<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[51][51][51][51];

int solve(int i,int j,int k,int l,int n,vector<vector<int>>& g){
     if(i>=n||k>=n||j>=n||l>=n||g[i][j]==-1||g[k][l]==-1) return INT_MIN;
     if(i>=n && j>=n)return 0;
     if(k==n-1 && l==n-1)return g[k][l];
     if(dp[i][j][k][l]!=-1)return dp[i][j][k][l];
     int ans=0;
     if(i==k && j==l)ans+=g[i][j];
     else{
        ans+=g[i][j];
        ans+=g[k][l];
     }
     int g1=solve(i,j+1,k,l+1,n,g);
    int h=solve(i+1,j,k+1,l,n,g);

     int i1=solve(i+1,j,k,l+1,n,g);

     int j1=solve(i,j+1,k+1,l,n,g);
    vector<int>a;
    a.push_back(g1);
      a.push_back(h);
      a.push_back(i1);
      a.push_back(j1);
    sort(a.rbegin(),a.rend());
     ans+=a[0];
     return dp[i][j][k][l]=ans;

}
    int cherryPickup(vector<vector<int>>& g) {
        int n=g.size();
        int r1=0,c1=0,r2=0,c2=0;
        memset(dp,-1,sizeof(dp));
        return max(0,solve(0,0,0,0,n,g));
    }
};
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