<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>>& a) {
        int n = a.size(), m = a[0].size();
        vector<vector<bool> > used(n, vector<bool> (m, true));
        priority_queue<vector<int> > q;
        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) continue;
                q.push({-a[i][j], i, j});
                used[i][j] = false;
            }
        }
        int answer = 0;
        while (!q.empty()) {
            int val = -q.top()[0], x = q.top()[1], y = q.top()[2];
            q.pop();
            for (int dx = -1; dx <= 1; ++dx) {
                for (int dy = -1; dy <= 1; ++dy) {
                    if (dx * dy != 0) continue;
                    if (dx == 0 && dy == 0) continue;
                    int nx = x + dx, ny = y + dy;
                    if (nx < 0 || ny < 0 || nx >= n || ny >= m) continue;
                    if (!used[nx][ny]) continue;
                    answer += max(0, val - a[nx][ny]);
                    used[nx][ny] = false;
                    q.push({-max(a[nx][ny], val), nx, ny});
                }
            }
        }
        return answer;
    }
};
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