<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 {
    int n;
public:
    int shortestPathLength(vector<vector<int>>& graph) {
        n = graph.size();
        if(!n)
            return 0;
        int ret = INT_MAX;
        for(int i = 0;i<n;i++){
            auto nex = bfs(graph, i);
            if(nex != -1){
           //     cout<<i<<" "<<nex<<endl;
                ret = min(ret, nex);
            }
                
        }
        return ret == INT_MAX?-1:ret;
    }
    
    int bfs(const vector<vector<int>>& graph, int start){
        int target = (1<<n)-1;
        vector<vector<int>>visited(n, vector<int>(target+1, 0));
        queue<pair<int,int>>to_visit;
        to_visit.push({start, 1<<start});
        visited[start][1<<start] = 1;//1<<start;
        int lvl = 0;
        while(!to_visit.empty()){
            auto s = to_visit.size();
            while(s--){
                auto [cur, state] = to_visit.front();
                to_visit.pop();
                if(state == target)
                    return lvl;
                for(const auto&nex:graph[cur]){
                    auto nex_state = state|(1<<nex);
                    if(!visited[nex][nex_state]){
                        to_visit.push({nex, nex_state});
                        visited[nex][nex_state] = 1;
                    }
                }
            }
            lvl++;
        }
        return -1;
    }
};
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