【Leetcode】803. Bricks Falling When Hit 803. 打砖块

解法

先把所有该消的都消掉，现在和第一行相通的点都是“稳定点”
然后把消除顺序反过来遍历一遍
当把(i,j)恢复成1之后，它会把它四个方向上的集合连通
假如连通完毕后“稳定点”所在的那个集合的的点的个数上涨了x个(x一定大于1)，说明消除(i,j)使得有x-1个点从“稳定点”变得不稳定了，那也就是这次消除后drop的数量

class Solution(object):
    def hitBricks(self, grid, hits):
        """
        :type grid: List[List[int]]
        :type hits: List[List[int]]
        :rtype: List[int]
        """
        from collections import defaultdict
        m = len(grid)
        n = len(grid[0])
        f = range(m*n+1)
        END = m*n
        sz = defaultdict(lambda:1)
        for i,j in hits:
            if grid[i][j]==1:
                grid[i][j]=2
                
        def find(x):
            r = x
            while f[r]!=r:
                r = f[r]
            while f[x]!=r:
                tmp = f[x]
                f[x] = r
                x = tmp
            return r
        
        def join(x,y):
            rx,ry = find(x),find(y)
            if rx!=ry:
                if rx<ry:
                    rx,ry = ry,rx
                f[ry] = rx
                sz[rx] += sz[ry]
        for i,j in itertools.product(xrange(m),xrange(n)):
            if grid[i][j]==1:
                x = i*n+j
                if i and grid[i-1][j]==1:
                    join(x-n,x)
                if j and grid[i][j-1]==1:
                    join(x-1,x)
                if i==0:
                    join(END,x)
        # print [find(x) for x in xrange(END+1)]
        prev = sz[END]
        ans = []
        for i,j in hits[::-1]:
            # print grid
            if grid[i][j]==2:
                x = i*n+j
                if i and grid[i-1][j]==1:join(x-n,x)
                if j and grid[i][j-1]==1:join(x-1,x)
                if i+1<m and grid[i+1][j]==1:join(x+n,x)
                if j+1<n and grid[i][j+1]==1:join(x+1,x)
                if i==0:
                    join(END,x)
                grid[i][j] = 1
                # print [find(x) for x in xrange(END+1)]
                if sz[END]>prev:
                    ans.append(sz[END]-prev-1)
                    prev = sz[END]
                    # print prev
                else:
                    ans.append(0)
            else:
                ans.append(0)
        return ans[::-1]