<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 f(int i1, int j1, int i2, int j2, vector<vector<int>>& a, vector<vector<vector<vector<int>>>>& dp) {
        int n = a.size();
        if (i1 >= n || i2 >= n || j1 >= n || j2 >= n || a[i1][j1] == -1 || a[i2][j2] == -1) return INT_MIN;
        if (i1 == n-1 && j1 == n-1) return a[i1][j1];
        
        if (dp[i1][j1][i2][j2] != -1) return dp[i1][j1][i2][j2];
        
        int cherries = a[i1][j1];
        if (i1 != i2 || j1 != j2) cherries += a[i2][j2];
        
        int maxCherries = max({
            f(i1+1, j1, i2+1, j2, a, dp),
            f(i1+1, j1, i2, j2+1, a, dp),
            f(i1, j1+1, i2+1, j2, a, dp),
            f(i1, j1+1, i2, j2+1, a, dp)
        });
        
        if (maxCherries == INT_MIN) return dp[i1][j1][i2][j2] = INT_MIN;
        
        dp[i1][j1][i2][j2] = cherries + maxCherries;
        return dp[i1][j1][i2][j2];
    }

    int cherryPickup(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<vector<vector<int>>>> dp(n, vector<vector<vector<int>>>(n, vector<vector<int>>(n, vector<int>(n, -1))));
        return max(0, f(0, 0, 0, 0, grid, dp));
    }
};

317459