hostel visit question (priority queue application ) - c++

Question

question: Dean of MAIT is going to visit Hostels of MAIT. As you know that he is a very busy person so he decided to visit only the first "K" nearest Hostels. Hostels are situated on a 2D plane. You are given the coordinates of hostels and you have to answer the Rocket distance of Kth nearest hostel from the origin ( Dean's place )
Input Format
The first line of input contains Q Total no. of queries and K There are two types of queries:
first type: 1 x y For query of 1st type, you came to know about the coordinates ( x, y ) of the newly constructed hostel. second type: 2 For query of 2nd type, you have to output the Rocket distance of Kth nearest hostel till now.
//The Dean will always stay at his place ( origin ). It is guaranteed that there will be at least k queries of type 1 before the first query of type 2.
Rocket distance between two points ( x2 , y2 ) and ( x1 , y1 ) is defined as (x2 - x1)2 + (y2 - y1)2
Constraints
1 < = k < = Q < = 10^5 -10^6 < = x , y < = 10^6
Output Format
For each query of type 2 output the Rocket distance of Kth nearest hostel from Origin.//
This is my code:
#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
class roomno
{
public:
int x;
int y;
roomno(int x,int y)
{
this->x=x;
this->y=y;
}
void print()
{
cout<<"location"<<"("<<x<<","<<y<<")"<<endl;
}
int distance ()
{
return (x*x+y*y);
}
};
class roomcompare
{
public:
bool operator() (roomno r1,roomno r2)
{
return r1.distance()>r2.distance();
}
};
int main()
{
int x[1000]},y[1000];
int l,k=0;
priority_queue<roomno,vector<roomno>,roomcompare> pq;
int n,i,j;
cin>>n>>l;
//cin>>n;
cin.ignore();
for( i=0;i<n;i++)
{
cin>>x[i];
}
cin.ignore();
for( j=0;j<n;j++)
{
cin>>y[j];
}
cin.ignore();
for(i=0;i<n;i++ )
{
roomno r1(x[i],y[i]);
pq.push(r1);
}
while(!pq.empty()&&k!=l)
{ k++;
roomno r2=pq.top();
r2.print();
pq.pop();
}
return 0;
}
Original link to code: https://codeshare.io/2j1bkA
What's wrong with my code?

Answer 1

Please consider adding what problem you are facing in your post. I've seen your code but I couldn't help with your code as I even saw a syntax error in your code and didn't know if it was the problem.
If I didn't misunderstand your question, std::set will fit better than priority queue as it is not possible to remove the largest item in a priority queue(min heap). The following code should work:
#include <iostream>
#include <set>
using namespace std;
using ll = long long;
multiset<ll> distances;
int main() {
ll n, k;
cin >> n >> k;
for(ll i = 0; i < n; ++i) {
ll query;
cin >> query;
if (query == 1) {
ll x, y;
cin >> x >> y;
distances.insert(x * x + y * y);
if (distances.size() > k) {
distances.erase(--distances.end());
}
} else {
cout << *distances.rbegin() << '\n';
}
}
return 0;
}