1869B - 2D Traveling - CodeForces Solution

#include <bits/stdc++.h>
#include <iostream>
#define ll long long int
#define pb push_back
#define pob pop_back
#define pf push_front
#define pof pop_front
#define gcd __gcd
#define int_string to_string
#define string_int stoi
#define mn(v) *min_element(v.begin(), v.end())
#define mx(v) *max_element(v.begin(), v.end())
#define index_character s.find('character')
#define countxchar count(s.begin(), s.end(), 'x')
#define index_ofX_vector find(v.begin(), v.end(), x) - v.begin()
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define n1 cout << "-1" << endl
#define sorted is_sorted(v.begin(), v.end())
#define nl << endl
#define sp << " "
#define mp make_pair
#define fi first
#define se second
#define Mx LLONG_MAX
#define Mn LLONG_MIN
#define mod %1000000007
const ll N = 2e5+5;
// freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
// BesidesDuplicateCharacterEraseInString s.erase(unique(s.begin(), s.end()), s.end());
// Upper/lower-> transform(s.begin(), s.end(), s.begin(), ::toupper/tolower);
using namespace std;
ll i, j, k, n, m;
const ll e=1e+9;
bool comp(pair<long double,ll> a,pair<long double,ll> b){
    if(a.fi==b.fi) return (a.se<b.se); return (a.fi>b.fi);}
// Don't get stuck on a single approach for long, think of multiple ways
// You will destroy your dream if you stay in your comfort zone
// **********************|| Main Code ||********************************
//

int main()
{ 
    ios::sync_with_stdio(0);
    cin.tie(nullptr);

    ll test = 1;
    cin >> test;    
    again: 
    while (test--)
    {         
        ll n,k,a,b;
        cin>>n>>k>>a>>b;
        vector<pair<ll,ll>> v(n);
        for(i=0;i<n;i++)
            cin>>v[i].fi>>v[i].se;
        ll x=Mx,y=Mx;        
        for(i=0;i<k;i++){
            ll d=abs(v[i].fi-v[a-1].fi)+abs(v[i].se-v[a-1].se);
            x=min(x,d);
        }
        for(i=0;i<k;i++){
            ll d=abs(v[i].fi-v[b-1].fi)+abs(v[i].se-v[b-1].se);
            y=min(y,d);
        }
        ll ans=abs(v[a-1].fi-v[b-1].fi)+abs(v[a-1].se-v[b-1].se);
        if(!k) cout << ans nl;
        else cout << min(ans,x+y) nl;
    }       
}