<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:
    int f(int i, int j, vector<vector<int>>& A, vector<vector<bool>>& visited){
        size_t n = A.size();
        size_t m = A[0].size();

        if(A[i][j] == 0) return 0;
        if(visited[i][j]) return 0;

        visited[i][j] = true;

        int left = A[i][j] + (j >= 1 ? f(i, j - 1, A, visited) : 0);
        int right = A[i][j] + (j < m-1 ? f(i, j + 1, A, visited) : 0);

        int top = A[i][j] + ( i >=1 ? f(i-1, j, A, visited) : 0);
        int down = A[i][j] + (i < n-1 ? f(i+1, j, A, visited) : 0);

        visited[i][j] = false;
        int paths[4] = {left, right, top, down};
        return *max_element(paths, paths + 4);
    }

    int getMaximumGold(vector<vector<int>>& grid) {
        size_t n = grid.size();
        size_t m = grid[0].size();
        int ans = INT_MIN;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                vector<vector<bool>> visited(n, vector<bool>(m, false));
                ans = max(ans, f(i, j, grid, visited));
            }
        }
        return ans;
    }
};
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