Facebook Hackercup 2016- Boomerang constellations problem solution

This problem is from the recently concluded Facebook Hackercup 2016 Qualification round. You can take a look at the problem statement in this link.
I am giving the abridged problem statement here. Given a set of coordinates in a 2D-plane, We have to find out the total number of pairs of lines with equal length and also share one end point.
For example, let us take the points {(0,0), (0,1), (1,0)}, there is one pair of lines, with a distance of 1. The line between (0,0),  (0,1) and the line between (0,0), (1,0).
The simplest way to solve this problem is this. Take a point (x,y) and try to calculate the distance to every other point. Maintain a map of distance to the number of other end points with that distance. This is O(n² log n) solution.
Here is the C++ code for the same.

	#include <iostream>
	#include <vector>
	#include <algorithm>
	#include <map>
	using namespace std;
	int main()
	{
	int t;
	cin >> t;
	for( int tc = 0; tc < t; tc++)
	{
	int n;
	cin >> n;
	int i, j;
	vector<pair<int,int> > pairs(n);
	for( i = 0; i < n; i++ )
	{
	int x,y;
	cin >> x >> y;
	pairs[i] = make_pair(x,y);
	}
	long long res = 0;
	map<int, int> dmap;
	for( i = 0; i < n; i++ )
	{
	dmap.clear();
	for( j = 0; j < n; j++ )
	{
	if( i != j )
	{
	int d = (pairs[i].first-pairs[j].first)*(pairs[i].first-pairs[j].first) +
	(pairs[i].second-pairs[j].second)*(pairs[i].second-pairs[j].second);
	res += dmap[d];
	dmap[d]++;
	}
	}
	}
	cout << "Case #" << (tc+1) << ": " << res<< endl;
	}
	return 0;
	}

view raw fbhacup2016_Q1.cpp hosted with by GitHub