<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();
        vector<vector<int>> dp(m, vector<int>(n, 0));
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
        for (int i = 0; i < n; i++)
        {
            pq.push({heightMap[0][i], 0, i});
            dp[0][i] = 1;
            pq.push({heightMap[m - 1][i], m - 1, i});
            dp[m - 1][i] = 1;
        }
        for (int i = 1; i < m - 1; i++)
        {
            pq.push({heightMap[i][0], i, 0});
            dp[i][0] = 1;
            pq.push({heightMap[i][n - 1], i, n - 1});
            dp[i][n - 1] = 1;
        }
        int ans = 0;
        while (!pq.empty())
        {
            auto it = pq.top();
            pq.pop();

            vector<int> X = {0, 0, -1, 1};
            vector<int> Y = {-1, 1, 0, 0};
            for (int i = 0; i < 4; i++)
            {
                int x = it[1] + X[i];
                int y = it[2] + Y[i];
                if (x >= 0 && y >= 0 && x < m && y < n)
                {
                    if (dp[x][y] == 0)
                    {
                        if (heightMap[x][y] < it[0])
                        {
                            ans += it[0] - heightMap[x][y];
                            pq.push({it[0], x, y});
                        }
                        else
                        {
                            pq.push({heightMap[x][y], x, y});
                        }
                        dp[x][y] = 1;
                    }
                }
            }
        }
        return ans;
    }
};
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