<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) {
        std::vector<std::vector<std::pair<int, int>>> graph(n);
        for (const auto& ele: times) {
            graph[ele[0] - 1].push_back(std::make_pair(ele[1], ele[2]));
        }

        std::vector<int> disTable(n, 10000);
        std::queue<std::pair<int, int>> q;
        for (const auto& ele: graph[k - 1]) {
            q.push(ele);
        }
        disTable[k - 1] = 0;

        while (!q.empty()) {
            auto ele = q.front();
            q.pop();
            const int index = ele.first - 1;
            const int distance = ele.second;
            disTable[index] = std::min(distance, disTable[index]);

            for (const auto& neighbor: graph[index]) {
                if (disTable[neighbor.first - 1] == 10000 || disTable[neighbor.first - 1] > disTable[index] + neighbor.second) {
                    q.push(std::make_pair(neighbor.first, neighbor.second + disTable[index]));
                }
                
            }
        }

        int res = 0;
        for (const auto ele: disTable) {
            if (ele == 10000) {
                return -1;
            }
            res = std::max(res, ele);
        }

        return res;
    }
};

/*
Constraint:
    1. k, n: 1 <= k, n <= 100
    2. time: 1 <= time <= 6000
    3. times size == 3
    4. all pair is unique

Case:
    1.
        Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
        res = 2

    2.
        Input: times = [[1,2,1]], n = 2, k = 1
        res = 1

    
Approach:
    1. Graph + topologic sort

        case 1:
            Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
                1
            2   ->  1
            |1
            3   ->  4
                1

            [1, 0, 1, m]
            {2: {[1, 1], [3, 1]}, 3: {[4, 1]}}

            q = [2, 3]

            complexity:
                time: O(n*m)
                space: O(n)
*/

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