<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>

#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

class Solution {
public:
    int cherryPickup(vector<vector<int>>& grid) 
    {
        int n = grid.size();
         // Possible moves
        vector<pair<int, int>> dirs = {{1, 0}, {0, 1}};
        vector<vector<vector<int>>> dp(n, vector<vector<int>>(n, vector<int>(n, INT_MIN)));
        queue<tuple<int, int, int, int, int>> q;

        // Initial state
        q.emplace(0, 0, 0, 0, grid[0][0]);
        dp[0][0][0] = grid[0][0];

        int maxCherries = 0;
        while (!q.empty()) 
        {
            auto [r1, c1, r2, c2, cherries] = q.front(); q.pop();

            // Reached the end
            if (r1 == n-1 && c1 == n-1) 
            {
                maxCherries = max(maxCherries, cherries);
                continue;
            }

            for (auto& [xoff1, yoff1] : dirs) 
            {
                for (auto& [xoff2, yoff2] : dirs) 
                {
                    int nr1 = r1 + xoff1;
                    int nc1 = c1 + yoff1;
                    int nr2 = r2 + xoff2;
                    int nc2 = c2 + yoff2;

                    if (nr1 < n && nc1 < n && nr2 < n && nc2 < n && grid[nr1][nc1] != -1 && grid[nr2][nc2] != -1) 
                    {
                        int newCherries = cherries;
                        if (nr1 == nr2 && nc1 == nc2)   newCherries += grid[nr1][nc1];
                        else                            newCherries += grid[nr1][nc1] + grid[nr2][nc2];

                        if (dp[nr1][nc1][nc2] < newCherries)
                        {
                            dp[nr1][nc1][nc2] = newCherries;
                            q.emplace(nr1, nc1, nr2, nc2, newCherries);
                        }
                    }
                }
            }
        }

        return max(maxCherries, 0);
    }
};

180084