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Problem solution · Python

ABC278 F — Shiritori

ABC278 F — Shiritori: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Depth-first search
Source
KATO-Hiro AtCoder Solutions
Length
43 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

Depth-first search

For ABC278 F — Shiritori, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 43 lines of Python from the credited upstream file abc278_f.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeABC278 F — Shiritori · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys    from functools import lru_cache     input = sys.stdin.readline    sys.setrecursionlimit(10**8)     n = int(input())    s = [input().rstrip() for _ in range(n)]     # しりとりに使う単語(=状態)が少ないことを利用して、bitDP    # ゲームは終局状態から考えるのが定石 + 再帰で表現     # メモ化再帰をデコレータで表現    @lru_cache(maxsize=None)    def rec(word_set, prev_word):        result = False         for i in range(n):            # i番目の単語を既に使っている            if (word_set >> i) & 1:                continue             # しりとりができない            if word_set != 0 and (s[i][0] != s[prev_word][-1]):                continue             result |= not rec(word_set | 1 << i, i)         return result     if rec(0, 0):  # 単語の集合、直前に使った単語        print("First")    else:        print("Second")  if __name__ == "__main__":    main()

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