<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 dp[10001][101];
	
    int dfs(vector<vector<pair<int, int>>>& graph, int src, int dst, int k) {
        if (src == dst) return 0;
        if (k <= -1) return INT_MAX;
        
        if (dp[src][k] != -1) return dp[src][k];
        int ans = INT_MAX;
        for (auto child : graph[src]) {
            int res = dfs(graph, child.first, dst, k-1);
        // cout<<src<<" "<<child.first<<" "<<res<<endl;
            if (res != INT_MAX)
                ans = min(ans, child.second +res);
        }
        cout<<ans<<endl;
        return dp[src][k] = ans;
    }
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) {
        memset(dp, -1, sizeof dp);
        // Build the graph
        vector<vector<pair<int, int>>> graph(n);
        for (auto vec : flights) {
            graph[vec[0]].emplace_back(make_pair(vec[1], vec[2]));
        }
        int ans = dfs(graph, src, dst, K);  // or make the dfs call
        if (ans == INT_MAX)
            return -1;
        return ans;
    }

    // int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int k) {
    //     vector<vector<pair<int,int>>> graphs(n);
    //     memset(dp, -1, sizeof(dp)); 

    //     for(int i=0;i<flights.size();i++)
    //     {
    //         pair<int,int>p(flights[i][1],flights[i][2]);
    //         int vertex=flights[i][0];
    //         if(graphs[vertex].size()==0)
    //         {
    //             vector<pair<int,int>>v;v.push_back(p);
    //             graphs[vertex]=v;
    //         }
    //         else
    //         {graphs[vertex].push_back(p);}
    //     }
        
    //      dfs( graphs,  src,  dst,  k+2);
    //      for(int k=0;k<101;k++){cout<< dp[dst][k]<<" ";}cout<<endl;
    //      return dp[dst][k+2];
    // }
};
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