結果
| 問題 |
No.3063 Circle Balancing
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-04-16 15:22:46 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
RE
|
| 実行時間 | - |
| コード長 | 1,255 bytes |
| コンパイル時間 | 284 ms |
| コンパイル使用メモリ | 82,304 KB |
| 実行使用メモリ | 67,328 KB |
| 最終ジャッジ日時 | 2025-04-16 15:23:56 |
| 合計ジャッジ時間 | 3,415 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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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
# Read input
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
# Read maze
maze = []
for _ in range(R):
row = sys.stdin.readline().strip()
maze.append(row)
# Initialize distance array and queue
dist = [[-1] * C for _ in range(R)]
q = deque()
dist[sy][sx] = 0
q.append((sy, sx))
# Directions: up, down, left, right
dx = [0, 0, -1, 1]
dy = [-1, 1, 0, 0]
# BFS loop
while q:
y, x = q.popleft()
current_dist = dist[y][x]
# Check if current position is the goal
if y == gy and x == gx:
print(current_dist)
sys.exit()
# Explore all four directions
for i in range(4):
ny = y + dy[i]
nx = x + dx[i]
# Check if the next cell is within bounds, not a wall, and not visited
if 0 <= ny < R and 0 <= nx < C:
if maze[ny][nx] == '.' and dist[ny][nx] == -1:
dist[ny][nx] = current_dist + 1
q.append((ny, nx))
# According to the problem statement, the goal is always reachable, so this line is theoretically unreachable
print(-1)
lam6er