<p>You are given a network of <code>n</code> nodes, labeled from <code>1</code> to <code>n</code>. You are also given <code>times</code>, a list of travel times as directed edges <code>times[i] = (u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>)</code>, where <code>u<sub>i</sub></code> is the source node, <code>v<sub>i</sub></code> is the target node, and <code>w<sub>i</sub></code> is the time it takes for a signal to travel from source to target.</p>

<p>We will send a signal from a given node <code>k</code>. Return <em>the <strong>minimum</strong> time it takes for all the</em> <code>n</code> <em>nodes to receive the signal</em>. If it is impossible for all the <code>n</code> nodes to receive the signal, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/23/931_example_1.png" style="width: 217px; height: 239px;" />
<pre>
<strong>Input:</strong> times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 1
<strong>Output:</strong> 1
</pre>

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

<pre>
<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 2
<strong>Output:</strong> -1
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= times.length &lt;= 6000</code></li>
	<li><code>times[i].length == 3</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><code>0 &lt;= w<sub>i</sub> &lt;= 100</code></li>
	<li>All the pairs <code>(u<sub>i</sub>, v<sub>i</sub>)</code> are <strong>unique</strong>. (i.e., no multiple edges.)</li>
</ul>

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        // Minimum Spanning Tree
        // initialization
        vector<vector<int>> g[n+1];
        for(auto edge : times){
            int from = edge[0], to = edge[1], w = edge[2];
            g[edge[0]].push_back({edge[1], w});
        }
        // Dijkstra's Algorithm
        int res = 0;
        vector<bool> visit(n+1, false);
        vector<int> path(n+1, INT_MAX);
        path[k] = 0;
        for(int i=0; i<n; i++){
            int cur = findmin(path, visit);
            if(cur == -1) return -1;
            res = max(res, path[cur]);
            visit[cur] = true;
            for(auto nxt : g[cur]){
                int node = nxt[0], w = nxt[1];
                if(!visit[node] && path[node] > path[cur]+w){
                    path[node] = path[cur] + w;
                }
            }
        }
        return res;
    }
    int findmin(vector<int> &path, vector<bool> &visit){
        int min_val = INT_MAX, min_idx = -1;
        for(int i=1; i<path.size(); i++){
            if(!visit[i] && min_val > path[i]){
                min_val = path[i];
                min_idx = i;
            }
        }
        return min_idx;
    }
};
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