Walls and Gates
要点：
- 同样是bfs，这题可以用渲染的方法（即全部gate进初始q），注意区别Shortest Distance from All Buildings。那道题要找到某个点到"all" buildings的距离，所以不能用渲染，因为不是到该点的一条路径。而本题类似surrounded region，最终的最佳路径只取一条路径
- bfs的方法全是在展开后入q前更新
- 剪枝：只要值更新了，就一定是最小值。不用入q了
错误点：
- 忘了更新距离。因为gate是0，所以距离可以统一更新为从哪来的+1
# You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where:
# Each 0 marks an empty land which you can pass by freely.
# Each 1 marks a building which you cannot pass through.
# Each 2 marks an obstacle which you cannot pass through.
# For example, given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2):
# 1 - 0 - 2 - 0 - 1
# | | | | |
# 0 - 0 - 0 - 0 - 0
# | | | | |
# 0 - 0 - 1 - 0 - 0
# The point (1,2) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is minimal. So return 7.
# Note:
# There will be at least one building. If it is not possible to build such house according to the above rules, return -1.
from collections import deque
import sys
class Solution(object):
def shortestDistance(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
def bfs(grid, m, n, x, y, total):
visited = [[False]*n for _ in xrange(m)] # error: don't use x,y
q = deque([(x,y)]) # error 1
count=0
dirs = [(1,0),(0,1),(-1,0),(0,-1)]
d = 0
while q:
ql = len(q) # error 2: use single object
d+=1
for _ in xrange(ql):
x,y = q.popleft()
for xd,yd in dirs:
x0,y0 = x+xd,y+yd
if 0<=x0<m and 0<=y0<n and not visited[x0][y0]:
visited[x0][y0]=True # error 3: dont forget
if grid[x0][y0]==0:
self.dist[x0][y0]+=d
self.reach[x0][y0]+=1
q.append((x0,y0))
if grid[x0][y0]==1:
count+=1
return count==total
m,n=len(grid),len(grid[0])
total = sum([val for line in grid for val in line if val==1])
self.dist = [[0 for y in xrange(n)] for x in xrange(m)]
self.reach = [[0 for y in xrange(n)] for x in xrange(m)]
for x in xrange(m):
for y in xrange(n):
if grid[x][y]==1:
if not bfs(grid, m, n, x, y, total):
return -1
shortest = sys.maxint
for x in xrange(m):
for y in xrange(n):
if self.reach[x][y]==total:
shortest = min(self.dist[x][y], shortest)
return shortest
sol = Solution()
assert sol.shortestDistance([[1,0,2,0,1],[0,0,0,0,0],[0,0,1,0,0]])==7, "must be 7"
assert sol.shortestDistance([[1]])==-1, "must be -1"