<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>

const int OO = 1e5;

class Solution {
private:
    int src;
    vector<vector<pair<int, int>>> graph;
    int memo[100+5][100+5];

    int bellman_ford(int dest, int edgesAllowed) {
        if (src == dest)
            return 0;

        if (edgesAllowed == 0)
            return OO;

        auto &ret = memo[dest][edgesAllowed];
        if (ret != INT_MIN)
            return ret;

        int res = OO;
        for (auto neighbour: graph[dest]) {
            int neighbourNode = neighbour.first;
            int edgeCost = neighbour.second;
            res = min(res, bellman_ford(neighbourNode, edgesAllowed - 1) + edgeCost);
        }

        return ret = res;
    }

public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        // preparation
        this->src = k - 1;
        vector<vector<pair<int, int>>> graph(n);
        for (auto edge: times) {
            int from = edge[0] - 1, to = edge[1] -1, weight = edge[2];
            graph[to].push_back({from, weight});
        }
        this->graph = graph;

        // dp initialization
        for (int i = 0; i < 105; i++)
            for (int j = 0; j < 105; j++)
                memo[i][j] = INT_MIN;
        
        // computing shortest distance to all nodes
        vector<int> dist = vector<int>(n, OO);
        for (int node = 0; node < n; node++)
            dist[node] = bellman_ford(node, n-1);

        int mx_dist = INT_MIN;
        for (auto val: dist)
            mx_dist = max(mx_dist, val);

        if (mx_dist == OO)
            return -1;
        else
            return mx_dist;
    }
};
49309