<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>

const int MOD = 1000000007;

class Solution {
public:
    int ways(vector<string>& pizza, int k) {
        int rows = pizza.size(), cols = pizza[0].size();
        vector<vector<int>> count(rows + 1, vector<int>(cols + 1));
        for (int i = rows - 1; i >= 0; i--) {
            for (int j = cols - 1; j >= 0; j--) {
                count[i][j] = count[i + 1][j] + count[i][j + 1] - count[i + 1][j + 1] + (pizza[i][j] == 'A');
            }
        }
        vector<vector<vector<int>>> dp(rows, vector<vector<int>>(cols, vector<int>(k + 1)));
        for (int p = 1; p <= k; p++) {
            for (int i = rows - 1; i >= 0; i--) {
                for (int j = cols - 1; j >= 0; j--) {
                    if (p == 1) {
                        dp[i][j][p] = (count[i][j] > 0);
                    } else {
                        int ways = 0;
                        for (int q = i + 1; q < rows; q++) {
                            if (count[i][j] - count[q][j] > 0) {
                                ways = (ways + dp[q][j][p-1]) % MOD;
                            }
                        }
                        for (int q = j + 1; q < cols; q++) {
                            if (count[i][j] - count[i][q] > 0) {
                                ways = (ways + dp[i][q][p-1]) % MOD;
                            }
                        }
                        dp[i][j][p] = ways;
                    }
                }
            }
        }
        return dp[0][0][k];
    }
};
10200