<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:
    def shortestPathLength(self, graph: List[List[int]]) -> int:
        n = len(graph)
        dist = [[inf]*n for _ in range(n)]
        
        for i, x in enumerate(graph): 
            dist[i][i] = 0
            for ii in x: dist[i][ii] = 1
        
        # floyd-warshall 
        for k in range(n): 
            for i in range(n): 
                for j in range(n): 
                    dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
        
        @cache 
        def fn(x, mask): 
            if mask == 0: return 0 
            ans = inf 
            for i in range(n): 
                if mask & (1 << i): 
                    ans = min(ans, dist[x][i] + fn(i, mask ^ (1<<i)))
            return ans 
        
        return min(fn(x, (1 << n)-1) for x in range(n))
51300