1569D - Inconvenient Pairs - CodeForces Solution

#include<iomanip>
#include <iostream>
#include<list>
#include<vector>
#include<algorithm>
#include<cstdlib>
#include<stack>
#include<set>
#include<deque>
#include<math.h>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<string.h>
#include<queue>
#define ll long long int

//ll p=999998727899999;
ll prime=998244353;
ll INF=1000000000000000000;
using namespace std;
ll nC2(ll a)
{
    return(a*(a-1)/(ll)2);
}
ll value(map<int,ll> m)
{
    ll t=0,z=0;
    for(auto p:m){
        //cout<<p.first<<"-->"<<p.second<<"&&&\n";
    z+=p.second;t-=nC2(p.second);
   // cout<<z<<"^^^\n";
    }
    t+=nC2(z);
    //cout<<z<<" "<<t<<"!!!!!\n";
    return(t);
}
ll count(vector<int> x,vector<int> y,int n,int m,vector<pair<int,int>> points,int k)
{
    sort(points.begin(),points.end());
    int i=0,j=0,cur=0;ll ans=0;
    while(cur<n-1)
    {
        int lb=x[cur],rb=x[cur+1];
        map<int,ll> m;
        while(j<k)
        {
            int cury=points[j].first;
            if(cury>rb)break;
            if(cury>lb&&cury<rb)
            {
                m[points[j].second]++;
            }
            j++;
        }
        ans+=value(m);
        cur++;
        i=j;
    }
    //cout<<ans<<"***\n";
    return(ans);
}
int main()
{
    int t;t=1;cin>>t;
     while(t--)
	{
        int n,m,k;cin>>n>>m>>k;
        vector<int> x(n),y(m);int j;
        for(j=0;j<n;j++)cin>>x[j];for(j=0;j<m;j++)cin>>y[j];
        vector<pair<int,int>> real,reflect;
        for(j=0;j<k;j++)
        {
            int a,b;cin>>a>>b;
            real.push_back({a,b});reflect.push_back({b,a});
        }
        cout<<count(x,y,n,m,real,k)+count(y,x,m,n,reflect,k)<<"\n";
	}
	return(0);
}