Challenge: The Road That Makes a Loop

Lab C++ Intermediate 20 points

Roads are built one by one between n towns. Print the first road (a b) that joins two towns that were already connected, or no loop.

Input

Line 1: n m. Then m lines a b in build order.

Output

a b or no loop.

Example

Input

Output

2 0

Sample input available Sample output available

Editor

C++

Hints

Before joining, check whether find(a) == find(b).

Keep reading all roads even after finding the answer.

Solution explanation

Union-find answers “already connected?” as roads are added.

Show solution

#include <iostream>
#include <numeric>
#include <vector>
using namespace std;

vector<int> parent;

int find(int x) {
    while (parent[x] != x) {
        parent[x] = parent[parent[x]];
        x = parent[x];
    }
    return x;
}

int main() {
    int n, m;
    cin >> n >> m;
    parent.resize(n);
    iota(parent.begin(), parent.end(), 0);
    int ansA = -1, ansB = -1;
    for (int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b;
        int ra = find(a), rb = find(b);
        if (ra == rb) {
            if (ansA == -1) { ansA = a; ansB = b; }
        } else {
            parent[rb] = ra;
        }
    }
    if (ansA == -1) cout << "no loop" << endl;
    else cout << ansA << " " << ansB << endl;
    return 0;
}

Output

Answer