<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)
        distance = [[0 for _ in range(N)] for _ in range(N)]
        def get_distance(i, j):
            q = deque()
            q.append([i,0])
            visited = set()
            visited.add(i)
            while q:
                cur, level = q.popleft()
                if cur == j:
                    return level
                for next in graph[cur]:
                    if next not in visited:
                        visited.add(next)
                        q.append([next, level+1])
            return 0

        for i in range(N):
            for j in range(i+1):
                distance[i][j] = distance[j][i] = get_distance(i, j)
        #print(distance)

        @cache
        def dp(cur, nexts):
            if nexts == 0:
                return 0
            res = float(inf)
            for i in range(N):
                if (1 << i) & nexts:
                    if i == cur:
                        continue
                    res = min(res, distance[cur][i] + dp(i, nexts & (~(1 << i))))
            #print("dp for cur = ", cur, ", nexts = ", bin(nexts), ", res = ", res)
            return res

        res = float(inf)
        for i in range(N):
            res = min(res, dp(i, ((1 << N) - 1) & (~(1 << i))))
        return res

51500