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Problem solution · Python

ABC428 E — Farthest Vertex

ABC428 E — Farthest Vertex: 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
55 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 ABC428 E — Farthest Vertex, 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

  • 55 lines of Python from the credited upstream file abc428_e.py.
  • The implementation visibly relies on sequence storage, ordered lookup, work queue.
  • No explicit loop blocks detected.

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 codeABC428 E — Farthest Vertex · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  from collections import deque  def calc_depth(vertex_count: int, graph, source: int):    PENDING = -1    depth = [PENDING for _ in range(vertex_count)]    depth[source] = 0    d = deque([source])     while d:        vertex = d.popleft()         for g in graph[vertex]:            if depth[g] == PENDING:                depth[g] = depth[vertex] + 1                d.append(g)     return depth  def main():    import sys     input = sys.stdin.readline     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)     # 木の直径の両端(s, t)を求める    # 同じ距離の場合は、(距離、頂点番号)の辞書順で比較    dist_0 = calc_depth(n, graph, 0)    s = max([(dist_0[i], i) for i in range(n)])[1]    dist_s = calc_depth(n, graph, s)    t = max([(dist_s[i], i) for i in range(n)])[1]    dist_t = calc_depth(n, graph, t)     for i in range(n):        ans = max((dist_s[i], s), (dist_t[i], t))[1] + 1        print(ans)  if __name__ == "__main__":    main()

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