<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 kin) {
        vector<vector<vector<long long>>> dp(
            pizza.size(),
            vector<vector<long long>>(
                pizza[0].size(),
                vector<long long>(kin+1, 0)
            )
        );
        vector<vector<int>> apples(
            pizza.size(),
            vector<int>(pizza[0].size(), 0)
        );
        for(int i=pizza.size()-1; i>=0; i--){
            for(int j=pizza[i].length()-1; j>=0; j--){
                if(pizza[i][j]=='A') apples[i][j] = 1;
                if(i==pizza.size()-1 && j<pizza[i].length()-1)
                    apples[i][j] += apples[i][j+1];
                else if(i<pizza.size()-1 && j==pizza[i].length()-1)
                    apples[i][j] += apples[i+1][j];
                else if(i<pizza.size()-1 && j<pizza[i].length()-1){
                    apples[i][j] += apples[i+1][j] + apples[i][j+1] - apples[i+1][j+1];
                }
            }
        }
        for(int i=pizza.size()-1; i>=0; i--){
            for(int j=pizza[i].length()-1; j>=0; j--){
                for(int k=1; k<=kin; k++){
                    if(k==1){
                        dp[i][j][k] = apples[i][j]>0 ? 1 : 0;
                    } else{
                        dp[i][j][k] = 0;
                        for(int h=i+1; h<pizza.size(); h++){
                            if(apples[i][j] - apples[h][j] > 0){
                                dp[i][j][k] += dp[h][j][k-1];
                                dp[i][j][k] %= 1000000007;
                            }
                        }
                        for(int v=j+1; v<pizza[i].length(); v++){
                            if(apples[i][j] - apples[i][v] > 0){
                                dp[i][j][k] += dp[i][v][k-1];
                                dp[i][j][k] %= 1000000007;
                            }
                        }
                    }
                }
            }
        }
        return dp[0][0][kin];
    }
};
10400