<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 ways(vector<string>& pizza, int k) {
        int mod = 1e9 + 7;
        int r = pizza.size();
        int c = pizza[0].size();

        vector<vector<int>> pref(r + 1, vector<int>(c + 1));
        for(int i = r - 1;i >= 0;i --) {
            for(int j = c - 1;j >= 0; j --) {
                if(pizza[i][j] == 'A') {
                    pref[i][j] ++;
                }
                pref[i][j] += pref[i + 1][j] + pref[i][j + 1] - pref[i + 1][j + 1];
            }
        }

        vector<vector<vector<int>>> dp(r, vector<vector<int>>(c, vector<int>(k + 1)));
        for(int q = 1; q <= k; q ++) {
            for(int i = r -1; i>= 0; i --) {
                for(int j = c - 1; j >= 0; j --) {
                    if(q == 1) {
                        if(pref[i][j] > 0){
                            dp[i][j][q] ++;
                        }
                    }
                    else{
                        int tot = 0;
                        for(int s = i + 1; s < r; s ++) {
                            if(pref[i][j] - pref[s][j] > 0) {
                                tot = (tot + dp[s][j][q - 1]) % mod;
                            }   
                        }
                        for(int s = j + 1; s < c; s ++) {
                            if(pref[i][j] - pref[i][s] > 0) {
                                tot = (tot + dp[i][s][q - 1]) % mod;
                            }   
                        }
                        dp[i][j][q] = tot;
                    }
                }
            }
        }


        return dp[0][0][k];
    }
};
10100