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Problem solution · Python

ABC394 E — Palindromic Shortest Path

ABC394 E — Palindromic Shortest Path: 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
67 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 ABC394 E — Palindromic Shortest Path, 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

  • 67 lines of Python from the credited upstream file abc394_e.py.
  • The implementation visibly relies on sequence storage, 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 codeABC394 E — Palindromic Shortest Path · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys    from collections import deque     input = sys.stdin.readline     n = int(input())    c = [list(input().rstrip()) for _ in range(n)]    inf = 10**9    dist = [[inf] * n for _ in range(n)]    q = deque()     def push(i, j, d):        if dist[i][j] != inf:            return         dist[i][j] = d        q.append((i, j))     # 回文は先頭からではなく、中央から両端を伸ばすように処理すると楽    # パスを頂点としたグラフと言い換え、多始点BFS の問題に帰着    # 初期化: 対角線上は距離0、与えられたパスの距離は1    for i in range(n):        push(i, i, 0)     for i in range(n):        for j in range(n):            if c[i][j] == "-":                continue             push(i, j, 1)     while q:        s, t = q.popleft()         # 両端を伸ばせる、かつ、それらの文字が同じ場合のみ頂点を追加        for ns in range(n):            for nt in range(n):                if c[ns][s] == "-":                    continue                if c[t][nt] == "-":                    continue                if c[ns][s] != c[t][nt]:                    continue                 push(ns, nt, dist[s][t] + 2)     for i in range(n):        ans = list()         for j in range(n):            d = dist[i][j]             if d == inf:                d = -1             ans.append(d)         print(*ans)  if __name__ == "__main__":    main()

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