Codeforces 1169B. Pairs Codeforces Round #562 (Div. 2)

Problem link

One of x,y is obviously one of a1,b1. For example, let's fix who is x from the first pair. Then you need to check that there exists some y such that all pairs that don't contain x, contain y. For this, you can remember in the array for each value how many pairs that don't contain x contain this value, and then you need to check that in the array there exists some value that is equal to the number of pairs left.

A solution in c++

	#pragma GCC optimize("Ofast,no-stack-protector")
	#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native"
	#include<bits/stdc++.h>

	using namespace std;

	/// Typedef
	typedef long long ll;

	#define sc1(a) scanf("%lld",&a)
	#define sc2(a,b) scanf("%lld %lld",&a,&b)

	#define pf1(a) printf("%lld\n", a)
	#define pf2(a,b) printf("%lld %lld\n",a,b)

	#define size1 300005

	ll first[size1], sec[size1], n, m;

	bool calculate(ll x, ll y){

	for(ll i = 0; i < m; i++){
	if(first[i] == x \|\| first[i] == y \|\| sec[i] == x \|\| sec[i] == y) continue;
	else return false;
	}
	return true;
	}

	int main() {

	//seive();
	//preCal();

	ll tc, num, t = 1, choose;

	//freopen("/opt/Coding/clion code/input.txt", "r", stdin);
	//freopen("/opt/Coding/clion code/output.txt", "w", stdout);

	sc2(n, m);

	for(ll i = 0; i < m; i++){
	sc2(first[i], sec[i]);
	}

	for(ll i = 1; i < m; i++){
	if(first[i] == first[0] \|\| first[i] == sec[0] \|\| sec[i] == first[0] \|\|sec[i] == sec[0]){
	continue;
	}


	if(calculate(first[i], first[0])) return cout << "YES" << endl,0;
	if(calculate(first[i], sec[0])) return cout << "YES" << endl,0;
	if(calculate(sec[i], first[0])) return cout << "YES" << endl,0;
	if(calculate(sec[i], sec[0])) return cout << "YES" << endl,0;
	return cout << "NO" << endl, 0;
	}

	cout << "YES" << endl;

	return 0;
	}