結果
問題 |
No.228 ゆきこちゃんの 15 パズル
|
ユーザー |
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提出日時 | 2025-04-15 23:41:42 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
|
実行時間 | - |
コード長 | 1,791 bytes |
コンパイル時間 | 181 ms |
コンパイル使用メモリ | 81,892 KB |
実行使用メモリ | 53,804 KB |
最終ジャッジ日時 | 2025-04-15 23:44:01 |
合計ジャッジ時間 | 1,685 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 16 WA * 1 |
ソースコード
def main(): import sys input = sys.stdin.read().split() grid = [list(map(int, input[i*4:(i+1)*4])) for i in range(4)] # Precompute the initial positions (0-based) initial_pos = {} for i in range(4): for j in range(4): num = i * 4 + j + 1 if num == 16: initial_pos[0] = (i, j) else: initial_pos[num] = (i, j) # Check Manhattan distance for each number possible = True for i in range(4): for j in range(4): num = grid[i][j] if num == 0: continue initial_i, initial_j = initial_pos[num] dx = abs(i - initial_i) dy = abs(j - initial_j) if dx + dy > 1: possible = False break if not possible: break if not possible: print("No") return # Check standard 15-puzzle solvability # Compute inversion count (excluding 0) flat = [] zero_row = 0 for i in range(4): for j in range(4): num = grid[i][j] if num == 0: zero_row = i else: flat.append(num) inv_count = 0 for i in range(len(flat)): for j in range(i + 1, len(flat)): if flat[i] > flat[j]: inv_count += 1 # For standard 15 puzzle, solvable if (inv_count + zero_row_from_bottom) is even # Here, zero_row is 0-based (0 is top row, 3 is bottom row) # So zero_row_from_bottom = 3 - zero_row zero_row_from_bottom = 3 - zero_row if (inv_count + zero_row_from_bottom) % 2 != 0: possible = False print("Yes" if possible else "No") if __name__ == "__main__": main()