CSES · 1680

Longest Flight Route

This C++ solution uses graph traversal for CSES 1680 Longest Flight Route. Read the reasoning, inspect the code, or try your own test case below.

1680Graphs & treesGraph traversalC++47 lines
Solution131of 248
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Approach

Graph traversal

Longest Flight Route DAG path from 1 to n and route reconstruction match.

Graphs & trees

Problem and code

Useful links.

Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

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Implementation

longest_flight_route.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
     
    int main(){
        ios::sync_with_stdio(0); cin.tie(0);
        int n, m; cin >> n >> m;
        vector<vector<int>> adj(n);
        vector<int> indegree(n);
        for(int i = 0; i < m; i++){
            int a, b; cin >> a >> b; a--; b--;
            adj[a].push_back(b);
            indegree[b]++;
        }
        queue<int> q;
        for(int i = 0; i < indegree.size(); i++){
            if(indegree[i] == 0) q.push(i);
        }
        vector<int> result;
        while(!q.empty()){
            int node = q.front(); q.pop();
            result.push_back(node);
            for(int nxt : adj[node]){
                indegree[nxt]--;
                if(indegree[nxt] == 0) q.push(nxt);
            }
        }
        vector<int> dist(n, -1e9);
        dist[0] = 1;
        vector<int> parent(n, -1);
        for(int u : result){
            if(dist[u] == -1e9) continue;
            for(int v : adj[u]){
                if(dist[u] + 1 > dist[v]){
                    parent[v] = u;
                    dist[v] = dist[u] + 1;
                }
            }
        }
        if(dist[n - 1] == -1e9){
            cout << "IMPOSSIBLE" << '\n';
        } else {
            cout << dist[n - 1] << '\n';
            vector<int> path;
            for(int i = n - 1; i != -1; i = parent[i]) path.push_back(i);
            for(int i = path.size() - 1; i >= 0; i--) cout << path[i] + 1 << (i == 0 ? "\n" : " ");
        }
    }
        

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