CPMagical CP Library

Problem solution · Python

ABC173 C — H and V

ABC173 C — H and V: 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
59 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 ABC173 C — H and V, 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

  • 59 lines of Python from the credited upstream file abc173_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 codeABC173 C — H and V · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def f(c, h_patterns, w_patterns, black_count):    # マスを塗りつぶしたかどうか    used = [[False for _ in range(len(w_patterns))]            for _ in range(len(h_patterns))]     for index, h_pattern in enumerate(h_patterns):        if h_pattern != 1:            continue         for w_dash in range(len(w_patterns)):            if not used[index][w_dash] and c[index][w_dash] == '#':                used[index][w_dash] = True                black_count -= 1     for index, w_pattern in enumerate(w_patterns):        if w_pattern != 1:            continue         for h_dash in range(len(h_patterns)):            if not used[h_dash][index] and c[h_dash][index] == '#':                used[h_dash][index] = True                black_count -= 1     return black_count  def main():    from itertools import product     h, w, k = map(int, input().split())    c = [list(input()) for _ in range(h)]    black_count = 0    ans = 0     # KeyInsight:    # 制約からbit全探索をすると楽に実装できる    # △: リストのコピーができていない=参照渡しの理解が不十分    # △: 愚直に塗りつぶす行と列を決めて、その中に黒マスが含まれていたら減らす方針に    for hi in range(h):        for wi in range(w):            if c[hi][wi] == '#':                black_count += 1     for h_patterns in product([1, 0], repeat=h):        for w_patterns in product([1, 0], repeat=w):            fi = f(c, h_patterns, w_patterns, black_count)             if fi == k:                ans += 1     print(ans)  if __name__ == '__main__':    main()

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗