<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:
    typedef tuple<int, int, int> TII;
    struct Compare{
        bool operator()(const TII& t1, const TII& t2){
            return get<0>(t1) > get<0>(t2);
        }
    };
    int trapRainWater(vector<vector<int>>& heightMap) {
        
        priority_queue<TII, vector<TII>, Compare> pq;

        int n = heightMap.size(), m = heightMap[0].size();

        if(n <= 2 || m <= 2) return 0; // 2行2列五法完成
        for(int i = 0; i < n; i++){
            pq.push(make_tuple(heightMap[i][0], i, 0));
            pq.push(make_tuple(heightMap[i][m - 1], i, m - 1));
        }
        for(int j = 1; j < m - 1; j++){
            pq.push(make_tuple(heightMap[0][j], 0, j));
            pq.push(make_tuple(heightMap[n - 1][j], n - 1, j));
        }

        vector<vector<int>> vis(n, vector<int>(m, 0));

        vector<pair<int, int>> dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

        int res = 0; // 存放最终结果
        while(!pq.empty()){
            auto [h, x, y] = pq.top();
            pq.pop();
            vis[x][y] = 1;

            for(int i = 0; i < 4; i++){
                int nx = x + dir[i].first, ny = y + dir[i].second;
                if(nx < 0 || nx >= n || ny < 0 || ny >= m) continue;
                if(vis[nx][ny]) continue;
                vis[nx][ny] = 1;

                if(heightMap[nx][ny] < h) res += h - heightMap[nx][ny]; // 存放水

                pq.push(make_tuple(max(h, heightMap[nx][ny]), nx, ny));
            }
        }
        

        return res;

    }
};
18881