<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>

#include<bits/stdc++.h>
using namespace std;
class Solution {
public:
    int trapRainWater(vector<vector<int>>& h) {
        int n=h.size();
        int m=h[0].size();
        vector<vector<int>> vis(n,vector<int>(m,0));
        priority_queue<pair<int,pair<int,int>>, vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>> 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({h[i][j],{i,j}});
                    vis[i][j]=1;
                }
            }
        }
        int mh=0;
        int ans=0;
        while(!pq.empty()){
            int ch=pq.top().first;
            int r=pq.top().second.first;
            int c=pq.top().second.second;
            pq.pop();
            mh=max(mh,ch);
            vector<int> rf={0,0,-1,1};
            vector<int> cf={1,-1,0,0};
            for(int i=0;i<4;i++){
                int nr=r+rf[i];
                int nc=c+cf[i];
                if(nr>=0 && nr<n && nc>=0 && nc<m && vis[nr][nc]==0){
                    pq.push({h[nr][nc],{nr,nc}});
                    vis[nr][nc]=1;
                    if(h[nr][nc]<mh){
                        ans+=mh-h[nr][nc];
                    }
                }
            }
        }
        return ans;
    }
};
28881