<p>Given a rectangular pizza represented as a <code>rows x cols</code>&nbsp;matrix containing the following characters: <code>&#39;A&#39;</code> (an apple) and <code>&#39;.&#39;</code> (empty cell) and given the integer <code>k</code>. You have to cut the pizza into <code>k</code> pieces using <code>k-1</code> cuts.&nbsp;</p>

<p>For each cut you choose the direction: vertical or horizontal, then you choose a cut position at the cell boundary and cut the pizza into two pieces. If you cut the pizza vertically, give the left part of the pizza to a person. If you cut the pizza horizontally, give the upper part of the pizza to a person. Give the last piece of pizza to the last person.</p>

<p><em>Return the number of ways of cutting the pizza such that each piece contains <strong>at least</strong> one apple.&nbsp;</em>Since the answer can be a huge number, return this modulo 10^9 + 7.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/04/23/ways_to_cut_apple_1.png" style="width: 500px; height: 378px;" /></strong></p>

<pre>
<strong>Input:</strong> pizza = [&quot;A..&quot;,&quot;AAA&quot;,&quot;...&quot;], k = 3
<strong>Output:</strong> 3 
<strong>Explanation:</strong> The figure above shows the three ways to cut the pizza. Note that pieces must contain at least one apple.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> pizza = [&quot;A..&quot;,&quot;AA.&quot;,&quot;...&quot;], k = 3
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> pizza = [&quot;A..&quot;,&quot;A..&quot;,&quot;...&quot;], k = 1
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= rows, cols &lt;= 50</code></li>
	<li><code>rows ==&nbsp;pizza.length</code></li>
	<li><code>cols ==&nbsp;pizza[i].length</code></li>
	<li><code>1 &lt;= k &lt;= 10</code></li>
	<li><code>pizza</code> consists of characters <code>&#39;A&#39;</code>&nbsp;and <code>&#39;.&#39;</code> only.</li>
</ul>

class Solution {
public:
    int m, n;
    int appless[55][55];
    long long mod = 1e9 + 7;
    vector<vector<vector<int>>> memo;

    int solve(int i, int j, int k) {
        if (appless[i][j] < k) return 0;
        if (k == 1) return appless[i][j] > 0;
        if (memo[i][j][k] != -1) return memo[i][j][k];

        int ans = 0;

        for (int h = i + 1; h < m; h++) {
            if (appless[i][j] - appless[h][j] > 0) {
                ans = (ans + solve(h, j, k - 1)) % mod;
            }
        }

        for (int v = j + 1; v < n; v++) {
            if (appless[i][j] - appless[i][v] > 0) {
                ans = (ans + solve(i, v, k - 1)) % mod;
            }
        }

        return memo[i][j][k] = ans;
    }

    int ways(vector<string>& pizza, int k) {
        m = pizza.size();
        n = pizza[0].size();
        memo = vector<vector<vector<int>>>(m, vector<vector<int>>(n, vector<int>(k + 1, -1)));
        
        memset(appless, 0, sizeof(appless));
        
        for (int i = m - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                appless[i][j] = (pizza[i][j] == 'A') + (i + 1 < m ? appless[i + 1][j] : 0) + (j + 1 < n ? appless[i][j + 1] : 0) - (i + 1 < m && j + 1 < n ? appless[i + 1][j + 1] : 0);
            }
        }

        return solve(0, 0, k);
    }
};

11000