Trapping Rain Water II

Given n x m non-negative integers representing an elevation map 2d where the area of each cell is 1 x 1, compute how much water it is able to trap after raining.

这是一道非常有意思的题目。如果明白了灌水类问题的核心思之后则可以明白，当前位置可灌的水为周围最低的bar(包括灌完水的）决定的．每次用堆找到一个最低的bar（弹出这个最低的bar,意思是处理过)之后该bar周围找点，如果没有加入过堆，且值小于堆顶值，则进行灌水．代码如下：

class Solution:
    # @param heights: a matrix of integers
    # @return: an integer
    def trapRainWater(self, heights):
        if not heights or not heights[0]:
            return 0
        import heapq
        heap = []
        m = len(heights)
        n = len(heights[0])
        visited = [[False] * n for i in xrange(m)]
        
        for i in xrange(n):
            visited[0][i] = True
            heapq.heappush(heap, (heights[0][i],(0,i)))
            if m > 1:
                visited[-1][i] = True
                heapq.heappush(heap, (heights[-1][i],(m-1,i)))
        for j in xrange(m):
            visited[j][0] = True
            heapq.heappush(heap, (heights[j][0], (j,0)))
            if n > 1:
                visited[j][-1] = True
                heapq.heappush(heap, (heights[j][-1], (j, n-1)))
        
        area = 0
        dx = [1, 0, -1, 0]
        dy = [0, -1, 0, 1]
        
        while heap:
            cur = heapq.heappop(heap)
            for i in xrange(4):
                x = cur[1][0] + dx[i]
                y = cur[1][1] + dy[i]
                if x >= 0 and y>= 0 and x < m and y < n and not visited[x][y]:
                    visited[x][y] = True
                    if heights[x][y] < cur[0]:
                        area += cur[0] - heights[x][y]
                        heapq.heappush(heap, (cur[0],(x,y)))
                    else:
                        heapq.heappush(heap, (heights[x][y],(x,y)))
                        
        return area

 注意不用担心斜对角线漏水～