Comet OJ - Contest # 15 (B: When we concentric together)

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Title Description

There on the plane  n n coordinate disparate point, may I ask how many of the group's three non-collinear points, three points  circumcenter  also in this  n in n points?

　

Enter a description

 

The first line has a positive integer  n representative of n points in the plane.

Then there are  n- n-row, among the first  i i line contains two integers  x_i, y_i, X i , Y i  represents the  i  coordinate points is  (x_i, y_i) ( X i , Y i ) .

1<=n<=2000

-10^9<=x,y<=10^9

!! If i = j, then (xi, yi) = (xj, yj);

Sample input

5
0 0
-2 0
0 2
-1 1
2 0

 Sample Output

2

　　

 

 

 　　　　When to get this title, thinking about violence, but involves four for, so a timeout. There is no time to engage in out of the game, is very regrettable. Once on shore, watching AC codes of others, and suddenly!

　　　　If the distance from a point to the other three points are equal, then this is certainly not the three points are collinear, so this does not need to manage a fundamentally, go directly from a point to another point a1, the number of occurrences with a map from the recording , if the number of occurrences of a certain distance K> = 3 is present, a1 is the circumcenter these points, due to the number of combinations is selected, it can be a CK 3, three of which are selected from the group of the number of occurrences, so CK 3 .

　　　　There are also map iterators wording, always forget

#include<cstring>
#include<iostream>
#include<map>
#include<algorithm>
using namespace std;
const int maxn = 2e3+10;
typedef long long ll;
ll num[maxn];
struct node
{
    ll x,y;
}st[maxn];
ll distance(ll x1,ll y1,ll x2, ll y2)
{
    return ((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main()
{
    int n ; 
    cin>>n;
    for(int i = 0 ; i< n ; i++)
    {
        cin>>st[i].x>>st[i].y;
    }
    ll sum = 0 ;
    for(int i = 0 ; i <n ;i ++)
    {
        map<ll,ll>mm;
        for(int j = 0 ; j<n  ; j++)
        {
            if(i!=j)
            {
                    ll s=distance(st[i].x,st[i].y,st[j].x,st[j].y);
                //    cout<<s<<endl;
                    mm[s]++;
            }
        }　　

        map<ll,ll>::iterator it;　　//遍历map写法

        for(it = mm.begin();it!=mm.end();it++)
        {
            if(it->second>=3)　　
            {
                ll k = it->second;
            //    cout<<k<<endl;
                sum+=k*(k-1)*(k-2)/6; 　　//C k 3　　
            }
        }
    }
    cout<<sum<<endl;
    return 0;
}