1263. 推箱子
难度:困难
时间:2023/05/08
「推箱子」是一款风靡全球的益智小游戏,玩家需要将箱子推到仓库中的目标位置。
游戏地图用大小为 m x n 的网格 grid 表示,其中每个元素可以是墙、地板或者是箱子。
现在你将作为玩家参与游戏,按规则将箱子 'B' 移动到目标位置 'T' :
- 玩家用字符
'S'表示,只要他在地板上,就可以在网格中向上、下、左、右四个方向移动。 - 地板用字符
'.'表示,意味着可以自由行走。 - 墙用字符
'#'表示,意味着障碍物,不能通行。 - 箱子仅有一个,用字符
'B'表示。相应地,网格上有一个目标位置'T'。 - 玩家需要站在箱子旁边,然后沿着箱子的方向进行移动,此时箱子会被移动到相邻的地板单元格。记作一次「推动」。
- 玩家无法越过箱子。
返回将箱子推到目标位置的最小 推动 次数,如果无法做到,请返回 -1。
示例 1:
输入:grid = [["#","#","#","#","#","#"],
["#","T","#","#","#","#"],
["#",".",".","B",".","#"],
["#",".","#","#",".","#"],
["#",".",".",".","S","#"],
["#","#","#","#","#","#"]]
输出:3
解释:我们只需要返回推箱子的次数。
示例 2:
输入:grid = [["#","#","#","#","#","#"],
["#","T","#","#","#","#"],
["#",".",".","B",".","#"],
["#","#","#","#",".","#"],
["#",".",".",".","S","#"],
["#","#","#","#","#","#"]]
输出:-1
示例 3:
输入:grid = [["#","#","#","#","#","#"],
["#","T",".",".","#","#"],
["#",".","#","B",".","#"],
["#",".",".",".",".","#"],
["#",".",".",".","S","#"],
["#","#","#","#","#","#"]]
输出:5
解释:向下、向左、向左、向上再向上。
提示:
m == grid.lengthn == grid[i].length1 <= m, n <= 20grid仅包含字符'.','#','S','T', 以及'B'。grid中'S','B'和'T'各只能出现一个。
解题思路:
优先级队列(双端队列)+BFS
class Solution:
def minPushBox(self, grid: List[List[str]]) -> int:
m, n = len(grid), len(grid[0])
def inArea(i: int, j: int):
return i >=0 and i < m and j >=0 and j < n and (grid[i][j] == '.' or grid[i][j] == 'T')
bij, sij = None, None
for i in range(m):
for j in range(n):
if grid[i][j] == 'S':
sij = [i, j]
grid[i][j] = '.'
if grid[i][j] == 'B':
bij = [i, j]
grid[i][j] = '.'
used = [[[[False] * n for _ in range(m)] for _ in range(n)] for _ in range(m)] # record by arr
used[sij[0]][sij[1]][bij[0]][bij[1]] = True
que = deque([[0] + sij + bij, ]) # use deque instead of heap for saving time
move = [[0, 1], [1, 0], [0, -1], [-1, 0]]
while que:
for _ in range(len(que)):
n_move, si, sj, bi, bj = que.popleft()
if grid[bi][bj] == 'T':
return n_move
for mi, mj in move:
si_, sj_ = si + mi, sj + mj
bi_, bj_, n_move_, lor = bi, bj, n_move, 0
if si_ == bi and sj_ == bj:
# move box
bi_, bj_, n_move_, lor = bi + mi, bj + mj, n_move + 1, 1
if not inArea(si_, sj_) or not inArea(bi_, bj_): continue
if not used[si_][sj_][bi_][bj_]:
if lor:
que.append([n_move_, si_, sj_, bi_, bj_])
else:
que.appendleft([n_move_, si_, sj_, bi_, bj_])
used[si_][sj_][bi_][bj_] = True
return -1