<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:
    typedef long long ll;
    ll n;
    vector<vector<int>> a;
    vector<vector<vector<ll>>> dp;
    ll dfs(ll r1, ll c1, ll r2)
    {
        ll c2 = r1 + c1 - r2;
        if(r1 == n - 1 && c1 == n - 1 && r2 == n - 1 && c2 == n - 1)
            return a[n - 1][n - 1];
        if(dp[r1][c1][r2] != -1)
            return dp[r1][c1][r2];
        
        vector<vector<ll>> dxdy = {
            {0, 1, 0, 1},
            {1, 0, 1, 0},
            {0, 1, 1, 0},
            {1, 0, 0, 1},
        };

        function<bool(ll, ll)> can = [&](ll i, ll j) {
            return 0 <= i && i < n && 0 <= j && j < n && a[i][j] != -1;
        };

        ll ans = -1e9;
        bool hasPath = false;
        for(vector<ll> i : dxdy)
        {
            if(can(r1 + i[0], c1 + i[1]) && can(r2 + i[2], c2 + i[3]))
            {
                ans = max(ans, dfs(r1 + i[0], c1 + i[1], r2 + i[2]));
            }
        }

        if(r1 == r2 && c1 == c2)
            ans += a[r1][c1];
        else
            ans += a[r1][c1] + a[r2][c2];
        
        return dp[r1][c1][r2] = ans;
    }
    int cherryPickup(vector<vector<int>> &_a)
    {
        a = _a, n = a.size();
        dp.resize(n, vector<vector<ll>>(n, vector<ll>(n, -1)));
        return max(dfs(0, 0, 0), 0LL);
    }
};
214428