<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>>& hts) {
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
        int r = hts.size(), c = hts[0].size();
        if (r == 0 || c == 0) return 0;
        vector<bool> vst(r * c, false);
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                if (i == 0 || i == r - 1 || j == 0 || j == c - 1) {
                    if (!vst[i * c + j]) q.push(pair<int, int>(hts[i][j], i * c + j));
                    vst[i * c + j] = true;
                }
            }
        }
        int dir[5] = {0, 1, 0, -1, 0};
        int res = 0;
        while (!q.empty()) {
            pair<int, int> t = q.top();
            q.pop();
            int x = t.second / c, y = t.second % c;
            for (int i = 0; i < 4; i++) {
                int nx = x + dir[i], ny = y + dir[i + 1];
                if (nx < 0 || nx > r - 1 || ny < 0 || ny >= c || vst[nx * c + ny]) continue;
                res += max(0, t.first - hts[nx][ny]);
                q.push(pair<int, int>(max(t.first, hts[nx][ny]), nx * c + ny));
                vst[nx * c + ny] = true;
            }
        }
        return res;
    }
};
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