链接：戳这里
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
思路：
先去重，题目要求的集合是一条线段。并且u,v是线段的两个端点。那么对于点集的每一条线段，都统计每一条的贡献
设线段上有k个点，那么贡献为即(C(k,2)+C(k,3)+...+C(k,k))=2^k-1-k。但是在多条线段的交点处如果恰好是g个重点存在
那么我们需要减去多算的，设该重点恰好在m条线段上，那么多算了(m-1)次。多算的贡献为(m-1)*((2^g)-g-1)
问题就变成了一条线段上有多少个点以及一个点出现在多少线段上。直接n*nlogn处理
注意卡精度，开long double 吧
代码：
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
