<p>You have an undirected, connected graph of <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>. You are given an array <code>graph</code> where <code>graph[i]</code> is a list of all the nodes connected with node <code>i</code> by an edge.</p>

<p>Return <em>the length of the shortest path that visits every node</em>. You may start and stop at any node, you may revisit nodes multiple times, and you may reuse edges.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/12/shortest1-graph.jpg" style="width: 222px; height: 183px;" />
<pre>
<strong>Input:</strong> graph = [[1,2,3],[0],[0],[0]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> One possible path is [1,0,2,0,3]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/12/shortest2-graph.jpg" style="width: 382px; height: 222px;" />
<pre>
<strong>Input:</strong> graph = [[1],[0,2,4],[1,3,4],[2],[1,2]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> One possible path is [0,1,4,2,3]
</pre>

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

<ul>
	<li><code>n == graph.length</code></li>
	<li><code>1 &lt;= n &lt;= 12</code></li>
	<li><code>0 &lt;= graph[i].length &lt;&nbsp;n</code></li>
	<li><code>graph[i]</code> does not contain <code>i</code>.</li>
	<li>If <code>graph[a]</code> contains <code>b</code>, then <code>graph[b]</code> contains <code>a</code>.</li>
	<li>The input graph is always connected.</li>
</ul>

class Solution {
public:
    int shortestPathLength(vector<vector<int>>& G) {
        int n = G.size(), N = 1 << n, res = INT_MAX;
        // cout<<N<<endl;
        for(int p = 0; p < n; ++p)
        {
            vector<vector<int>> v(n, vector<int>(N));
            queue<array<int, 2>> qu;
            int t = 1 << p, L = 0;
            qu.push({p, t});
            v[p][t] = 1;
            while(!qu.empty())
            {
                for(int k = qu.size(); k > 0; --k)
                {
                    auto [i, m] = qu.front();
                    qu.pop();
                    // cout<<m<<' ';
                    if(m == N - 1)
                    {
                        res = L;
                        break;
                    }
                    for(int j : G[i])
                    {
                        int m2 = m | 1 << j;
                        if(!v[j][m2]++)
                            qu.push({j, m2});
                    }
                }
                if(++L >= res) break;
            }
        }
        return res;
    }
};
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