<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 stepsAway[12][12];
    void calc(vector<vector<int>>& graph,int node){
        unordered_set<int> visited{node};
        deque<int> dq{std::begin(graph[node]),std::end(graph[node])};
        int steps=0;
        while(!dq.empty()){
            steps++;
            int N=dq.size();
            for(int i=0;i<N;i++){
                int neb=dq.front();
                dq.pop_front();
                if(visited.count(neb))continue;
                visited.insert(neb);
                stepsAway[node][neb]=steps;
                for(int v:graph[neb]){
                    if(visited.count(v)==0)dq.push_back(v);
                    //visited.insert(v);
                }

            }
        }
    }
    int end;
    int dp[4096][12];
    int recurse(int visited,int curr, int n){
        if(dp[visited][curr]!=-1)return dp[visited][curr];
        if(visited==end)return 0;
        int best=INT_MAX;
        for(int i=0;i<n;i++){
            if(((visited>>i)&1)==0){
                visited|=(1<<i);
                best=min(best,stepsAway[curr][i]+recurse(visited,i,n));
                visited^=(1<<i);
            }
        }
        dp[visited][curr]=best;
        return best;
    }
    int shortestPathLength(vector<vector<int>>& graph) {
        memset(dp,-1,sizeof(dp));
        int n=graph.size();
        for(int i=0;i<n;i++)calc(graph,i);
        // for(int i=0;i<n;i++){
        //     for(int j=0;j<n;j++)cout<<stepsAway[i][j]<<" ";
        //     cout<<endl;
        // }
        end=1<<n;
        end-=1;
        int best=INT_MAX;
        for(int i=0;i<n;i++){
            best=min(best,recurse((1<<i),i,n));
        }
        return best;
    }
};
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