<p>A city is represented as a <strong>bi-directional connected</strong> graph with <code>n</code> vertices where each vertex is labeled from <code>1</code> to <code>n</code> (<strong>inclusive</strong>). The edges in the graph are represented as a 2D integer array <code>edges</code>, where each <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> denotes a bi-directional edge between vertex <code>u<sub>i</sub></code> and vertex <code>v<sub>i</sub></code>. Every vertex pair is connected by <strong>at most one</strong> edge, and no vertex has an edge to itself. The time taken to traverse any edge is <code>time</code> minutes.</p>

<p>Each vertex has a traffic signal which changes its color from <strong>green</strong> to <strong>red</strong> and vice versa every&nbsp;<code>change</code> minutes. All signals change <strong>at the same time</strong>. You can enter a vertex at <strong>any time</strong>, but can leave a vertex <strong>only when the signal is green</strong>. You <strong>cannot wait </strong>at a vertex if the signal is <strong>green</strong>.</p>

<p>The <strong>second minimum value</strong> is defined as the smallest value<strong> strictly larger </strong>than the minimum value.</p>

<ul>
	<li>For example the second minimum value of <code>[2, 3, 4]</code> is <code>3</code>, and the second minimum value of <code>[2, 2, 4]</code> is <code>4</code>.</li>
</ul>

<p>Given <code>n</code>, <code>edges</code>, <code>time</code>, and <code>change</code>, return <em>the <strong>second minimum time</strong> it will take to go from vertex </em><code>1</code><em> to vertex </em><code>n</code>.</p>

<p><strong>Notes:</strong></p>

<ul>
	<li>You can go through any vertex <strong>any</strong> number of times, <strong>including</strong> <code>1</code> and <code>n</code>.</li>
	<li>You can assume that when the journey <strong>starts</strong>, all signals have just turned <strong>green</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/09/29/e1.png" style="width: 200px; height: 250px;" /> &emsp; &emsp; &emsp; &emsp;<img alt="" src="https://assets.leetcode.com/uploads/2021/09/29/e2.png" style="width: 200px; height: 250px;" />
<pre>
<strong>Input:</strong> n = 5, edges = [[1,2],[1,3],[1,4],[3,4],[4,5]], time = 3, change = 5
<strong>Output:</strong> 13
<strong>Explanation:</strong>
The figure on the left shows the given graph.
The blue path in the figure on the right is the minimum time path.
The time taken is:
- Start at 1, time elapsed=0
- 1 -&gt; 4: 3 minutes, time elapsed=3
- 4 -&gt; 5: 3 minutes, time elapsed=6
Hence the minimum time needed is 6 minutes.

The red path shows the path to get the second minimum time.
- Start at 1, time elapsed=0
- 1 -&gt; 3: 3 minutes, time elapsed=3
- 3 -&gt; 4: 3 minutes, time elapsed=6
- Wait at 4 for 4 minutes, time elapsed=10
- 4 -&gt; 5: 3 minutes, time elapsed=13
Hence the second minimum time is 13 minutes.      
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/09/29/eg2.png" style="width: 225px; height: 50px;" />
<pre>
<strong>Input:</strong> n = 2, edges = [[1,2]], time = 3, change = 2
<strong>Output:</strong> 11
<strong>Explanation:</strong>
The minimum time path is 1 -&gt; 2 with time = 3 minutes.
The second minimum time path is 1 -&gt; 2 -&gt; 1 -&gt; 2 with time = 11 minutes.</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>n - 1 &lt;= edges.length &lt;= min(2 * 10<sup>4</sup>, n * (n - 1) / 2)</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li>There are no duplicate edges.</li>
	<li>Each vertex can be reached directly or indirectly from every other vertex.</li>
	<li><code>1 &lt;= time, change &lt;= 10<sup>3</sup></code></li>
</ul>

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt,arch=pantherlake")

using u16 = uint16_t;

class Solution {
public:
    static bitset<10001> expandCircle(const vector<vector<u16>>& adjacencies, const bitset<10001>& circle) __attribute__ ((__target__ ("avx2,bmi,bmi2,lzcnt,popcnt,arch=pantherlake"))) {
        bitset<10001> new_circle;
        int n = adjacencies.size();
        for (int i = 1; i < n; ++i) {
            if (circle.test(i)) {
                for (auto adj : adjacencies[i]) {
                    new_circle.set(adj);
                }
            }
        }
        return new_circle;
    }
    int secondMinimum(int n, vector<vector<int>>& edges, int time, int change) __attribute__ ((__target__ ("avx2,bmi,bmi2,lzcnt,popcnt,arch=pantherlake"))) {
        vector<vector<u16>> adjacencies(n + 1);
        for (const auto& e : edges) {
            u16 x = e.front(), y=e[1];
            adjacencies[x].push_back(y);
            adjacencies[y].push_back(x);
        }
        for (auto& a : adjacencies) {
            sort(a.begin(), a.end());
        }
        int n_expansions = 0;
        int o_expansions = 0;
        bitset<10001> ncircle, ocircle;
        ncircle.set(n);
        ocircle.set(1);
        int found = 0;
        while (found < 2) {
            if (ncircle.count() < ocircle.count()) {
                ++n_expansions;
                ncircle = expandCircle(adjacencies, ncircle);
            } else {
                ++o_expansions;
                ocircle = expandCircle(adjacencies, ocircle);
            }
            if ((ncircle & ocircle).count()) ++found;
        }
        int hops = n_expansions + o_expansions;
        int ans = 0;
        int cycle_offset = 0;
        while (hops) {
            if ((cycle_offset / change) & 1) {
                // wait for green
                ans += change - (cycle_offset % change);
                cycle_offset = 0;
            }
            int hops_before_red = min(hops, change / time);
            hops -= hops_before_red;
            ans += hops_before_red * time;
            cycle_offset += hops_before_red * time;
            cycle_offset %= (change << 1);
            if (hops && (cycle_offset < change)) {
                // 1 more hop into the light change.
                --hops;
                ans += time;
                cycle_offset += time;
            }
        }
        return ans;
    }
};
auto fast = (ios::sync_with_stdio(0), cout.tie(0), cin.tie(0), true);
/****
// first move: [0;946]
// light is red at 183, green at 366, green at 732, red at 915, green at 1098
// 2nd move: [1098;2044]
// light is green at 1830, red at 2013, green at 2196
// 3rd move: [2196:3142]

****/
174406