Submission #69907462

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Source Code Expand

#include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
/*
    Problem Statement

*/


/*
    Observation
    

    so each node must reach all other node wxcept its diametricaly opposite node

    and we can arrange labels in such a way such that its diametrical opposite always give const


    for n=12
    0...............11


    from 0 every other node must be at distance <=2 except 11
    from 1 every other node must be at distance <=2 except 10
    ....
    and so on

    since the structure would feel like symmetrical
    so,

    (N) - l1 - l2 - (n-N-1)

    so no.of nodes at l1 and l2 would be equal
     which is (n-2)/2


     if n-2 is not even then not poss
     so, n==1 poss
     n>=4

     0 1 2 3
     1 0 1 3

     not poss beacause 1 and 2 are at dist 1

     Each node must be connected with x other nodes
     i.e degree of all nodes must be same

     there must be (x-2)/2  would be deg of each node
*/


/*
    Claims

*/
void solve()
{
    // Add your solution here
     int n;
    cin >> n;
    if (n % 2 == 1 ||n<6) {
        cout << -1 << endl;
        return ;
    }

    int half = n / 2;
    set<pair<int,int>> edges;

    for (int i = 0; i < n; i++) {
        for (int j = 1; j < half; j++) {
            int v = (i + j) % n;
            if (v == (i + half) % n) continue; // skip diametric opposite
            int u = i;
            if (u > v) swap(u, v);
            edges.insert({u, v});
        }
    }

    cout <<  edges.size() << endl;
    for (auto it : edges){
        cout << it.first << " " << it.second << "\n";

    }
}
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

Submission Info

Judge Result