結果

問題 No.1949 足し算するだけのパズルゲーム(2)
ユーザー lam6er
提出日時 2025-03-26 15:47:33
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 1,554 bytes
コンパイル時間 297 ms
コンパイル使用メモリ 82,304 KB
実行使用メモリ 137,756 KB
最終ジャッジ日時 2025-03-26 15:48:54
合計ジャッジ時間 18,221 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
other AC * 22 WA * 4
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ソースコード

diff #
プレゼンテーションモードにする

import heapq
def main():
    import sys
    input = sys.stdin.read
    data = input().split()
    
    idx = 0
    H = int(data[idx])
    W = int(data[idx+1])
    Y = int(data[idx+2]) - 1  # converting to 0-based index
    X = int(data[idx+3]) - 1
    idx +=4
    
    A = []
    for i in range(H):
        row = list(map(int, data[idx:idx+W]))
        A.append(row)
        idx += W
    
    current_power = A[Y][X]
    visited = [[False for _ in range(W)] for _ in range(H)]
    visited[Y][X] = True
    remaining = H * W - 1  # exclude starting position
    
    heap = []
    # Check initial position's neighbors
    directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
    for dy, dx in directions:
        ny = Y + dy
        nx = X + dx
        if 0 <= ny < H and 0 <= nx < W and not visited[ny][nx]:
            heapq.heappush(heap, (A[ny][nx], ny, nx))
    
    while remaining > 0:
        if not heap:
            print("No")
            return
        
        attack, y, x = heapq.heappop(heap)
        if visited[y][x]:
            continue
        
        if attack > current_power:
            print("No")
            return
        
        current_power += attack
        visited[y][x] = True
        remaining -= 1
        
        # Check neighbors of the current cell
        for dy, dx in directions:
            ny = y + dy
            nx = x + dx
            if 0 <= ny < H and 0 <= nx < W and not visited[ny][nx]:
                heapq.heappush(heap, (A[ny][nx], ny, nx))
    
    print("Yes")
if __name__ == "__main__":
    main()
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