<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>

// BFS
// O(n*n*2^n)
// n*2^n max phan tu trong queue -> O(n*2^n) (link moi node voi max 2^n mask value, co n node)
// voi moi phan tu(node): co the visit n-1 canh -> O(n)
// => O(n*2^n) * O(n)=O(n^2*2^n)=12^2 * 2^12=589824=1e6 (<1sec)

// note:
// neu giai hoan vi+bfs
// vi du voi hoan vi: 1023 (tim min path 1-0 + min path 0-2 + min path 2 -3)
// hoan vi: N!
// min path(i-j): O(N) (bfs)
// N-1 lan tinh min path(i-j): O(N)
// O(N!*n*n) -> TLE voi N=12
class Solution {
public:
    int shortestPathLength(vector<vector<int>>& a) {
        // a[i] ->{j,l,k}: canh [i,j], [i,l], [i,k]
        int n=a.size();
        queue<tuple<int,int,int>> q;// node, distance, mark: mark cac node da di qua
        vector<vector<bool>> vi(n,vector<bool>(1<<n));
        for(int i=0;i<n;i++){
            q.push({i,0,(1<<i)});
            vi[i][1<<i]=true;
        }
        while(!q.empty()){
            auto cur=q.front();
            q.pop();
            int i=get<0>(cur), d=get<1>(cur), mask=get<2>(cur);
            if (mask==(1<<n)-1) return d;
            for(auto &x:a[i]){
                int newmask=mask|(1<<x);
                if (vi[x][newmask]) continue;
                q.push({x,d+1,newmask});
                vi[x][newmask]=true;
            }
        }
        return 0;
    }
};
12906