<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 {
    using p = pair<int, pair<int, int>>;

public:
    int trapRainWater(vector<vector<int>>& height) {
        int n = height.size();
        int m = height[0].size();
        priority_queue<p, vector<p>, greater<p>> pq;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (i == 0 || j == 0 || i == n - 1 || j == m - 1) {
                    pq.push({height[i][j], {i, j}});
                }
            }
        }
        int ans = 0;
        int row[] = {0, 0, 1, -1};
        int col[] = {1, -1, 0, 0};
        vector<vector<int>> vis(n, vector<int>(m, false));
        while (!pq.empty()) {
            auto front = pq.top();
            pq.pop();
            int h = front.first;
            int x = front.second.first;
            int y = front.second.second;
            if (vis[x][y])
                continue;
            ans += h - height[x][y];
            vis[x][y] = 1;
            for (int k = 0; k < 4; ++k) {
                int dr = row[k] + x;
                int dc = col[k] + y;
                if (dr < n && dc < m && dr >= 0 && dc >= 0 && !vis[dr][dc]) {
                    pq.push({max(h, height[dr][dc]), {dr, dc}});
                }
            }
        }
        return ans;
    }
};
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