結果
| 問題 |
No.1949 足し算するだけのパズルゲーム(2)
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-03-31 17:31:33 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,431 bytes |
| コンパイル時間 | 125 ms |
| コンパイル使用メモリ | 82,544 KB |
| 実行使用メモリ | 126,944 KB |
| 最終ジャッジ日時 | 2025-03-31 17:32:17 |
| 合計ジャッジ時間 | 12,154 ms |
|
ジャッジサーバーID (参考情報) |
judge5 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 18 WA * 8 |
ソースコード
import heapq
def main():
H, W, Y, X = map(int, input().split())
Y -= 1 # convert to 0-based
X -= 1
grid = []
for _ in range(H):
grid.append(list(map(int, input().split())))
visited = [[False]*W for _ in range(H)]
visited[Y][X] = True
current_power = grid[Y][X]
heap = []
directions = [ (-1,0), (1,0), (0,-1), (0,1) ]
# Initialize the heap with adjacent cells
for dy, dx in directions:
ny = Y + dy
nx = X + dx
if 0 <= ny < H and 0 <= nx < W:
if not visited[ny][nx] and grid[ny][nx] <= current_power:
heapq.heappush(heap, (grid[ny][nx], ny, nx))
required = H * W - 1 # since starting cell is not counted
count = 0
while heap:
attack, i, j = heapq.heappop(heap)
if visited[i][j]:
continue
if current_power < attack:
print("No")
return
current_power += attack
visited[i][j] = True
count += 1
# Check adjacent cells
for dy, dx in directions:
ny = i + dy
nx = j + dx
if 0 <= ny < H and 0 <= nx < W:
if not visited[ny][nx] and grid[ny][nx] <= current_power:
heapq.heappush(heap, (grid[ny][nx], ny, nx))
if count == required:
print("Yes")
else:
print("No")
if __name__ == "__main__":
main()
lam6er