<p>Given an <code>m x n</code> integer matrix <code>heightMap</code> representing the height of each unit cell in a 2D elevation map, return <em>the volume of water it can trap after raining</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/08/trap1-3d.jpg" style="width: 361px; height: 321px;" />
<pre>
<strong>Input:</strong> heightMap = [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> After the rain, water is trapped between the blocks.
We have two small ponds 1 and 3 units trapped.
The total volume of water trapped is 4.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/08/trap2-3d.jpg" style="width: 401px; height: 321px;" />
<pre>
<strong>Input:</strong> heightMap = [[3,3,3,3,3],[3,2,2,2,3],[3,2,1,2,3],[3,2,2,2,3],[3,3,3,3,3]]
<strong>Output:</strong> 10
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == heightMap.length</code></li>
	<li><code>n == heightMap[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>0 &lt;= heightMap[i][j] &lt;= 2 * 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int trapRainWater(vector<vector<int>>& height) {
        int m = height.size(), n = height[0].size();
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>>
            pq;
        for (int i = 0; i < m; i++) {
            visited[i][0] = true;
            visited[i][n - 1] = true;
            pq.push({height[i][0], i, 0});
            pq.push({height[i][n - 1], i, n - 1});
        }
        for (int j = 1; j < n - 1; j++) {
            visited[0][j] = true;
            visited[m - 1][j] = true;
            pq.push({height[0][j], 0, j});
            pq.push({height[m - 1][j], m - 1, j});
        }
        vector<int> xd = {0, 0, 1, -1};
        vector<int> yd = {1, -1, 0, 0};
        int minH = 0;
        int ans = 0;
        while (!pq.empty()) {
            int i = pq.top()[1], j = pq.top()[2];
            minH = max(minH, pq.top()[0]);
            pq.pop();
            for (int d = 0; d < 4; d++) {
                int ni = i + xd[d], nj = j + yd[d];
                if (ni < 0 || ni >= m || nj < 0 || nj >= n || visited[ni][nj])
                    continue;
                int diff = minH - height[ni][nj];
                if (diff >= 0)
                    ans += diff;
                pq.push({height[ni][nj], ni, nj});
                visited[ni][nj] = true;
            }
        }
        return ans;
    }
};
23881