offset]: index[offset][l] = set() index[offset][l].add(word) offset += 1for i in range(1, N+1): testcase = f.readline().strip() state = False offset = 0 result = None for l in testcase: if l == '(': state = True s = set() elif l == ')': state = False offset += 1 if result is None: result = s else: result &= s del s else: wordset = index[offset].get(l, set()) if state : s |= wordset else: if not wordset: #wordset is an empty set print("Case #%d: 0" % i) break if result is None: result = wordset.copy() else: result &= wordset if not result: print("Case #%d: 0" % i) break offset += 1 else: print("Case #%d: %d" % (i, len(result)))f.close()