<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 calc(int src,vector<vector<int>>&g){
    int n=g.size();
    int dp[n][1<<n];
    int vis[n][1<<n];
    memset(vis,0,sizeof(vis));
    for(int i=0;i<n;i++){
        for(int j=0;j<(1<<n);j++)dp[i][j]=1e9;
    }
    dp[src][(1<<src)]=0;
    set<pair<int,pair<int,int>>>s;
    s.insert({0,{src,(1<<src)}});
    while(!s.empty()){
        auto node=*s.begin();
        s.erase(s.begin());
        int v=node.second.first;
        int mask=node.second.second;
        if(vis[v][mask])continue;
        vis[v][mask]=1;
        for(auto &c:g[v]){
            int new_mask=mask|(1<<c);
            if(dp[c][new_mask]>dp[v][mask]+1){
                dp[c][new_mask]=dp[v][mask]+1;
                s.insert({dp[c][new_mask],{c,new_mask}});
            }
        }
    }
    int mini=1e9;
    for(int i=0;i<n;i++){
        mini=min(dp[i][(1<<n)-1],mini);
    }
    return mini;
}
    int shortestPathLength(vector<vector<int>>&graph) {
        int n=graph.size();
        int ans=1e9;
        for(int i=0;i<n;i++){
           ans=min(ans,calc(i,graph));
        }
        return ans;
    }
};
53270