<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>

#include <bits/stdc++.h>
#define ss second
#define ff first
#define pii pair<int,int>
#define pb push_back
#define ll long long
#define vi vector<int>
#define vll vector<ll>	
#define pll pair<ll,ll>
#define vpll vector<pll>
#define ld long double
#define all(x) x.begin(),x.end()
#define inf 1000000000000000


class Solution {
public:
    ll vis[105],par[105],dis[105];

    void dijk(ll sr,vector<vector<pll>>& gr){
        
        set<pll> st;
        st.insert({0,sr});
        dis[sr]=0;

        while(st.size()>0){
            ll u=st.begin()->ss;
            st.erase(st.begin());

            if(vis[u])continue;
            vis[u]=1;

            for(auto x:gr[u]){
                ll v=x.ff;
                ll w=x.ss;
                if(dis[u]+w<dis[v]){
                    dis[v]=dis[u]+w;
                    st.insert({dis[v],v});
                }
            }
        }
    }
     
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        memset(vis,0,sizeof(vis));
        // memset(dis,inf,sizeof(dis));
        std::fill(dis, dis + 105, inf);

        vector<vector<pll>> gr(n+1,vector<pll>(0));
        for(auto x:times){
            gr[x[0]].pb({x[1],x[2]});
        }

        dijk(k,gr);
        ll res=0;
        for(ll i=1;i<=n;i++){

            cout<<i<<" "<<dis[i]<<endl;
            if(dis[i]==inf)return -1;
            res=max(res,dis[i]);
        }
        return res;
    }

};


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