<p>You are given an <code>n x n</code> <code>grid</code> representing a field of cherries, each cell is one of three possible integers.</p>

<ul>
	<li><code>0</code> means the cell is empty, so you can pass through,</li>
	<li><code>1</code> means the cell contains a cherry that you can pick up and pass through, or</li>
	<li><code>-1</code> means the cell contains a thorn that blocks your way.</li>
</ul>

<p>Return <em>the maximum number of cherries you can collect by following the rules below</em>:</p>

<ul>
	<li>Starting at the position <code>(0, 0)</code> and reaching <code>(n - 1, n - 1)</code> by moving right or down through valid path cells (cells with value <code>0</code> or <code>1</code>).</li>
	<li>After reaching <code>(n - 1, n - 1)</code>, returning to <code>(0, 0)</code> by moving left or up through valid path cells.</li>
	<li>When passing through a path cell containing a cherry, you pick it up, and the cell becomes an empty cell <code>0</code>.</li>
	<li>If there is no valid path between <code>(0, 0)</code> and <code>(n - 1, n - 1)</code>, then no cherries can be collected.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/14/grid.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,-1],[1,0,-1],[1,1,1]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The player started at (0, 0) and went down, down, right right to reach (2, 2).
4 cherries were picked up during this single trip, and the matrix becomes [[0,1,-1],[0,0,-1],[0,0,0]].
Then, the player went left, up, up, left to return home, picking up one more cherry.
The total number of cherries picked up is 5, and this is the maximum possible.
</pre>

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

<pre>
<strong>Input:</strong> grid = [[1,1,-1],[1,-1,1],[-1,1,1]]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>grid[i][j]</code> is <code>-1</code>, <code>0</code>, or <code>1</code>.</li>
	<li><code>grid[0][0] != -1</code></li>
	<li><code>grid[n - 1][n - 1] != -1</code></li>
</ul>

class Solution 
{
public:
    int dp[60][60][60][60];
    int find(vector<vector<int>>&mat,int r1,int c1,int r2,int c2){
        if(r1>=mat.size()||c1>=mat[0].size()||r2>=mat.size()||c2>=mat[0].size()||mat[r1][c1]==-1||mat[r2][c2]==-1){
            return INT_MIN;
        }
        if(r1==mat.size()-1&&c1==mat[0].size()-1){
            return mat[r1][c1];
        }
        if(r2==mat.size()-1&&c2==mat[0].size()-1){
            return mat[r2][c2];
        }
        if(dp[r1][c1][r2][c2]!=-1){
            return dp[r1][c1][r2][c2];
        }
        int c=0;
        if(r1==r2&&c1==c2){
            c+=mat[r1][c1];
        }
        else{
            c+=mat[r1][c1]+mat[r2][c2];
        }
        int f1=find(mat,r1,c1+1,r2,c2+1);
        int f2=find(mat,r1+1,c1,r2+1,c2);
        int f3=find(mat,r1+1,c1,r2,c2+1);
        int f4=find(mat,r1,c1+1,r2+1,c2);
        c+=max({f1,f2,f3,f4});
        return dp[r1][c1][r2][c2]=c;
    }
    int cherryPickup(vector<vector<int>>& mat){
        memset(dp,-1,sizeof(dp));
        int n=mat.size();
        int m=mat[0].size();
        int val=find(mat,0,0,0,0);
        return max(0,val);
    }
};
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