<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 cherryPickup(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<vector<int>>> dp(n, vector<vector<int>>(n, vector<int>(n, -1)));
        queue<vector<int>> q;
        q.push({0, 0, 0});
        int x[] = {-1, 1, 0, 0}, y[] = {0, 0, -1, 1};
        int ans = 0;
        dp[0][0][0] = grid[0][0];
        while (!q.empty()) {
            int x1 = q.front()[0];
            int y1 = q.front()[1];
            int y2 = q.front()[2];
            q.pop();
            int x2 = x1 + y1 - y2;
            int curr = dp[x1][y1][y2];
            if (x1 == n - 1 && y1 == n - 1) {
                ans = max(ans, dp[n - 1][n - 1][n - 1]);
                continue;
            }

            // Consider all 4 possible moves
            int dx[] = {1, 0};
            int dy[] = {0, 1};

            for (int i = 0; i < 2; ++i) {
                int nx1 = x1 + dx[i];
                int ny1 = y1 + dy[i];
                if (nx1 < n && ny1 < n && grid[nx1][ny1] >= 0) {
                    for (int j = 0; j < 2; ++j) {
                        int nx2 = x2 + dx[j];
                        int ny2 = y2 + dy[j];
                        if (nx2 < n && ny2 < n && grid[nx2][ny2] >= 0) {
                            int cherries = grid[nx1][ny1] + grid[nx2][ny2];
                            if (nx1 == nx2 && ny1 == ny2) {
                                cherries -= grid[nx1][ny1];
                            }
                            if (dp[nx1][ny1][ny2] < curr + cherries) {
                                dp[nx1][ny1][ny2] = curr + cherries;
                                q.push({nx1, ny1, ny2});
                            }
                        }
                    }
                }
            }
        }
        return max(0, dp[n - 1][n - 1][n - 1]);
    }
};

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