CSES Flight Discount Problem | Dijkstra's Algorithm | Complete Walkthrough in C++

[@U-DAY]

The core logic remains the same, but eliminating constant time operations is necessary to fix the TLE. I’m sharing this code to provide better readability and clarity of my approach. Feel free to review it! #include using namespace std; #define ll long long vector dijkstra_algo(ll st, vectoradj[], ll n) { priority_queuepq; vectorvis(n, LONG_MAX); pq.push({0, st}); vis[st] = 0; while(!pq.empty()) { auto it = pq.top(); pq.pop(); ll x = it.first; ll y = it.second; for(auto i: adj[y]) { ll curr = x + i.second; if(curr < vis[i.first]) { vis[i.first] = curr; pq.push({curr, i.first}); } } } return vis; } int main() { ll m, n; cin>>n>>m; vectoradj[n]; vectorradj[n]; vectoredges; for(int i=0; i>x>>y>>z; x--, y--; adj[x].push_back({y, z}); radj[y].push_back({x, z}); edges.push_back({x, y, z}); } vectorvis1 = dijkstra_algo(0, adj, n); vectorvis2 = dijkstra_algo(n-1, radj, n); ll ans = 1e18; for(auto it: edges) { ll curr = vis1[it[0]] + vis2[it[1]] + it[2]/2; ans = min(ans, curr); } cout

[@mahmoudabdellah5332]

Good Work

[@4rt15t]

thank you!!!

[@deshdeepakkant524]

nice explaintion

[@varshith3740]

very good explanation bro