<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 n=pizza.size(), m=pizza[0].size();
        const int mod = 1e9+7;
        // int apples = 0;
        // for(auto &i:pizza){
        //     for(auto j:i) if(j=='A') apples++;
        // }
        vector<vector<vector<int>>> dp(n,vector<vector<int>>(m,vector<int>(k,-1)));
        function<int(int,int,int)> solve = [&](int r, int c, int cuts){
            if(r>=n || c>=m) return 0;
            if(dp[r][c][cuts]!=-1) return dp[r][c][cuts];
            int apple = 0;
            for(int i=r;i<n;i++){
                for(int j=c;j<m;j++) if(pizza[i][j]=='A') apple++;
            }
            if(apple==0) return dp[r][c][cuts]=0;
            if(cuts==0) return dp[r][c][cuts]=1;
            dp[r][c][cuts]=0;
            for(int i=r+1;i<n;i++){
                int a=0;
                for(int x=i;x<n;x++){
                    for(int y=c;y<m;y++) if(pizza[x][y]=='A') a++;
                }
                if(a<apple) dp[r][c][cuts] = (dp[r][c][cuts] + solve(i,c,cuts-1)) % mod;
            }
            for(int j=c+1;j<m;j++){
                int a=0;
                for(int x=r;x<n;x++){
                    for(int y=j;y<m;y++) if(pizza[x][y]=='A') a++;
                }
                if(a<apple) dp[r][c][cuts] = (dp[r][c][cuts] + solve(r,j,cuts-1)) % mod;
            }
            cout<<"DP "<<r<<" "<<c<<" "<<cuts<<" "<<dp[r][c][cuts]<<"\n";
            return dp[r][c][cuts];

        };


        return solve(0,0,k-1);
    }
};
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