Number of Possible Sets of Closing Branches
<p>There is a company with <code>n</code> branches across the country, some of which are connected by roads. Initially, all branches are reachable from each other by traveling some roads.</p>

<p>The company has realized that they are spending an excessive amount of time traveling between their branches. As a result, they have decided to close down some of these branches (<strong>possibly none</strong>). However, they want to ensure that the remaining branches have a distance of at most <code>maxDistance</code> from each other.</p>

<p>The <strong>distance</strong> between two branches is the <strong>minimum</strong> total traveled length needed to reach one branch from another.</p>

<p>You are given integers <code>n</code>, <code>maxDistance</code>, and a <strong>0-indexed</strong> 2D array <code>roads</code>, where <code>roads[i] = [u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>]</code> represents the <strong>undirected</strong> road between branches <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with length <code>w<sub>i</sub></code>.</p>

<p>Return <em>the number of possible sets of closing branches, so that any branch has a distance of at most </em><code>maxDistance</code><em> from any other</em>.</p>

<p><strong>Note</strong> that, after closing a branch, the company will no longer have access to any roads connected to it.</p>

<p><strong>Note</strong> that, multiple roads are allowed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2959.Number%20of%20Possible%20Sets%20of%20Closing%20Branches/images/example11.png" style="width: 221px; height: 191px;" />
<pre>
<strong>Input:</strong> n = 3, maxDistance = 5, roads = [[0,1,2],[1,2,10],[0,2,10]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The possible sets of closing branches are:
- The set [2], after closing, active branches are [0,1] and they are reachable to each other within distance 2.
- The set [0,1], after closing, the active branch is [2].
- The set [1,2], after closing, the active branch is [0].
- The set [0,2], after closing, the active branch is [1].
- The set [0,1,2], after closing, there are no active branches.
It can be proven, that there are only 5 possible sets of closing branches.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2959.Number%20of%20Possible%20Sets%20of%20Closing%20Branches/images/example22.png" style="width: 221px; height: 241px;" />
<pre>
<strong>Input:</strong> n = 3, maxDistance = 5, roads = [[0,1,20],[0,1,10],[1,2,2],[0,2,2]]
<strong>Output:</strong> 7
<strong>Explanation:</strong> The possible sets of closing branches are:
- The set [], after closing, active branches are [0,1,2] and they are reachable to each other within distance 4.
- The set [0], after closing, active branches are [1,2] and they are reachable to each other within distance 2.
- The set [1], after closing, active branches are [0,2] and they are reachable to each other within distance 2.
- The set [0,1], after closing, the active branch is [2].
- The set [1,2], after closing, the active branch is [0].
- The set [0,2], after closing, the active branch is [1].
- The set [0,1,2], after closing, there are no active branches.
It can be proven, that there are only 7 possible sets of closing branches.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 1, maxDistance = 10, roads = []
<strong>Output:</strong> 2
<strong>Explanation:</strong> The possible sets of closing branches are:
- The set [], after closing, the active branch is [0].
- The set [0], after closing, there are no active branches.
It can be proven, that there are only 2 possible sets of closing branches.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10</code></li>
	<li><code>1 &lt;= maxDistance &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= roads.length &lt;= 1000</code></li>
	<li><code>roads[i].length == 3</code></li>
	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li><code>1 &lt;= w<sub>i</sub> &lt;= 1000</code></li>
	<li>All branches are reachable from each other by traveling some roads.</li>
</ul>
Bit Manipulation; Graph; Enumeration; Shortest Path; Heap (Priority Queue)
class Solution {
public:
    int numberOfSets(int n, int maxDistance, vector<vector<int>>& roads) {
        int ans = 0;
        for (int mask = 0; mask < 1 << n; ++mask) {
            int g[n][n];
            memset(g, 0x3f, sizeof(g));
            for (auto& e : roads) {
                int u = e[0], v = e[1], w = e[2];
                if ((mask >> u & 1) & (mask >> v & 1)) {
                    g[u][v] = min(g[u][v], w);
                    g[v][u] = min(g[v][u], w);
                }
            }
            for (int k = 0; k < n; ++k) {
                if (mask >> k & 1) {
                    g[k][k] = 0;
                    for (int i = 0; i < n; ++i) {
                        for (int j = 0; j < n; ++j) {
                            g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
                        }
                    }
                }
            }
            int ok = 1;
            for (int i = 0; i < n && ok == 1; ++i) {
                for (int j = 0; j < n && ok == 1; ++j) {
                    if ((mask >> i & 1) & (mask >> j & 1) && g[i][j] > maxDistance) {
                        ok = 0;
                    }
                }
            }
            ans += ok;
        }
        return ans;
    }
};