<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<vector<vector<int>>>createGraph(vector<vector<int>>&edges, int n){
        vector<vector<vector<int>>>graph(n+1);
        for(int i=0;i<=n; i++){
            graph.push_back({{}});
        }
        for(vector<int>edge: edges){
            int source = edge[0];
            int dest = edge[1];
            int cost = edge[2];
            graph[source].push_back({dest,cost});
        }
        return graph;
    }

    int minTime(vector<vector<vector<int>>>&graph, int n, int src){
        vector<int>cost(n+1,0);
        cost[src]=0;
        vector<bool>vis(n+1,false);

        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int, int>>>pq;
        pq.push({0,src});

        while(!pq.empty()){
            pair<int,int>p = pq.top();
            int currNode = p.second;
            int currCost = p.first;
            pq.pop();
            if(vis[currNode])  continue;

            vis[currNode] = true;
            cost[currNode] = currCost;

            for(int i=0; i<graph[currNode].size(); i++){
                int neighbor = graph[currNode][i][0];
                int weight = graph[currNode][i][1];

                if(vis[neighbor])  continue;
                pq.push({currCost+weight, neighbor});
            }
        }
        for(int i=1; i<=n; i++){
            if(vis[i]==0)  return -1;
        }
        int ans = 0;
        for(int x:cost) ans = max(ans,x);
        return ans;
    }

    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        vector<vector<vector<int>>>graph = createGraph(times,n);
        return minTime(graph,n,k);
    }
};
57590