<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 {
    int dp[55][55][55][55];
public:

int f(int r1, int r2, int c1, int c2, int n, vector<vector<int>>& grid) {

    if (r1 >= n || r2 >= n || c1 >= n || c2 >= n || grid[r1][c1] == -1 || grid[r2][c2] == -1)
        return INT_MIN;

    if (r1 == n-1 && c1 == n-1)
        return grid[r1][c1];

    if (dp[r1][r2][c1][c2] != -1)
        return dp[r1][r2][c1][c2];

    int ans = INT_MIN;

    for (int k1 = 0; k1 <= 1; k1++) {
        for (int k2 = 0; k2 <= 1; k2++) {

            int val = 0;
            if (r1 == r2 && c1 == c2) {
                val += grid[r1][c1];
            } else {
                val = grid[r1][c1] + grid[r2][c2];
            }

            ans = max(ans, val + f(r1 + k1, r2 + k2, c1 + 1 - k1, c2 + 1 - k2, n, grid));

        }
    }
    return dp[r1][r2][c1][c2] = ans;
}

int cherryPickup(vector<vector<int>>& grid) {

    int n = grid.size();
    if (grid[0][0] == -1 || grid[n-1][n-1] == -1)
        return 0;

    memset(dp, -1, sizeof(dp));

    return max(0, f(0, 0, 0, 0, n, grid));
}
};

49578