<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>

class Solution {
public:
    static const int N = 10010, M = 4 * N, INF = 0x3f3f3f3f;
    int idx = 0;
    int h[N], e[M], ne[M];
    int dis1[N], dis2[N];

    void add(int a, int b){
        e[idx] = b;
        ne[idx] = h[a];
        h[a] = idx++;
    }
    int secondMinimum(int n, vector<vector<int>>& edges, int time, int change) {
        memset(dis1, INF, sizeof(dis1));
        memset(dis2, INF, sizeof(dis2));
        memset(h, -1, sizeof(h));

        for(auto &e : edges){
            add(e[0], e[1]);
            add(e[1], e[0]);
        }

        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;

        q.push({0, 1});
        dis1[1] = 0;
        while(!q.empty()){
            auto t  = q.top();
            q.pop();
            int u = t.second, step = t.first;
            for(int i = h[u]; i != -1; i = ne[i]){
                int j = e[i];
                int a = step / change, b = step % change;
                int wait = a % 2 == 0 ? 0 : change -b;
                int dist = step + time + wait;
                if(dis1[j] > dist){
                    dis2[j] = dis1[j];
                    dis1[j] = dist;
                    q.push({dis1[j], j});
                    q.push({dis2[j], j});
                }else if(dis1[j] < dist && dist < dis2[j]){
                    dis2[j] = dist;
                    q.push({dis2[j], j});
                }
            }
        }
        return dis2[n];
    }
};
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