遍历每个i，把剩余点与i的距离做key放入map，value是具有相同距离的点的个数

#include <map>
#include <queue>
#include <set>
#include <algorithm>
using namespace std;
class Solution
{
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points)
    {
        int res = 0;
        for(int i = 0; i < points.size(); i++)
        {
            map<int,int> mymap;
            for(int j = 0; j<points.size(); j++)
            {
                if(j!=i)
                    mymap[dis(points[i],points[j])]++;
            }
            for(map<int,int>::iterator iter = mymap.begin(); iter != mymap.end(); iter++)
            {
                  if(iter->second>=2)
                    res += (iter->second) * (iter->second - 1);
            }
        }
        return res;
    }
private:
    int dis(pair<int,int> &x,pair<int,int> &y)
    {
        return (x.first-y.first)*(x.first-y.first)+(x.second-y.second)*(x.second-y.second);
    }
};