Given an m x n matrix of positive integers representing the height of each unit cell in a 2D elevation map, compute the volume of water it is able to trap after raining.

Note:
Both m and n are less than 110. The height of each unit cell is greater than 0 and is less than 20,000.

Example:

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Given the following 3x6 height map:
[
  [1,4,3,1,3,2],
  [3,2,1,3,2,4],
  [2,3,3,2,3,1]
]
Return 4.


The above image represents the elevation map [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]] before the rain.


After the rain, water are trapped between the blocks. The total volume of water trapped is 4.

主要思想就是从外围向内灌水，每次找到外围最小的点，如果它旁边有没访问过，且比当前值小的点，可以灌水，否则不可以，然后更新外围，用新的点的坐标和max(x, nx)做为值.

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class Cell implements Comparable<Cell> {
    public int x, y, h;
    public Cell(int x, int y, int h) {
        this.x = x;
        this.y = y;
        this.h = h;
    }
    @Override
    public int compareTo(Cell that) {
        return this.h - that.h;
    }
}
int[] dx = {0, 0, 1, -1};
int[] dy = {1, -1, 0, 0};
public int trapRainWater(int[][] heightMap) {
    if (heightMap == null || heightMap.length == 0)
        return 0;
    if (heightMap[0] == null || heightMap[0].length == 0)
        return 0;
    int m = heightMap.length;
    int n = heightMap[0].length;
    PriorityQueue<Cell> pq = new PriorityQueue<>();
    boolean[][] visited = new boolean[m][n];
    int ans = 0;
    for (int j = 0; j < n; j++) {
        visited[0][j] = true;
        pq.add(new Cell(0, j, heightMap[0][j]));
        visited[m - 1][j] = true;
        pq.add(new Cell(m - 1, j, heightMap[m - 1][j]));
    }
    for (int i = 1; i < m - 1; i++) {
        visited[i][0] = true;
        pq.add(new Cell(i, 0, heightMap[i][0]));
        visited[i][n - 1] = true;
        pq.add(new Cell(i, n - 1, heightMap[i][n - 1]));
    }
    while (!pq.isEmpty()) {
        Cell min = pq.poll();
        for (int i = 0; i < 4; i++) {
            int nx = min.x + dx[i];
            int ny = min.y + dy[i];
            if (nx >= 0 && nx < m && ny >= 0 && ny < n && visited[nx][ny] == false) {
                int h = heightMap[nx][ny];
                visited[nx][ny] = true;
                pq.offer(new Cell(nx, ny, Math.max(h, min.h)));
            	ans += Math.max(0, min.h - h);
            }
        }
    }
    return ans;
}