<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 rows, cols;
    vector<vector<int>> grid;

    int memo[55][55][55];

    int dp(int col, int forward, int backward){
        
        if (col == cols){
            return 0;
        }
        if (memo[col][forward][backward] != -1){
            return memo[col][forward][backward];
        }
        int ans = INT_MIN;
        unordered_map<int, int> rc;
        for (int f = forward; f < rows; ++f){
            if (grid[f][col] == -1){
                break;
            }
            if (grid[f][col] == 1){
                ++rc[f];
            }
            if (col == cols-1 && f != rows - 1){
                continue;
            }
            auto rc1 = rc;
            for (int b = backward; b < rows; ++b){
                if (grid[b][col] == -1){
                    break;
                }
                if (grid[b][col] == 1){
                    ++rc1[b];
                }
                if (col == cols - 1 && b != rows - 1){
                    continue;
                }
                if (b < f){
                    continue;
                }
                ans = max(ans, dp(col+1, f, b) + int(rc1.size()));                
            }



        }
        memo[col][forward][backward] = ans;
        //cout << col << " " << backward << " " << forward << " " << ans << endl;
        return ans;
    }

    int cherryPickup(vector<vector<int>>& grid_) {
        memset(memo, -1, sizeof(memo));
        grid = move(grid_);
        rows = grid.size();
        cols = grid[0].size();
        return max(dp(0, 0, 0), 0);
    }
};
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