<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:
    def cherryPickup(self, grid: List[List[int]]) -> int:
        import collections
        n = len(grid)
        f_value = lambda x, y: grid[y][x]
        is_valid = lambda x, y: (f_value(x, y) >= 0) if ((0<=x<=(n-1)) and (0<=y<=(n-1))) else False

        if not is_valid(0, 0) or not is_valid(n-1, n-1):
            return 0
        
        if n==1:
            return f_value(0, 0)
        cur_states = [(0, 0, 0, 0, f_value(0, 0))]
        while True:
            new_state_map = collections.defaultdict(list)
            for x1, y1, x2, y2, val in cur_states:
                # right, right
                if is_valid(x1+1, y1) and is_valid(x2+1, y2):
                    reward = f_value(x1+1, y1)
                    if not ((x1+1==x2+1) and (y1==y2)):
                        reward += f_value(x2+1, y2)
                    new_state_map[(x1+1, y1, x2+1, y2)].append(val + reward)
                # right, down
                if is_valid(x1+1, y1) and is_valid(x2, y2+1):
                    reward = f_value(x1+1, y1)
                    if not ((x1+1==x2) and (y1==y2+1)):
                        reward += f_value(x2, y2+1)
                    new_state_map[(x1+1, y1, x2, y2+1)].append(val + reward)
                # down, right
                if is_valid(x1, y1+1) and is_valid(x2+1, y2):
                    reward = f_value(x1, y1+1)
                    if not ((x1==x2+1) and (y1+1==y2)):
                        reward += f_value(x2+1, y2)
                    new_state_map[(x1, y1+1, x2+1, y2)].append(val + reward)
                # down, down
                if is_valid(x1, y1+1) and is_valid(x2, y2+1):
                    reward = f_value(x1, y1+1)
                    if not ((x1==x2) and (y1+1==y2+1)):
                        reward += f_value(x2, y2+1)
                    new_state_map[(x1, y1+1, x2, y2+1)].append(val + reward)
            cur_states = [(x1, y1, x2, y2, max(values)) for (x1, y1, x2, y2), values in new_state_map.items()]
            if not cur_states:
                return 0
            x1, y1, x2, y2, val = cur_states[0]
            # print(x1+y1, x2+y2, val)
            if (x1==(n-1)) and (y1==(n-1)):
                return val
        return 0



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