結果

問題 No.1719 Tree and Permutation
ユーザー zkouzkou
提出日時 2021-09-23 17:17:52
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 3,986 bytes
コンパイル時間 169 ms
コンパイル使用メモリ 81,820 KB
実行使用メモリ 108,320 KB
最終ジャッジ日時 2023-10-24 11:29:12
合計ジャッジ時間 3,424 ms
ジャッジサーバーID
(参考情報)
judge15 / judge12
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 WA -
testcase_01 WA -
testcase_02 WA -
testcase_03 WA -
testcase_04 WA -
testcase_05 WA -
testcase_06 WA -
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ソースコード

diff #

import sys
input = sys.stdin.buffer.readline
read = sys.stdin.buffer.read


class dsu:
    """Data structures and algorithms for disjoint set union problems.

    Given an undirected graph, it processes the following queries in O(alpha(n)) time (amortized).

    >   Edge addition

    >   Deciding whether given two vertices are in the same connected component

    Each connected component internally has a representative vertex.

    When two connected components are merged by edge addition, 
    one of the two representatives of these connected components becomes the representative of the new connected component.
    """

    __slots__ = ["n", "parent_or_size"]

    def __init__(self, n):
        """It creates an undirected graph with n vertices and 0 edges.

        Constraints
        -----------

        >   0 <= n <= 10 ** 8

        Complexity
        ----------

        >   O(n)
        """
        self.n = n
        self.parent_or_size = [-1] * n

    def merge(self, a, b):
        """It adds an edge (a, b).

        If the vertices a and b were in the same connected component, 
        it returns the representative of this connected component. 
        Otherwise, it returns the representative of the new connected component.

        Constraints
        -----------

        >   0 <= a < n 

        >   0 <= b < n

        Complexity
        ----------

        >   O(alpha(n)) amortized
        """
        # assert 0 <= a < self.n
        # assert 0 <= b < self.n
        x = self.leader(a)
        y = self.leader(b)
        if x == y:
            return x
        if self.parent_or_size[y] < self.parent_or_size[x]:
            x, y = y, x
        self.parent_or_size[x] += self.parent_or_size[y]
        self.parent_or_size[y] = x
        return x

    def same(self, a, b):
        """It returns whether the vertices a and b are in the same connected component.

        Constraints
        -----------

        >   0 <= a < n

        >   0 <= b < n

        Complexity
        ----------

        >   O(alpha(n)) amortized
        """
        # assert 0 <= a < self.n
        # assert 0 <= b < self.n
        return self.leader(a) == self.leader(b)

    def leader(self, a):
        """It returns the representative of the connected component that contains the vertex a.

        Constraints
        -----------

        >   0 <= a < n

        Complexity
        ----------

        >   O(alpha(n)) amortized
        """
        # assert 0 <= a < self.n
        path = []
        while self.parent_or_size[a] >= 0:
            path.append(a)
            a = self.parent_or_size[a]
        for child in path:
            self.parent_or_size[child] = a
        return a

    def size(self, a):
        """It returns the size of the connected component that contains the vertex a.

        Constraints
        -----------

        >   0 <= a < n

        Complexity
        ----------

        >   O(alpha(n)) amortized
        """
        # assert 0 <= a < self.n
        return -self.parent_or_size[self.leader(a)]

    def groups(self):
        """It divides the graph into connected components and returns the list of them.

        More precisely, it returns the list of the "list of the vertices in a connected component". 
        Both of the orders of the connected components and the vertices are undefined.

        Complexity
        ----------

        >   O(n)
        """
        result = [[] for _ in range(self.n)]
        for i in range(self.n):
            result[self.leader(i)].append(i)
        return [g for g in result if g]



T = int(input())
assert 1 <= T <= 10 ** 5
N_sum = 0
for _ in range(T):
    N = int(input())
    N_sum += N
    assert 3 <= N <= 10 ** 5
    uf = dsu(N)
    for _ in range(N - 1):
        u, v = map(int, input().split())
        u -= 1; v -= 1
        assert u != v
        assert 0 <= u < N
        assert 0 <= v < N
        uf.merge(u, v)
    assert len(uf.groups()) == 1

assert N_sum <= 5 * 10 ** 5
0