CSES :: Graph Algorithms :: Round Trip

Problem : Please find the problem here.

Explanation : We need to find out the path in the graph where start and end node is the same (a cycle).
This can be easily checked with simple Graph Traveral, check if any of the node's children node, during traversal, is already visited by some other parent.

Code : Used DFS to check cycles in the graph and backtracking the path using the nodes parent array maintained over traversal.

#include <bits/stdc++.h>

using namespace std;

const int mxN = 1e5;

void DFS(int u, int pu, vector<int> (&adj)[mxN], vector<int> &visited, vector<int> &parent) {
    visited[u] = 1;
    parent[u] = pu;

    for (int v : adj[u]) {
        if (!visited[v]) {
            DFS(v, u, adj, visited, parent);
        } else {
            if (v == pu) {
                continue;
            }
            else {
                int u2 = u;
                vector<int> ans;
                while (u^v) {
                    ans.emplace_back(u);
                    u = parent[u];
                }

                ans.push_back(v);
                ans.push_back(u2);

                cout << ans.size() << '\n';
                for (auto a : ans) {
                    cout << a+1 << ' ';
                }
                exit(0);
            }

        }
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int n, m;
    cin >> n >> m;

    vector<int> adj[mxN], visited(n, 0), parent(n, 0);


    for (int i = 0, u, v; i < m; i++) {
        cin >> u >> v; u--, v--;
        adj[u].emplace_back(v);
        adj[v].emplace_back(u);
    }

    for (int i = 0; i < n; i++) {
        if (!visited[i]) DFS(i, -1, adj, visited, parent);
    }

    cout << "IMPOSSIBLE";

    return 0;
}