ABC264 E — Blackout 2

1# -*- coding: utf-8 -*-2 3 4class UnionFind:5    '''Represents a data structure that tracks a set of elements partitioned6       into a number of disjoint (non-overlapping) subsets.7    Landau notation: O(α(n)), where α(n) is the inverse Ackermann function.8    See:9    https://www.youtube.com/watch?v=zV3Ul2pA2Fw10    https://en.wikipedia.org/wiki/Disjoint-set_data_structure11    https://atcoder.jp/contests/abc120/submissions/444494212    '''13 14    def __init__(self, number_count: int):15        '''16        Args:17            number_count: The size of elements (greater than 2).18        '''19        self.parent_numbers = [-1 for _ in range(number_count)]20 21    def find_root(self, number: int) -> int:22        '''Follows the chain of parent pointers from number up the tree until23           it reaches a root element, whose parent is itself.24        Args:25            number: The trees id (0-index).26        Returns:27            The index of a root element.28        '''29        if self.parent_numbers[number] < 0:30            return number31 32        self.parent_numbers[number] = self.find_root(self.parent_numbers[number])33        return self.parent_numbers[number]34 35    def get_group_size(self, number: int) -> int:36        '''37        Args:38            number: The trees id (0-index).39        Returns:40            The size of group.41        '''42        return -self.parent_numbers[self.find_root(number)]43 44    def is_same_group(self, number_x: int, number_y: int) -> bool:45        '''Represents the roots of tree number_x and number_y are in the same46           group.47        Args:48            number_x: The trees x (0-index).49            number_y: The trees y (0-index).50        '''51        return self.find_root(number_x) == self.find_root(number_y)52 53    def merge_if_needs(self, number_x: int, number_y: int) -> bool:54        '''Uses find_root to determine the roots of the tree number_x and55           number_y belong to. If the roots are distinct, the trees are combined56           by attaching the roots of one to the root of the other.57        Args:58            number_x: The trees x (0-index).59            number_y: The trees y (0-index).60        '''61        x = self.find_root(number_x)62        y = self.find_root(number_y)63 64        if x == y:65            return False66 67        if self.get_group_size(x) >= self.get_group_size(y):68            self.parent_numbers[x] += self.parent_numbers[y]69            self.parent_numbers[y] = x70        else:71            self.parent_numbers[y] += self.parent_numbers[x]72            self.parent_numbers[x] = y73        return True74 75 76def main():77    import sys78 79    input = sys.stdin.readline80 81    n, m, e = map(int, input().split())82    edges = list()83    84    for _ in range(e):85        ai, bi = map(int, input().split())86        ai -= 187        bi -= 188 89        # 少なくとも一つの発電所につながっていればよい = 発電所を区別せず一つにまとめる90        # 問題文から、都市: [1, n - 1]、n: 発電所91        ai = min(ai, n)92        bi = min(bi, n)93    94        edges.append((ai, bi))95    96    # クエリ先読み97    q = int(input())98    x = [0] * q99    cut = [False] * e100 101    for i in range(q):102        xi = int(input())103        xi -= 1104 105        x[i] = xi106        cut[xi] = True107    108    # 電線が切れていない部分をまとめる109    uf = UnionFind(n + 1)110 111    for i in range(e):112        if not cut[i]:113            ui, vi = edges[i]114            uf.merge_if_needs(ui, vi)115    116    ans = [0] * q117 118    # 電線を切断 = クエリを逆順に連結119    for qi in range(q - 1, -1, -1):120        xi = x[qi]121        ui, vi = edges[xi]122 123        ans[qi] = uf.get_group_size(n) - 1  # 発電所を含む連結成分から、発電所を除外124        uf.merge_if_needs(ui, vi)125    126    print(*ans, sep="\n")127 128 129if __name__ == "__main__":130    main()131