<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>

const int INF = 0x3f3f3f3f;
const int N = 10003;
vector<int> g[N];
int dis[N][2];
int vst[N];

class Solution {
    void bfs(int u, int t, int c) {
        vst[u] = 1;
        queue<int> q;
        q.push(u);
        while(!q.empty()) {
            int v = q.front();
            q.pop();
            int curt = dis[v][0];
            if(curt / c & 1) { curt = curt + c - curt % c; }
            curt = curt + t;
            int sec = dis[v][1];
            if(sec == INF) { sec = -1; }
            else {
                if(sec / c & 1) { sec = sec + c - sec % c; }
                sec = sec + t;
            }
            for(int i = 0; i < g[v].size(); ++i) {
                int w = g[v][i];
                if(-1 == sec) {
                    if(dis[w][0] == INF && dis[w][1] == INF) {
                        dis[w][0] = curt;
                        vst[w] = 0;
                    }
                    else if(dis[w][0] != INF && dis[w][1] == INF) {
                        if(curt != dis[w][0]) {vst[w] = 0; dis[w][1] = curt; }
                    }
                    else {
                        if(curt != dis[w][0] && curt != dis[w][1]) {
                            if(curt < dis[w][1]) {vst[w] = 0; dis[w][1] = curt; }
                            if(dis[w][1] < dis[w][0]) { swap(dis[w][0], dis[w][1]); }
                        }
                    }
                }
                else {
                    if(dis[w][0] == INF && dis[w][1] == INF) {
                        dis[w][0] = curt;
                        dis[w][1] = sec;
                        vst[w] = 0;
                    }
                    else if(dis[w][0] != INF && dis[w][1] == INF) {
                        if(curt != dis[w][0]) {vst[w] = 0; dis[w][1] = curt; }
                        else {
                            vst[w] = 0;
                            dis[w][1] = sec;
                        }
                    }
                    else {
                        if(curt != dis[w][0] && curt != dis[w][1]) {
                            if(curt < dis[w][1]) {vst[w] = 0; dis[w][1] = curt; }
                            if(dis[w][1] < dis[w][0]) { swap(dis[w][0], dis[w][1]); }
                        }
                        else if(sec != dis[w][0] && sec != dis[w][1]) {
                            if(sec < dis[w][1]) {vst[w] = 0; dis[w][1] = sec; }
                            if(dis[w][1] < dis[w][0]) { swap(dis[w][0], dis[w][1]); }
                        }
                    }
                }
                
                // printf("dealing %d, %d, %d\n", w, dis[w][0], dis[w][1]);
                if(0 == vst[w]) {
                    vst[w] = 1;
                    q.push(w);
                }
            }
        }
        return;
    }
    int cal(int n, int t, int c) {
        int curt = dis[n][0];
        if(curt / c & 1) { curt = curt + c - curt % c; }
        curt = curt + t;
        if(curt / c & 1) { curt = curt + c - curt % c; }
        return curt + t;
    }
public:
    int secondMinimum(int n, vector<vector<int>>& edges, int time, int change) {
        for(int i = 0; i < n; ++i) { g[i] = vector<int>(); }
        memset(dis, 0x3f, sizeof(dis));
        memset(vst, 0, sizeof(vst));
        dis[1][0] = 0;
        dis[1][1] = INF;
        
        for(int i = 0; i < edges.size(); ++i) {
            int u = edges[i][0];
            int v = edges[i][1];
            g[u].push_back(v);
            g[v].push_back(u);
        }
        bfs(1, time, change);
        int ans = dis[n][1];
        int second = cal(n, time, change);
        // printf("%d, %d, %d, %d\n", dis[n][0], dis[n][1], second, min(dis[n][1], second));
        return min(dis[n][1], second);
    }
};
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