<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:
    void dijkstra(vector<vector<int>>matrix,int start,int V,vector<int>&dist,vector<bool>&visited){
        //做V次
        dist[start]=0;
        for(int i=0;i<V;i++){
            int cur_min=INT_MAX;int cur_node=0;
            for(int j=0;j<dist.size();j++){
                //每次從visited為false中挑出dist最小的
                if(cur_min>dist[j]&&visited[j]==false){
                    cur_node=j;cur_min=dist[j];
                }
            }
            visited[cur_node]=true;
            //從cur_node往外走
            for(int k=0;k<matrix[cur_node].size();k++){
                if(matrix[cur_node][k]!=-1&&dist[cur_node]!=INT_MAX&&dist[cur_node]+matrix[cur_node][k]<dist[k]){
                    dist[k]=dist[cur_node]+matrix[cur_node][k];
                }
            }
        }
    }
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        //這邊的times都是edges
        //用Dijkstra時轉成鄰接矩陣
        //有向圖
        vector<vector<int>>matrix(n,vector<int>(n,-1));
        vector<int>dist(n,INT_MAX);
        vector<bool>visited(n,false);
        for(int i=0;i<times.size();i++){
            matrix[times[i][0]-1][times[i][1]-1]=times[i][2];
        }
        
        dijkstra(matrix,k-1,n,dist,visited);
        int cur_max=0;
        for(int i=0;i<dist.size();i++){
            if(dist[i]>cur_max)cur_max=dist[i];
        }
        if(cur_max==INT_MAX||cur_max==0)return -1;
        else return cur_max;
    }
};
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