HDU5020（判断三点共线）BestCoder Round #10（1003）

Revenge of Collinearity

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 184    Accepted Submission(s): 51

Problem Description

In geometry, collinearity is a property of a set of points, specifically, the property of lying on a single line. A set of points with this property is said to be collinear (often misspelled as colinear).
---Wikipedia

Today, Collinearity takes revenge on you. Given a set of N points in two-dimensional coordinate system, you have to find how many set of <P _i, P _j, P _k> from these N points are collinear. Note that <P _i, P _j, P _k> cannot contains same point, and <P _i, P _j, P _k> and <P _i, P _k, P _j> are considered as the same set, i.e. the order in the set doesn’t matter.

 

Input

The first line contains a single integer T, indicating the number of test cases. 

Each test case begins with an integer N, following N lines, each line contains two integers Xi and Yi, describing a point.

[Technical Specification]
1. 1 <= T <= 33
2. 3 <= N <= 1 000
3. -1 000 000 000 <= Xi, Yi <= 1 000 000 000, and no two points are identical.
4. The ratio of test cases with N > 100 is less than 25%.

 

Output

For each query, output the number of three points set which are collinear.

 

Sample Input

    
    
    
    

     
     
     
     2
3
1 1
2 2
3 3
4
0 0
1 0
0 1
1 1
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     1
0
    
    
    
    

题意：给出n个点，找出三点贡献的点的集合

思路：n=1000，我比赛的时候实在没思路，虽然知道暴力n^3一定会超时，但还也写完意思的交了一发，平时几乎不碰几何党的飘过~

            赛后看了某神犇的代码，思路瞬间开阔了

            就是将n个点按x，y从小到大排序，x为第一关键字，y为第二关键字

            排完序以后就是两层for暴搜了，因为已经排好序，就可以按斜率将所有斜率一样的点归为一个集合

            假设当前要判断的点为i，有p个点与之共线，因为i是一定要选的，所以只需在剩下的p个点中选择2个就好了，为p*(p-1)/2

            累加这些就是答案了