Codeforces Global Round 19 E. Best Pair

Best Pair
题意：cntx 为x出现次数 f(x,y) = (cntx+cnty)*(x+y); 求不被禁用的最大f(x,y);
思路：按cnt （出现次数）分类 ，最多有 根号n类 再去枚举x，y

//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
#include
#define int long long
#define fi first
#define se second
#define pb push_back
#define pii pair
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
const int inf=8e18;
const int maxn=3e5+100;
int a[maxn];
vector cnt[maxn];
bool cmp(int a,int b)
{
	return a>b;
}
signed main()
{
	IOS
	int tt;
	cin>>tt;
	while(tt--)
	{
		int n,m;
		mapmp;
		mapmp2;
		cin>>n>>m;
		for(int i=1; i<=n; i++)
		{
			cin>>a[i];
			mp[a[i]]++;
		}
		vectorv;
		int mx=0;
		for(int i=1; i<=m; i++)
		{
			int x,y;
			cin>>x>>y;
			mp2[ {x,y}]=1;
			mp2[ {y,x}]=1;
		}
		for(auto it:mp)
		{
			if(cnt[it.se].empty())
			{
				v.pb(it.se);
			}
			cnt[it.se].pb(it.fi);
		}

		for(int i=1;i<=n;i++) sort(cnt[i].begin(),cnt[i].end(),cmp);

		
		for(int i=0; imx)
						{
							mx=(it1+it2)*(v[i]+v[j]);
						}
						break;
					}
				} 
				//cout<