CCF Python题解(100分)201809-3 元素选择器

CCF Python题解(100分)201809-3 元素选择器

n, m = map(int, input().split())
textlist = []
for i in range(n):
    text = input()
    level = text.count('.') // 2
    tag = ""
    id = ""
    if len(text.split()) == 1:  # 标签
        tag = text
    else:
        tag, id = text.split()
    tag = tag.strip('.')
    pline = -1
    for j in range(i - 1, -1, -1):
        if textlist[j]['level'] == level - 1:
            pline = j + 1
            break
    textlist.append({'tag': tag, 'id': id, 'level': level, 'pline': pline, 'line': i + 1})


def f(pline, i, count):
    if pline - 1 < 0:
        return False
    text = textlist[pline - 1]
    pline = text['pline']

    if select[i].upper() == text['tag'].upper() or select[i] == text['id']:  # 匹配到
        count += 1
        if i == 0 and count == len(select):
            return True
        else:
            if f(pline, i - 1, count):
                return True
    else:
        if f(pline, i, count):
            return True
        else:
            return False


for i in range(m):
    select = input()
    if len(select.split()) == 1:  # 标签或id
        count = 0
        str1 = ""
        for text in textlist:
            if text['id'] == select or text['tag'].upper() == select.upper():
                count += 1
                str1 += " " + str(text['line'])
        print(count, end="")
        print(str1)
    else:  # 后代
        select = select.split()
        list = []

        j = len(select) - 1
        for text in textlist[::-1]:
            if text['tag'].upper() == select[j].upper() or text['id'] == select[j]:
                line = text['line']
                pline = text['pline']

                if f(pline, j - 1, 1):
                    list.append(line)
        print(len(list), ' '.join(map(str, list[::-1])))