<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:
    static bool comp(pair<int,int> a, pair<int,int> b){
        return a.second > b.second;
    }
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        
        vector<vector<int>> adj(n, vector<int>(n,0));
        for(int i = 0; i < times.size();i++){
            adj[times[i][0]-1][times[i][1]-1] = times[i][2]+1;
        }
        vector<int> d(n,0);
        vector<int> visited(n,0);

        for(int i = 0; i < n;i++){
            d[i] = INT_MAX;
        }
        d[k-1] = 0;
        

        for(int i = 0; i < n;i++){
            for(int j = 0; j < n;j++){
                cout << adj[i][j] << " ";
            }
            cout << endl;
        }
        
        priority_queue<pair<int,int>,vector<pair<int,int>>, greater<pair<int,int>>> q;
        q.push(make_pair(0,k-1));
        // visited[k-1] = 1;
        while(q.size()>0){
            
            
            int curr = q.top().second;
            // visited[curr] = 1;
            q.pop();
            for(int i = 0; i < n;i++){
                if(adj[curr][i] > 0){
                    if(adj[curr][i]-1+d[curr] < d[i]){
                        d[i] = adj[curr][i]-1+d[curr];
                        q.push(make_pair(adj[curr][i]-1,i));
                    }
                }
            }
        }
        int minL = 0;
        for(int i = 0; i < n;i++){
            if(d[i] == INT_MAX){
                return -1;
            }
            minL = max(d[i],minL);
            
        }
        return minL;
        
        
    }
};
41359