結果
問題 |
No.86 TVザッピング(2)
|
ユーザー |
![]() |
提出日時 | 2025-06-12 12:50:54 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
|
実行時間 | - |
コード長 | 1,181 bytes |
コンパイル時間 | 160 ms |
コンパイル使用メモリ | 82,292 KB |
実行使用メモリ | 69,260 KB |
最終ジャッジ日時 | 2025-06-20 14:02:28 |
合計ジャッジ時間 | 2,493 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 24 WA * 6 |
ソースコード
n, m = map(int, input().split()) grid = [input().strip() for _ in range(n)] cells = [] for i in range(n): for j in range(m): if grid[i][j] == '.': cells.append((i, j)) k = len(cells) if k % 2 != 0: print("NO") exit() # Check if all cells are in the same row or column same_row = True row = cells[0][0] for i, j in cells: if i != row: same_row = False break same_col = True col = cells[0][1] for i, j in cells: if j != col: same_col = False break if same_row or same_col: print("NO") exit() # Check connectivity using BFS from collections import deque visited = [[False] * m for _ in range(n)] start_i, start_j = cells[0] queue = deque([(start_i, start_j)]) visited[start_i][start_j] = True count = 1 directions = [(-1, 0), (1, 0), (0, -1), (0, 1)] while queue: i, j = queue.popleft() for di, dj in directions: ni, nj = i + di, j + dj if 0 <= ni < n and 0 <= nj < m: if grid[ni][nj] == '.' and not visited[ni][nj]: visited[ni][nj] = True count += 1 queue.append((ni, nj)) print("YES" if count == k else "NO")