边工作边刷题：70天一遍leetcode: day 85-4

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

https://repl.it/CfGu/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"