<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) {
        vector<int> visited(n+1, 0);
        vector<int> distance(n+1, 1e8);
        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> pq;
        vector<vector<int>> adj(n+1);
        map<pair<int,int>, int> mii;
        int no_of_nodes = times.size();
        for(int i = 0; i<no_of_nodes; i++){
            int u = times[i][0];
            int v = times[i][1];
            int w = times[i][2];
            adj[u].push_back(v);
            mii[{u,v}] = w;
        }
        pq.push({0, k});
        distance[k] = 0;
        while(!pq.empty()){
            pair<int,int> curr = pq.top();
            pq.pop();
            if(visited[curr.second] == 1){
                continue;
            }
            visited[curr.second] = 1;
            distance[curr.second] = curr.first;
            for(auto it : adj[curr.second]){
                if(distance[it] > curr.first + mii[{curr.second, it}]){
                    pq.push({curr.first + mii[{curr.second, it}], it});
                }
            }
        }
        bool is_all = true;
        for(int i = 1; i<=n; i++){
            if(visited[i] == 0){
                is_all = false;
            }
        }
        if(is_all == false){
            return -1;
        }
        else{
            int maxi = 0;
            for(int i = 1; i<=n; i++){
                maxi = max(maxi, distance[i]);
            } 
            return maxi;
        }
    }
};
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