- Give each element a component representative.
- Merge representatives when a connection is accepted.
- Answer connectivity or component queries from the compressed representatives.
Code notes
- 127 lines of Python from the credited upstream file abc328_f.py.
- The implementation visibly relies on sequence storage, ordered lookup.
- No explicit loop blocks detected.
Complexity
Account for every find and union operation; with path compression and ranked merging, the amortized cost is nearly constant per operation.
Check the problem constraints before deciding whether this complexity will pass.
Use this to learn the idea, then write your own version.
12 3 4class WeightedUnionFind: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 9 See:10 https:www.youtube.com/watch?v=zV3Ul2pA2Fw11 https:en.wikipedia.org/wiki/Disjoint-set_data_structure12 https:atcoder.jp/contests/abc120/submissions/444494213 https:qiita.com/drken/items/cce6fc5c579051e64fab14 """15 16 def __init__(self, number_count: int):17 """18 Args:19 number_count: The size of elements (greater than 2).20 """21 self.parent_numbers = [i for i in range(number_count)]22 self.rank = [0 for _ in range(number_count)]23 self.diff_weight = [0 for _ in range(number_count)]24 25 def find_root(self, number: int) -> int:26 """Follows the chain of parent pointers from number up the tree until27 it reaches a root element, whose parent is itself.28 Args:29 number: The trees id (0-index).30 Returns:31 The index of a root element.32 """33 if self.parent_numbers[number] == number:34 return number35 else:36 parent_number = self.parent_numbers[number]37 root = self.find_root(parent_number)38 self.diff_weight[number] += self.diff_weight[parent_number]39 self.parent_numbers[number] = root40 return root41 42 def calc_weight(self, number: int) -> int:43 """Calculate the weight of the node.44 Args:45 number: The trees id (0-index).46 Returns:47 The weight of the node.48 """49 self.find_root(number)50 return self.diff_weight[number]51 52 def is_same_group(self, number_x: int, number_y: int) -> bool:53 """Represents the roots of tree number_x and number_y are in the same54 group.55 Args:56 number_x: The trees x (0-index).57 number_y: The trees y (0-index).58 """59 return self.find_root(number_x) == self.find_root(number_y)60 61 def merge_if_needs(self, number_x: int, number_y: int, weight: int) -> bool:62 """Uses find_root to determine the roots of the tree number_x and63 number_y belong to. If the roots are distinct, the trees are64 combined by attaching the roots of one to the root of the other.65 Args:66 number_x: The trees x (0-index).67 number_y: The trees y (0-index).68 weight : Cost between nodes.69 """70 71 72 weight += self.calc_weight(number_x)73 weight -= self.calc_weight(number_y)74 root_x, root_y = self.find_root(number_x), self.find_root(number_y)75 76 if root_x == root_y:77 return False78 79 if self.rank[root_x] < self.rank[root_y]:80 root_x, root_y = root_y, root_x81 weight = -weight82 if self.rank[root_x] == self.rank[root_y]:83 self.rank[root_x] += 184 85 self.parent_numbers[root_y] = root_x86 self.diff_weight[root_y] = weight87 return True88 89 def calc_cost(self, from_x: int, to_y: int) -> int:90 """Calculate cost between nodes.91 Args:92 from_x: The trees x (0-index).93 to_y : The trees y (0-index).94 Returns:95 Cost between nodes.96 """97 return self.calc_weight(to_y) - self.calc_weight(from_x)98 99 100def main():101 import sys102 103 input = sys.stdin.readline104 105 n, q = map(int, input().split())106 uf = WeightedUnionFind(n)107 ans = list()108 109 for i in range(q):110 ai, bi, di = map(int, input().split())111 ai -= 1112 bi -= 1113 114 if not uf.is_same_group(ai, bi):115 uf.merge_if_needs(ai, bi, di)116 117 cost = uf.calc_cost(ai, bi)118 119 if cost == di:120 ans.append(i + 1)121 122 print(*ans)123 124 125if __name__ == "__main__":126 main()127