题解 | #H25 数据分类处理#

方法1-使用列表构建输出

s1 = input().split()[1:]
s2 = list(map(str, sorted(set(map(int, input().split()[1:])))))
ret = []
for num in s2:
    tmp = []
    for i, sub in enumerate(s1):
        if num in sub:
            tmp.extend([str(i), sub])
    if tmp:
        ret.extend([num, str(len(tmp) // 2)] + tmp)
print(str(len(ret)) + " " + " ".join(ret))

方法2-直接构建字符串

I = input().split()[1:]
R = list(map(str, sorted(map(int, set(input().split()[1:])))))

out = ""  # 最终输出
total = 0  # 统计总计数
for r in R:
    s = ''
    count = 0  # 统计I中包含r的元素个数
    for i, v in enumerate(I):
        if r in v:
            s += str(i) + " " + v + " "
            total += 2
            count += 1
    if count:
        s = r + " " + str(count) + " " + s
        total += 2
        out += s
print((str(total) + " " + out).strip())

方法3-使用字典

I = input().split()[1:]
R = sorted(map(int, set(input().split()[1:])))

# 构建字典
dic = {}
for r in R:
    for idx, item in enumerate(I):
        if str(r) in item:
            if r not in dic:
                dic[r] = [(str(idx), item)]
            else:
                dic[r].append((str(idx), item))

# 计算total
total = len(dic) * 2 + sum(len(v) for v in dic.values()) * 2

# 构建输出列表
ret = []
for k, v in sorted(dic.items()):
    count = str(len(v))
    ret.extend([str(k), count] + [elem for pair in v for elem in pair])

# 打印结果
print(f'{total} ' + ' '.join(ret))