<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 getMaximumGold(vector<vector<int>>& grid) {
        vector<int> sums = {0};
        for(int i = 0; i < grid.size(); i++) {
            for(int j = 0; j < grid[0].size(); j++) {
                if(grid[i][j] > 0) {
                    doit(sums, grid, i, j, 0);
                }
            }
        }
        sort(sums.rbegin(), sums.rend());
        return sums[0];
    }
    void doit(vector<int> &sum, vector<vector<int>> &grid, int r, int c, int cur) {
        cur += grid[r][c]; int row = grid.size(), col = grid[0].size(); int temp = grid[r][c];
        grid[r][c] = 0; bool done = false;
        if(r-1 >= 0 && grid[r-1][c] > 0) {
            done = true;
            doit(sum, grid, r-1, c, cur);
        }
        if(c-1 >= 0 && grid[r][c-1] > 0) {
            done = true;
            doit(sum, grid, r, c-1, cur);
        }
        if(r+1 < row && grid[r+1][c] > 0) {
            done = true;
            doit(sum, grid, r+1, c, cur);
        }
        if(c+1 < col && grid[r][c+1] > 0) {
            done = true;
            doit(sum, grid, r, c+1, cur);
        }
        if(!done) {
            sum.push_back(cur);
        }
        grid[r][c] = temp;
    }
};
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