Cherry Pickup II
<p>You are given a <code>rows x cols</code> matrix <code>grid</code> representing a field of cherries where <code>grid[i][j]</code> represents the number of cherries that you can collect from the <code>(i, j)</code> cell.</p>

<p>You have two robots that can collect cherries for you:</p>

<ul>
	<li><strong>Robot #1</strong> is located at the <strong>top-left corner</strong> <code>(0, 0)</code>, and</li>
	<li><strong>Robot #2</strong> is located at the <strong>top-right corner</strong> <code>(0, cols - 1)</code>.</li>
</ul>

<p>Return <em>the maximum number of cherries collection using both robots by following the rules below</em>:</p>

<ul>
	<li>From a cell <code>(i, j)</code>, robots can move to cell <code>(i + 1, j - 1)</code>, <code>(i + 1, j)</code>, or <code>(i + 1, j + 1)</code>.</li>
	<li>When any robot passes through a cell, It picks up all cherries, and the cell becomes an empty cell.</li>
	<li>When both robots stay in the same cell, only one takes the cherries.</li>
	<li>Both robots cannot move outside of the grid at any moment.</li>
	<li>Both robots should reach the bottom row in <code>grid</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1463.Cherry%20Pickup%20II/images/sample_1_1802.png" style="width: 374px; height: 501px;" />
<pre>
<strong>Input:</strong> grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
<strong>Output:</strong> 24
<strong>Explanation:</strong> Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (3 + 2 + 5 + 2) = 12.
Cherries taken by Robot #2, (1 + 5 + 5 + 1) = 12.
Total of cherries: 12 + 12 = 24.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1463.Cherry%20Pickup%20II/images/sample_2_1802.png" style="width: 500px; height: 452px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0,0,0,0,1],[2,0,0,0,0,3,0],[2,0,9,0,0,0,0],[0,3,0,5,4,0,0],[1,0,2,3,0,0,6]]
<strong>Output:</strong> 28
<strong>Explanation:</strong> Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (1 + 9 + 5 + 2) = 17.
Cherries taken by Robot #2, (1 + 3 + 4 + 3) = 11.
Total of cherries: 17 + 11 = 28.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>rows == grid.length</code></li>
	<li><code>cols == grid[i].length</code></li>
	<li><code>2 &lt;= rows, cols &lt;= 70</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 100</code></li>
</ul>
Array; Dynamic Programming; Matrix
class Solution {
public:
    int cherryPickup(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int f[m][n][n];
        memset(f, -1, sizeof(f));
        f[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
        for (int i = 1; i < m; ++i) {
            for (int j1 = 0; j1 < n; ++j1) {
                for (int j2 = 0; j2 < n; ++j2) {
                    int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
                    for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
                        for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
                            if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] != -1) {
                                f[i][j1][j2] = max(f[i][j1][j2], f[i - 1][y1][y2] + x);
                            }
                        }
                    }
                }
            }
        }
        int ans = 0;
        for (int j1 = 0; j1 < n; ++j1) {
            for (int j2 = 0; j2 < n; ++j2) {
                ans = max(ans, f[m - 1][j1][j2]);
            }
        }
        return ans;
    }
};