题目链接：

Eureka

Time Limit: 8000/4000 MS (Java/Others)  

  Memory Limit: 65536/65536 K (Java/Others)

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

题意：

给 n个点,大于等于2个在同一条直线上的点可以构成一个集合,问你现在有多少个集合;

思路：

先把给的这些点按坐标排序,然后按顺序选一个点i,这个点i作为一定选到集合里面的点,然后再选枚举这个点之后的点j，形成一条直线,再看这条直线上没有被访问过的点k(i<k<n&&k!=j)有多少;假设有num个,那么就可以形成包含点i的集合2^^num-1,同时这些点里面有重合的点,还有就是为降低复杂度,要用极角排序,但是最后判断的时候极角排序的精度好像又不太够,所有我就直接用原来的坐标判断的;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
    for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + ‘0‘);
    putchar(‘\n‘);
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=500+10;
const double eps=1e-9;

int n,vis[1010];
LL fx,fy,f[1010];
struct node
{
    double ang;
    LL x,y;
}po[1010],temp[1010];
int cmp1(node a,node b)
{
    return a.ang<b.ang;
}
int cmp(node a,node b)
{
    if(a.y==b.y)return a.x<b.x;
    return a.y<b.y;
}

int main()
{
        int t;
        read(t);
        f[0]=1;
        For(i,1,1008)
        {
            f[i]=f[i-1]*2%mod;
        }
        while(t--)
        {
            read(n);
            For(i,1,n)
            {
                read(po[i].x);read(po[i].y);
            }
            sort(po+1,po+n+1,cmp);
            LL ans=0;
            For(i,1,n-1)
            {
                int cnt=0,s=0;
                For(j,i+1,n)
                {
                    if(po[j].x==po[i].x&&po[j].y==po[i].y){s++;continue;}
                    temp[++cnt].ang=atan2(po[j].y-po[i].y,po[j].x-po[i].x);
                    temp[cnt].x=po[j].x;
                    temp[cnt].y=po[j].y;
                }
                sort(temp+1,temp+cnt+1,cmp1);
                fx=po[i].x,fy=po[i].y;
				int d=0;
                for(int j=1;j<=cnt;)
                {
                    int k,num=s+1;
                    for(k=j+1;k<=cnt;k++)
                    {
                        if((temp[k].y-fy)*(temp[j].x-fx)!=(temp[j].y-fy)*(temp[k].x-fx))break;
                        num++;
                    }
                    j=k;
                    ans=(ans+f[num]-1+mod)%mod;
                    d++;
                }
                ans=(ans-(LL)(d-1)*(f[s]-1+mod)%mod+mod)%mod;
            }
            cout<<ans<<endl;
        }
        return 0;
}