<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>

struct test {
    int dist,node,stops;
};
class cmp {
    public:
    bool operator()(test &a,test &b) {
        return a.dist>b.dist;
    }
};
#define pp pair<int,int>
class Solution {
public:
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int k) {
        vector <list<pp>> graph(n,list<pp>());
        for(auto i:flights) {
            int a=i[0],b=i[1],wt=i[2];
            graph[a].push_back({b,wt});
        }
        priority_queue <test,vector <test>,cmp> pq;
        vector <vector <bool>> vis(n,vector <bool> (k+2,false));
        vector <vector <int>> distarr(n,vector <int> (k+2,INT_MAX));
        pq.push({0,src,0});
        while(pq.size()) {
            auto curr=pq.top();
            int node=curr.node,myDist=curr.dist,stops=curr.stops;
            
            if(vis[node][stops] || stops>k+1) {
                pq.pop();
                continue;
            }
            // cout<<node<<" "<<myDist<<" "<<stops<<endl;
            vis[node][stops]=true;
            distarr[node][stops]=myDist;
            for(auto it:graph[node]) {
                int wt=it.second;
                int neigh=it.first;
                if(stops+1>k+1)continue;
                if(!vis[neigh][stops+1] && myDist+wt<=distarr[neigh][stops+1]) {
                    // cout<<neigh<<" "<<stops+1<<" "<<myDist+wt<<endl;
                    pq.push({myDist+wt,neigh,stops+1});
                    distarr[neigh][stops+1]=myDist+wt;
                }
            }
        }
        int ans=INT_MAX;
        for(int i=0;i<k+2;i++) {
            ans=min(ans,distarr[dst][i]);
            // cout<<distarr[dst][i]<<endl;
        }
        if(ans==INT_MAX)ans=-1;
        return ans;
    }
};
19155