<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 
{
    struct YT
    { int A, u, b, h; };
    
public:
    int numSubmat(vector<vector<int>>& m) 
    {
        int out{};
        vector t(size(m)+1, vector(size(m[0])+1, YT{}));
        for(int i{}; i<size(m); ++i)
            for(int j{}; j<size(m[0]); ++j)
                if(m[i][j]) 
                {
                    int b  = t[i+1][j].b,
                        u  = t[i][j+1].u, 
                        h  = t[i][j+1].h, 
                        A0 = t[i][j]  .A,
                        A1 = t[i][j+1].A,
                        A2 = t[i+1][j].A;
                    
                    out += (t[i+1][j+1] = !u or u>b ? YT{(b+1)*(h+1), b+1, b+1, h+1} : YT{A1+A2-A0+1, u, b+1, h+1}).A;
                }
        return out;
    }
};
20700