- Give each element a component representative.
- Merge representatives when a connection is accepted.
- Answer connectivity or component queries from the compressed representatives.
Code notes
- 122 lines of Python from the credited upstream file abc157_d.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 UnionFind(object):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, k = map(int, input().split())82 uf = UnionFind(n)83 friends = [0 for _ in range(n)]84 blocks = [[] for _ in range(n)]85 86 for i in range(m):87 ai, bi = map(int, input().split())88 ai -= 189 bi -= 190 friends[ai] += 191 friends[bi] += 192 93 uf.merge_if_needs(ai, bi)94 95 for i in range(k):96 ci, di = map(int, input().split())97 ci -= 198 di -= 199 100 blocks[ci].append(di)101 blocks[di].append(ci)102 103 ans = [0 for _ in range(n)]104 105 for i in range(n):106 size = uf.get_group_size(i)107 size -= friends[i]108 size -= 1109 110 ans[i] = size111 112 for i in range(n):113 for block in blocks[i]:114 if uf.is_same_group(i, block):115 ans[i] -= 1116 117 print(" ".join(map(str, ans)))118 119 120if __name__ == "__main__":121 main()122