<p>In a gold mine <code>grid</code> of size <code>m x n</code>, each cell in this mine has an integer representing the amount of gold in that cell, <code>0</code> if it is empty.</p>

<p>Return the maximum amount of gold you can collect under the conditions:</p>

<ul>
	<li>Every time you are located in a cell you will collect all the gold in that cell.</li>
	<li>From your position, you can walk one step to the left, right, up, or down.</li>
	<li>You can&#39;t visit the same cell more than once.</li>
	<li>Never visit a cell with <code>0</code> gold.</li>
	<li>You can start and stop collecting gold from <strong>any </strong>position in the grid that has some gold.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,6,0],[5,8,7],[0,9,0]]
<strong>Output:</strong> 24
<strong>Explanation:</strong>
[[0,6,0],
 [5,8,7],
 [0,9,0]]
Path to get the maximum gold, 9 -&gt; 8 -&gt; 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
<strong>Output:</strong> 28
<strong>Explanation:</strong>
[[1,0,7],
 [2,0,6],
 [3,4,5],
 [0,3,0],
 [9,0,20]]
Path to get the maximum gold, 1 -&gt; 2 -&gt; 3 -&gt; 4 -&gt; 5 -&gt; 6 -&gt; 7.
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 15</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 100</code></li>
	<li>There are at most <strong>25 </strong>cells containing gold.</li>
</ul>

class Solution {
public:
    const vector<tuple<int, int>> dxy = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

    int get_max(vector<vector<int>>& grid, const int i, const int j)
    {
        if (grid[i][j] <= 0)
        {
            return 0;
        }

        // ! matrix如果只有一个元素的话根本就不会触发for loop，
        // ! 初始化为0是unsafe
        // int ans = 0;
        int ans = grid[i][j];
        grid[i][j] *= -1;
        for (const auto [dx, dy] : dxy)
        {
            const int nx = i + dx;
            const int ny = j + dy;

            if (0 <= nx and nx < grid.size() and 0 <= ny and ny < grid[0].size())
            {
                ans = max(ans, get_max(grid, nx, ny) - grid[i][j]);
            }
        }

        grid[i][j] *= -1;
        return ans;
    }

    int getMaximumGold(vector<vector<int>>& grid) {
        int ans = 0;
        for (int i = 0; i < grid.size(); ++i)
        {
            for (int j = 0; j < grid[0].size(); ++j)
            {
                ans = max(ans, get_max(grid, i, j));
            }
        }

        return ans;
    }
};
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