<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:
 vector<pair<int, int>> G[6001];
    
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
         int visited[6001] = {0};
        priority_queue<pair<int,int> > PQ;
        for(int i=0;i<=n;i++) {
            visited[i] = 1e7;
        }
        for(int i=0;i<times.size();i++) {
            G[times[i][0]].push_back({times[i][2], times[i][1]});
        }
        int cnt = 1;
        PQ.push({0 ,k});
        int ans = 0;
        visited[k] = 0;
        while(!PQ.empty()) {
            pair<int, int> u = PQ.top();
            PQ.pop();
            for(int i=0;i<G[u.second].size();i++) {
                pair<int, int> v = G[u.second][i];
                int val = (-1 * u.first) + v.first;
                if(visited[v.second]>val) {
                    visited[v.second] = val;
                    PQ.push({-1 * val, v.second});
                }
            }
        }
        for(int i=1;i<=n;i++) {
            if(k==i) continue;
            if(visited[i] == 1e7) continue;
            ans = max(ans, visited[i]);
            cnt++;
        }
        if(cnt == n) {
            return ans;
        }
        return -1;
    }
};
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