<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>

using pt = pair<int, int>;
using ppt = pair<pt, int>;

const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};

// int dp[1 << 25];

class Solution {
public:
    int getMaximumGold(vector<vector<int>>& grid) {
        vector<pt> gpositions;
        vector<int> weight;
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> index(n, vector<int>(m, -1));
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (grid[i][j]) {
                    index[i][j] = gpositions.size();
                    gpositions.emplace_back(i, j);
                    weight.push_back(grid[i][j]);
                }
            }
        }
        int g = gpositions.size();
        vector<vector<int>> gr(g);
        for (int i = 0; i < g; ++i) {
            for (int dir = 0; dir < 4; ++dir) {
                int x = gpositions[i].first + dx[dir];
                int y = gpositions[i].second + dy[dir];
                if (!(0 <= x && x < n && 0 <= y && y < m)) continue;
                if (index[x][y] == -1) continue;
                gr[i].push_back(index[x][y]);
            }
        }

        // memset(dp, 0, sizeof(dp));
        const int fmask = (1 << g) - 1;
        queue<ppt> q;
        int ans = 0;
        for (int i = 0; i < g; ++i) {
            int bit = 1 << i;
            int mask = fmask & (~bit);
            const auto& pos = gpositions[i];
            // dp[mask] |= bit;
            q.emplace(pt{mask, i}, weight[i]);
            ans = max(ans, weight[i]);
        }
        while (!q.empty()) {
            ppt p = q.front();
            int mask = p.first.first;
            int fm = p.first.second;
            int w = p.second;
            q.pop();
            for (int to : gr[fm]) {
                int bit = 1 << to;
                if ((mask & bit) == 0) continue;
                int nmask = mask & (~bit);
                // if ((dp[nmask] & bit) == 0) {
                //    dp[nmask] |= bit;
                    int nw = w + weight[to];
                    q.emplace(pt{nmask, to}, nw);
                    ans = max(ans, nw);
                // }
                
            }
        }
        return ans;
    }
};
78200