HDU 5738 Eureka

Eureka

Problem Description

Professor Zhang draws n points on the plane, which are conveniently labeled by 1,2,...,n. The i-th point is at (xi,yi). Professor Zhang wants to know the number of best sets. As the value could be very large, print it modulo 109+7.

A set P (P contains the label of the points) is called best set if and only if there are at least one best pair in P. Two numbers u and v (u,v∈P,u≠v) are called best pair, if for every w∈P, f(u,v)≥g(u,v,w), where f(u,v)=(xu−xv)2+(yu−yv)2−−−−−−−−−−−−−−−−−−√ and g(u,v,w)=f(u,v)+f(v,w)+f(w,u)2.

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤1000) -- then number of points.

Each of the following n lines contains two integers xi and yi (−109≤xi,yi≤109) -- coordinates of the i-th point.

 

Output

For each test case, output an integer denoting the answer.

 

Sample Input

3

 

 

3

1 1

1 1

1 1

 

3

0 0

0 1

1 0

 

1

0 0

 

Sample Output

4

3

0

 

#include<cstdio>
#include<iostream>
#include<map>
#include<cmath>
#include<algorithm>
#define fi first
#define se second
using namespace std;
typedef long long LL;
const int MOD=1e9+7;
struct point
{
    LL x,y;
    bool operator == (const point &t)const
    {
        return x==t.x&&y==t.y;
    }
    bool operator < (const point &t)const
    {
        return x==t.x?y<t.y:x<t.x;
    }
}p[1005];
map<pair<LL,LL>,int>f;
LL gcd(LL a,LL b)
{
    return b==0?a:gcd(b,a%b);
}
pair<LL,LL> get_slope(point a,point b)
{
    LL up=b.y-a.y;
    LL dw=b.x-a.x;
    LL t=gcd(up,dw);
    up/=t,dw/=t;
    return make_pair(up,dw);
}
LL quick_mod(int a,int b)
{
    LL res=1,t=a;
    while(b)
    {
        if(b&1)res=(res*t)%MOD;
        t=(t*t)%MOD;
        b>>=1;
    }
    return res;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        LL ans=0;
        for(int i=0;i<n;i++)
            scanf("%lld%lld",&p[i].x,&p[i].y);
        sort(p,p+n);
        for(int i=0;i<n;i++)
        {
            f.clear();
            int res=0;
            for(int j=i+1;j<n;j++)
            {
                if(p[i]==p[j])res++;
                else f[get_slope(p[i],p[j])]++;
            }
            (ans+=quick_mod(2,res)-1)%=MOD;
            for(map<pair<LL,LL>,int>::iterator iter=f.begin();iter!=f.end();iter++)
                (ans+=quick_mod(2,res)*(quick_mod(2,iter->se)-1))%=MOD;

        }
        printf("%lld
",ans);
    }
}
/*
1
6
-1 -1
1 -1
-1 1
1 1
0 0
0 0

25
*/