<p>There are <code>n</code> cities connected by some number of flights. You are given an array <code>flights</code> where <code>flights[i] = [from<sub>i</sub>, to<sub>i</sub>, price<sub>i</sub>]</code> indicates that there is a flight from city <code>from<sub>i</sub></code> to city <code>to<sub>i</sub></code> with cost <code>price<sub>i</sub></code>.</p>

<p>You are also given three integers <code>src</code>, <code>dst</code>, and <code>k</code>, return <em><strong>the cheapest price</strong> from </em><code>src</code><em> to </em><code>dst</code><em> with at most </em><code>k</code><em> stops. </em>If there is no such route, return<em> </em><code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-3drawio.png" style="width: 332px; height: 392px;" />
<pre>
<strong>Input:</strong> n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1
<strong>Output:</strong> 700
<strong>Explanation:</strong>
The graph is shown above.
The optimal path with at most 1 stop from city 0 to 3 is marked in red and has cost 100 + 600 = 700.
Note that the path through cities [0,1,2,3] is cheaper but is invalid because it uses 2 stops.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-1drawio.png" style="width: 332px; height: 242px;" />
<pre>
<strong>Input:</strong> n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
<strong>Output:</strong> 200
<strong>Explanation:</strong>
The graph is shown above.
The optimal path with at most 1 stop from city 0 to 2 is marked in red and has cost 100 + 100 = 200.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-2drawio.png" style="width: 332px; height: 242px;" />
<pre>
<strong>Input:</strong> n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
<strong>Output:</strong> 500
<strong>Explanation:</strong>
The graph is shown above.
The optimal path with no stops from city 0 to 2 is marked in red and has cost 500.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>0 &lt;= flights.length &lt;= (n * (n - 1) / 2)</code></li>
	<li><code>flights[i].length == 3</code></li>
	<li><code>0 &lt;= from<sub>i</sub>, to<sub>i</sub> &lt; n</code></li>
	<li><code>from<sub>i</sub> != to<sub>i</sub></code></li>
	<li><code>1 &lt;= price<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
	<li>There will not be any multiple flights between two cities.</li>
	<li><code>0 &lt;= src, dst, k &lt; n</code></li>
	<li><code>src != dst</code></li>
</ul>

class Solution {
public:
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int k) {
        // stops,nodes,dist
        vector<vector<pair<int,int>>> g(n);
        for(int i=0;i<flights.size();i++){
            g[flights[i][0]].push_back({flights[i][1],flights[i][2]});
            // cout<<i<<" ";
        }
        queue<vector<int>> q;
        vector<int> dist(n,INT_MAX);
        q.push({0,src,0});
        while(!q.empty()){
            vector<int> v=q.front();
            q.pop();
            if(v[0]>k+1) continue;
            int stops=v[0];
            int node=v[1];
            int dis=v[2];
            cout<<stops<<" "<<node<<" "<<dis<<endl;
            if(dist[node]>dis){
                dist[node]=dis;
                for(auto x:g[node]){
                    q.push({stops+1,x.first,dist[node]+x.second});
                }
            }
        }
        return dist[dst]==INT_MAX ? -1 : dist[dst];


        // we can also do it directly using bfs
        
        // vector<int> dist(n, INT_MAX);
        // dist[src] = 0;

        // for(int i = 0;i<=k;i++){
        //     vector<int> temp(dist);
        //     for(const auto& flight:flights){
        //         if(dist[flight[0]] != INT_MAX){
        //             temp[flight[1]] = min(temp[flight[1]], dist[flight[0]]+flight[2]);
        //         }
        //     }
        //     dist = temp;
        // }
        // return dist[dst] == INT_MAX ? -1 : dist[dst];

        // vector<vector<pair<int,int>>> g(n);
        // for(int i=0;i<flights.size();i++){
        //     g[flights[i][0]].push_back({flights[i][1],flights[i][2]});
        // }
        // set<pair<int,pair<int,int>>> s;
        // s.insert({0,{src,0}});
        // int ans=INT_MAX;
        // vector<vector<int>> dist(n,vector<int>(k+2,INT_MAX));
        // dist[src][0]=0;
        // while(!s.empty()){
        //     auto it=s.begin();
        //     int cost=(*it).first;
        //     int stat=(*it).second.first;
        //     int cnt=(*it).second.second;
        //     s.erase(it);
        //     if(cost > dist[stat][cnt])continue;
        //     else dist[stat][cnt]=cost;
        //     if(cnt==k+1 and stat!=dst) continue;
        //     if(stat==dst){
        //         ans=min(ans,cost);
        //         continue;
        //     }
        //     for(auto x:g[stat]){
        //         s.insert({dist[stat][cnt]+x.second,{x.first,cnt+1}});
        //     }
        // }
        // if(ans==INT_MAX) return -1;
        // return ans;
    }
};
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