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Problem solution · Python

ABC220 F — Distance Sums 2

ABC220 F — Distance Sums 2: a Python solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Breadth-first search
Source
KATO-Hiro AtCoder Solutions
Length
78 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Breadth-first search

For ABC220 F — Distance Sums 2, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 78 lines of Python from the credited upstream file abc220_f.py.
  • The implementation visibly relies on sequence storage, ordered lookup, work queue.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from KATO-Hiro AtCoder Solutions by KATO-Hiro and is used under the CC0-1.0 licence.

Open problem ↗Original source ↗
Full codeABC220 F — Distance Sums 2 · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def calc_depth(vertex_count: int, graph):    from collections import deque     PENDING = -1    depth = [PENDING for _ in range(vertex_count)]    parent = [PENDING for _ in range(vertex_count)]    depth[0], parent[0] = 0, 0    d = deque()    d.append(0)     while d:        vertex = d.popleft()         for g in graph[vertex]:            if depth[g] == PENDING:                depth[g] = depth[vertex] + 1                parent[g] = vertex                d.append(g)     return depth  def main():    import sys     input = sys.stdin.readline    sys.setrecursionlimit(10 ** 7)     n = int(input())    graph = [[] for _ in range(n)]        for _ in range(n - 1):        ai, bi = map(int, input().split())        ai -= 1        bi -= 1            graph[ai].append(bi)        graph[bi].append(ai)     # dist(1, j)を求める    depth = calc_depth(n, graph)    tree_size = [0] * n    ans = [0] * n    ans[0] = sum(depth)     # 部分木のサイズを求める    def dfs(vertex, parent=-1):        tree_size[vertex] = 1         for to in graph[vertex]:            if to == parent:                continue             tree_size[vertex] += dfs(to, vertex)                return tree_size[vertex]     dfs(0)     # 頂点1とそれ以外の頂点との差分の更新    def dfs2(vertex, parent=-1):        for to in graph[vertex]:            if to == parent:                continue             ans[to] = ans[vertex] - tree_size[to] + (n - tree_size[to])            dfs2(to, vertex)        dfs2(0)    print(*ans, sep='\n')  if __name__ == "__main__":    main()

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