Shortest Path Visiting All Nodes
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0847.Shortest%20Path%20Visiting%20All%20Nodes/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0847.Shortest%20Path%20Visiting%20All%20Nodes/images/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>
Bit Manipulation; Breadth-First Search; Graph; Dynamic Programming; Bitmask
class Solution {
public:
    int shortestPathLength(vector<vector<int>>& graph) {
        int n = graph.size();
        queue<pair<int, int>> q;
        bool vis[n][1 << n];
        memset(vis, false, sizeof(vis));
        for (int i = 0; i < n; ++i) {
            q.emplace(i, 1 << i);
            vis[i][1 << i] = true;
        }
        for (int ans = 0;; ++ans) {
            for (int k = q.size(); k; --k) {
                auto [i, st] = q.front();
                q.pop();
                if (st == (1 << n) - 1) {
                    return ans;
                }
                for (int j : graph[i]) {
                    int nst = st | 1 << j;
                    if (!vis[j][nst]) {
                        vis[j][nst] = true;
                        q.emplace(j, nst);
                    }
                }
            }
        }
    }
};