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Problem solution · Python

ABC319 C — False Hope

ABC319 C — False Hope: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Direct simulation
Source
KATO-Hiro AtCoder Solutions
Length
72 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

Direct simulation

For ABC319 C — False Hope, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 72 lines of Python from the credited upstream file abc319_c.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeABC319 C — False Hope · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys    from itertools import permutations     input = sys.stdin.readline     cells = list()     # グリッドを1次元にすることで、順列との対応づけがしやすくなる    for _ in range(3):        ci = list(map(int, input().split()))        cells += ci     # print(cells)     # 縦・横・斜めのマス(0-8)の組を用意    trio = [        (0, 1, 2),        (3, 4, 5),        (6, 7, 8),        (0, 3, 6),        (1, 4, 7),        (2, 5, 8),        (2, 4, 6),        (0, 4, 8),    ]     count, count_all = 0, 0     for order in permutations(range(9)):        count_all += 1        ok = True         # がっかり = 最初の2つが同じ数字で、3つ目の数字が最後に登場        # ペアが3種類        for a, b, c in trio:            if (                (cells[a] == cells[b])                and (order[a] < order[c])                and (order[b] < order[c])            ):                ok = False                break             if (                (cells[b] == cells[c])                and (order[b] < order[a])                and (order[c] < order[a])            ):                ok = False                break             if (                (cells[c] == cells[a])                and (order[c] < order[b])                and (order[a] < order[b])            ):                ok = False                break         if ok:            count += 1     print(count / count_all)  if __name__ == "__main__":    main()

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