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On a 2-dimensional grid
, there are 4 types of squares:
1
represents the starting square. There is exactly one starting square.2
represents the ending square. There is exactly one ending square.0
represents empty squares we can walk over.-1
represents obstacles that we cannot walk over.
Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.
Example 1:
Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)
Example 2:
Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)
Example 3:
Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.
Note:
1 <= grid.length * grid[0].length <= 20
思路:BFS+用整数压缩状态
class Solution:
def uniquePathsIII(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
n,m = len(grid),len(grid[0])
si = sj = -1
cnt = 0
for i in range(n):
for j in range(m):
if grid[i][j] == 1:
si,sj = i,j
elif grid[i][j] == 0:
cnt += 1
res = 0
# vis = set()
q = [(si,sj,0,0)]
dirs = [[1,0],[-1,0],[0,1],[0,-1]]
while q:
si,sj,s,c = q.pop()
for di,dj in dirs:
ti,tj=si+di,sj+dj
if 0<=ti<n and 0<=tj<m:
if grid[ti][tj]==0 and s&(1<<(m*ti+tj))==0:
q.append((ti,tj,s|(1<<(m*ti+tj)),c+1))
elif grid[ti][tj]==2 and c==cnt:
res+=1
return res