<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<vector<int>>  graph(n + 1, vector<int>(n + 1, INT_MAX));
        for (auto &time : times) {
            graph[time[0]][time[1]] = time[2];
        }
        vector<int> spath(n+1, INT_MAX);
        vector<bool> visited(n+1, false);
        auto cmp = [&spath](int a, int b) {
            return spath[a] > spath[b]; // Min-heap based on the values vector
        };
        priority_queue<int, vector<int>, decltype(cmp)> vq(cmp);
        spath[k] = 0;
        vq.push(k);
        

        while(!vq.empty()) {
            int curr = vq.top();
            vq.pop();
            if (visited[curr]) {
                continue;
            }
            visited[curr] = true;

            for(int i = 1; i<=n; i++){
                if(graph[curr][i]!=INT_MAX) {
                    int nwt = spath[curr]+graph[curr][i];
                    //spath[i] = min(spath[i],nwt);
                    if(spath[i] > nwt) {
                        spath[i] = nwt;
                        vq.push(i);
                    }
                    
                }
            }
        }
        //spath[k] = INT_MAX;
        int maxTime = 0;
        for (int i = 1; i <= n; i++) {
            if (spath[i] == INT_MAX) {
                return -1; // If any node is unreachable, return -1
            }
            maxTime = max(maxTime, spath[i]);
        }

        return maxTime; // Return the maximum time
    }
};
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