結果
| 問題 |
No.3063 Circle Balancing
|
| コンテスト | |
| ユーザー |
gew1fw
|
| 提出日時 | 2025-06-12 18:15:31 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
RE
|
| 実行時間 | - |
| コード長 | 1,303 bytes |
| コンパイル時間 | 246 ms |
| コンパイル使用メモリ | 82,560 KB |
| 実行使用メモリ | 65,664 KB |
| 最終ジャッジ日時 | 2025-06-12 18:16:20 |
| 合計ジャッジ時間 | 3,317 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| sample | RE * 2 |
| other | RE * 27 |
ソースコード
import sys
from collections import deque
R, C = map(int, sys.stdin.readline().split())
sy, sx = map(int, sys.stdin.readline().split())
gy, gx = map(int, sys.stdin.readline().split())
# Convert to 0-based indices
sy -= 1
sx -= 1
gy -= 1
gx -= 1
maze = []
for _ in range(R):
row = sys.stdin.readline().strip()
maze.append(list(row))
# Initialize distance matrix with -1 (unvisited)
dist = [[-1 for _ in range(C)] for _ in range(R)]
dist[sy][sx] = 0
queue = deque()
queue.append((sy, sx))
# Directions: up, down, left, right
directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
while queue:
y, x = queue.popleft()
current_distance = dist[y][x]
for dy, dx in directions:
ny = y + dy
nx = x + dx
# Check if within bounds
if 0 <= ny < R and 0 <= nx < C:
# Check if the cell is passable and not visited
if maze[ny][nx] == '.' and dist[ny][nx] == -1:
dist[ny][nx] = current_distance + 1
# Check if reached the goal
if ny == gy and nx == gx:
print(dist[ny][nx])
sys.exit()
queue.append((ny, nx))
# According to the problem statement, the goal is always reachable, so this line is theoretically unreachable
print(-1)
gew1fw