<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>>& heightMap) {
        int m = heightMap.size(), n = heightMap[0].size();
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
        for(int i = 0; i<m; i++) {
            for(int j = 0; j<n; j++) {
                if(i==0 || j==0 || i==m-1 || j==n-1) pq.push({heightMap[i][j], i, j});
            }
        }
        int water = 0;
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        while(!pq.empty()) {
            int i = pq.top()[1];
            int j = pq.top()[2];
            int h = pq.top()[0];
            pq.pop();
            if(visited[i][j]) continue;
            visited[i][j] = true;
            water += (h - heightMap[i][j]);
            vector<int> neighs = {i, j-1, i, j+1, i+1, j, i-1, j};
            for(int idx = 0; idx<7; idx+=2) {
                int r = neighs[idx];
                int c = neighs[idx+1];
                if(r<0 || r>=m || c<0 || c>=n) continue;
                if(!visited[r][c]) pq.push({max(h, heightMap[r][c]), r, c});
            }
        }
        return water;
    }
};
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