- Give each element a component representative.
- Merge representatives when a connection is accepted.
- Answer connectivity or component queries from the compressed representatives.
Code notes
- 112 lines of Python from the credited upstream file abc287_c.py.
- The implementation visibly relies on hash lookup, 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: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 from collections import Counter78 import sys79 80 input = sys.stdin.readline81 82 n, m = map(int, input().split())83 uf = UnionFind(n)84 c = Counter()85 count = n86 87 for _ in range(m):88 ai, bi = map(int, input().split())89 ai -= 190 bi -= 191 92 c[ai] += 193 c[bi] += 194 95 if uf.is_same_group(ai, bi):96 print("No")97 exit()98 else:99 uf.merge_if_needs(ai, bi)100 count -= 1101 102 d = Counter(c.values())103 104 if count == 1 and (d[1] == 2 and d[2] == n - 2):105 print("Yes")106 else:107 print("No")108 109 110if __name__ == "__main__":111 main()112