ABC259 D — Circumferences

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 is_on_circle(xi, yi, ri, x, y):77    if (x - xi) ** 2 + (y - yi) ** 2 == ri ** 2:78        return True79    else:80        return False81 82 83def is_hit(xi, yi, ri, xj, yj, rj):84    dist = abs(xi - xj) ** 2 + abs(yi - yj) ** 285 86    if (max(ri, rj) - min(ri, rj)) ** 2 <= dist <= (ri + rj) ** 2:87        return True88    else:89        return False90 91 92def main():93    from collections import deque94    import sys95 96    input = sys.stdin.readline97 98    n = int(input())99    sx, sy, tx, ty = map(int, input().split())100    xyr = [tuple(map(int, input().split())) for _ in range(n)]101    cs, ct = -1, -1 # s, tがどの円に属するか102    uf = UnionFind(n)103 104    # 始点・終点がどの円に属するか判定105    for i in range(n):106        xi, yi, ri = xyr[i]107 108        if is_on_circle(xi, yi, ri, sx, sy):109            cs = i110        if is_on_circle(xi, yi, ri, tx, ty):111            ct = i112    113    # 円同士が接するか判定114    for i in range(n):115        xi, yi, ri = xyr[i]116 117        for j in range(i + 1, n):118            xj, yj, rj = xyr[j]119 120            if is_hit(xi, yi, ri, xj, yj, rj):121                uf.merge_if_needs(i, j)122    123    if uf.is_same_group(cs, ct):124        print("Yes")125    else:126        print("No")127 128 129if __name__ == "__main__":130    main()131