hdu5738（极角排序—+组合计数+数论）
Eureka
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3188 Accepted Submission(s): 899
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.
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.
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
推导一下可以知道best set一定是一些共线的点, 于是问题变成问有多少个子集共线. 首先, 把所有点按照(x,y)(x,y)(x,y)双关键字排序, 然后枚举最左边的点iii, 那么其他点jjj一定满足j>ij > ij>i. 把在这个点右边的点都做下极角排序(按照1gcd(dx,dy)(dx,dy)\frac{1}{gcd(dx, dy)}(dx, dy)gcd(dx,dy)1(dx,dy)排序), 统计下共线的就好了. 需要注意下对重点的处理.
分析：
对于集合P满足题目的公式的话，换句话说就是P集合中的所有的点都共线。
这题的最关键点就是对与重复点和重复边的处理
先把给的这些点按坐标逆时针排序,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（对于和i相同的点要特殊处理），计算出每个点j和i之间的向量存在q[]数组里面，然后使每个向量变成最简（即除以最大公约数），然后排序所有的向量那么相等的向量即在同一条直线上。
有n个元素的集合的子集的数目为2^n个
cnt记录i之后和i相同的点的数目
q[]为化简后的向量数组，q[x]~q[y]相等
所以对于这个点集合组成的集合有这些部分：
1：重点和i自身组成的集合，i算一个点，其余cnt个元素集合取出一个非空的集合即可：： pw[cnt]-1
2：在y-x个点组成的集合中取出一个非空子集:pw[y-x]-1;在cnt个重点集合中取出一个子集（可为空）:pw[cnt];两者之极即为所有组合
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