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Merge pull request #141 from AnjaliPai16/flight_routes
Flight routes graph algorithms solution added
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/*
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Problem: Flight Routes (CSES Problem Set 1196)
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Category: Graph Algorithms
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Difficulty: Medium
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Time Complexity: O(m * log(n * k))
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Space Complexity: O(n * k)
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Approach:
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This is a variant of Dijkstra’s algorithm where we need the k shortest paths
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from the source (1) to destination (n), allowing repeated nodes in routes.
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We maintain for each node a max-heap of up to k shortest distances found so far.
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We process nodes in increasing order of distance (via a min-heap).
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If a new smaller distance is found, we push it into that node's heap.
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If the heap exceeds k, we remove the largest (keeping only k smallest).
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This avoids sorting every time (unlike a vector-based approach),
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resulting in much faster performance for large graphs.
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Key Insights:
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- Standard Dijkstra only tracks one best distance per node; here we track up to k.
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- Using a max-heap per node allows O(log k) insertion/removal.
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- We only store k distances per node → O(n * k) memory.
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- Since k ≤ 10, the algorithm is very efficient.
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Edge Cases:
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- Multiple edges between the same nodes.
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- Repeated visits to the same node with different costs.
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- Large weights (use long long).
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*/
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#include <bits/stdc++.h>
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using namespace std;
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#define int long long
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#define pii pair<long long, int>
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int32_t main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int n, m, k;
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cin >> n >> m >> k;
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vector<vector<pair<int,int>>> adj(n + 1);
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for (int i = 0; i < m; i++) {
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int a, b, c;
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cin >> a >> b >> c;
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adj[a].push_back({b, c});
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}
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// Max-heaps to store up to k shortest distances per node
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vector<priority_queue<long long>> dist(n + 1);
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// Min-heap for Dijkstra: {distance, node}
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priority_queue<pair<long long,int>, vector<pair<long long,int>>, greater<pair<long long,int>>> pq;
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pq.push({0, 1});
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dist[1].push(0);
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while (!pq.empty()) {
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auto [currDist, node] = pq.top();
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pq.pop();
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// Skip if this path is not among the k smallest for this node
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if (dist[node].top() < currDist && (int)dist[node].size() == k)
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continue;
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for (auto &[next, wt] : adj[node]) {
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long long newDist = currDist + wt;
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// If less than k distances stored or this one is smaller than the largest
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if ((int)dist[next].size() < k || newDist < dist[next].top()) {
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dist[next].push(newDist);
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pq.push({newDist, next});
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if ((int)dist[next].size() > k)
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dist[next].pop(); // Keep only k smallest distances
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}
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}
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}
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// Collect all distances from max-heap of destination node and sort ascending
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vector<long long> ans;
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while (!dist[n].empty()) {
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ans.push_back(dist[n].top());
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dist[n].pop();
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}
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sort(ans.begin(), ans.end());
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for (auto &d : ans)
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cout << d << " ";
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cout << "\n";
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return 0;
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}

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