Educational Codeforces R#65 (Div. 2) - C. News Distribution

問題概要

グループに所属する数を数えよ
Problem - C - Codeforces

解法

サイズ付きUnionFindを用いる。
が、(10^5) * 5はPythonだと相当きつい。
PyPy3を用いたが、Input関数を止めてsys.stdin.readlineで高速化しない無理だった。。。

参考

www.kumilog.net

提出コード

# coding: UTF-8
import sys

def find(x):
    if node[x] < 0:
        return x
    else:
        node[x] = find(node[x])
        return node[x]

def unite(x, y):
    x = find(x)
    y = find(y)
    ret = False
    if x != y:
        ret = True
        if rank[x] > rank[y]:
            node[x] += node[y]
            node[y] = x
        else:
            node[y] += node[x]
            node[x] = y
            if rank[x] == rank[y]:
                rank[y] += 1
    return ret

def is_same(x, y):
    return find(x) == find(y)

def size(x):
    return -node[find(x)]

readl= lambda: list(map(int, sys.stdin.readline().split()))
readt= lambda: tuple(map(int, sys.stdin.readline().split()))
read = lambda: sys.stdin.readline().rstrip()
readi = lambda: int(read())
readmi = lambda: map(int, sys.stdin.readline().split())
readms = lambda: map(str, sys.stdin.readline().split())

n, m = readmi()
node = [-1 for i in range(n + 1)]
rank = [0 for _ in range(n + 1)]
query = [readt() for _ in range(m)]
for a in query:
    if a[0] >= 2:
        for j in a[2:]:
            unite(a[1], j)
for i in range(1, n + 1):
    print(size(i), end=" ")