题解 | #[NOIP2017]图书管理员#

python题解 给出的测试用例包含重复数据 考虑了去重，存储了之前已经计算过的需求码

from collections import defaultdict

while True:
    try:
        n, q = map(int, input().split())
        book_fcode = [input() for _ in range(n)]
        require_code = defaultdict(int)
        require_list = []
        for _ in range(q):
            a, b = input().split()
            require_code[b] = -1
            require_list.append(b)
            
        for req_code in require_code.keys():
            fcode_list = []
            for fcode in book_fcode:
                if fcode[::-1].startswith(req_code[::-1]):
                    fcode_list.append(fcode)
            if len(fcode_list):
                require_code[req_code] = min([int(s) for s in fcode_list])
        
        for req_code in require_list:
            print(require_code[req_code])

    except:
        break