<p>Given an <code>m x n</code> binary matrix <code>mat</code>, <em>return the number of <strong>submatrices</strong> that have all ones</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/10/27/ones1-grid.jpg" style="width: 244px; height: 245px;" />
<pre>
<strong>Input:</strong> mat = [[1,0,1],[1,1,0],[1,1,0]]
<strong>Output:</strong> 13
<strong>Explanation:</strong> 
There are 6 rectangles of side 1x1.
There are 2 rectangles of side 1x2.
There are 3 rectangles of side 2x1.
There is 1 rectangle of side 2x2. 
There is 1 rectangle of side 3x1.
Total number of rectangles = 6 + 2 + 3 + 1 + 1 = 13.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/10/27/ones2-grid.jpg" style="width: 324px; height: 245px;" />
<pre>
<strong>Input:</strong> mat = [[0,1,1,0],[0,1,1,1],[1,1,1,0]]
<strong>Output:</strong> 24
<strong>Explanation:</strong> 
There are 8 rectangles of side 1x1.
There are 5 rectangles of side 1x2.
There are 2 rectangles of side 1x3. 
There are 4 rectangles of side 2x1.
There are 2 rectangles of side 2x2. 
There are 2 rectangles of side 3x1. 
There is 1 rectangle of side 3x2. 
Total number of rectangles = 8 + 5 + 2 + 4 + 2 + 2 + 1 = 24.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= m, n &lt;= 150</code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
    
    typedef long long ll;
public:
    int numSubmat(vector<vector<int>>& mat) {
        ll n = mat.size();
        ll m = mat[0].size();
        vector<vector<ll>> dp(m, vector<ll>(n + 1, 0));
        ll ans =0;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                int ref = i;
                for(int k = i; k >= 0 && mat[k][j] == 1; k--){
                    ref = k - 1;
                    if(j == 0){
                        dp[j][i - k + 1] = 1;  
                    } else {
                        dp[j][i - k + 1] = dp[j - 1][i - k + 1] + 1LL;
                    }
                    //cout << i << ' ' << j << ' ' << i - k + 1 << ' ' << dp[i][j][i - k + 1] << '\n';
                    ans += dp[j][i - k + 1];
                }

                for(int k = ref; k>= 0; k--) dp[j][i - k + 1] = 0;
            }
        }
    
        return ans;
    }
};
19000