枚举遍历在PAT上超时了一个测试点

n = int(input())

# info = []

# for i in range(n):
#     t = []
#     for j in range(6):
#         t.append(input())
#     info.append(t)

info = [[input() for __ in range(6)] for _ in range(n)] # 这两种写法等价

m = int(input())

def query(seq,key):
    seq = int(seq)
    ans = []
    for i in range(n):
        if seq == 3:# 要特判一下seq等于3的情况
            keys = info[i][seq].split()
            for x in keys:
                if x == key:
                    ans.append(info[i][0])
                    break
        elif info[i][seq] == key:
            ans.append(info[i][0])

    if not ans:
        print("Not Found")
        return
    ans.sort()
    for x in ans:
        print(x)

for i in range(m):
    q = input()
    print(q)
    seq,key = q.split(':')
    key = key.strip()
    query(seq,key)

方法二：可以使用多个字典

from collections import defaultdict

book = defaultdict(set)
author = defaultdict(set)
key_word = defaultdict(set)
publisher = defaultdict(set)
year = defaultdict(set)

n = int(input())

for i in range(n):
    _id = input()
    _book = input()
    book[_book].add(_id)

    _author = input()
    author[_author].add(_id)

    _key_word = input().split()
    for item in _key_word:
        key_word[item].add(_id)

    _publisher = input()
    publisher[_publisher].add(_id)

    _year = input()
    year[_year].add(_id)

m = int(input())


def query(seq, key):
    seq = int(seq)
    ans = []

    if seq == 1:
        for x in book[key]:
            ans.append(x)
    elif seq == 2:
        for x in author[key]:
            ans.append(x)
    elif seq == 3:
        for x in key_word[key]:
            ans.append(x)
    elif seq == 4:
        for x in publisher[key]:
            ans.append(x)
    elif seq == 5:
        for x in year[key]:
            ans.append(x)

    if not ans:
        print("Not Found")
        return
    ans.sort()
    for x in ans:
        print(x)


for i in range(m):
    q = input()
    print(q)
    seq, key = q.split(':')
    key = key.strip()
    query(seq, key)