CodeForces - 552D Vanya and Triangles （数学几何求三角形个数）水

CodeForces - 552D

Vanya and Triangles

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Description

Vanya got bored and he painted n distinct points on the plane. After that he connected all the points pairwise and saw that as a result many triangles were formed with vertices in the painted points. He asks you to count the number of the formed triangles with the non-zero

Input

The first line contains integer n (1 ≤ n ≤ 2000) — the number of the points painted on the plane.

Next n lines contain two integers each x_i, y_i ( - 100 ≤ x_i, y_i ≤ 100) — the coordinates of the i-th point. It is guaranteed that no two given points coincide.

Output

In the first line print an integer — the number of triangles with the non-zero area among the painted points.

Sample Input

Input

40 01 12 02 2

Output

3

Input

30 01 12 0

Output

1

Input

11 1

Output

0

Hint

Note to the first sample test. There are 3 triangles formed: (0, 0) - (1, 1) - (2, 0); (0, 0) - (2, 2) - (2, 0); (1, 1) - (2, 2) - (2, 0).

Note to the second sample test. There is 1 triangle formed: (0, 0) - (1, 1) - (2, 0).

Note to the third sample test. A single point doesn't form a single triangle.

Source

Codeforces Round #308 (Div. 2)

//题意：

给你n个点，问可以组成多少个三角形？

//思路：

先求出来所有的情况为sum=C(n,3)，再找出来不能组成的三角形的个数（三点在一条直线上）C(kk,3)，相减即为所求。

在求kk时会用到去重公式：kk=(sqrt(1+8*cnt)+1)/2;不明白的自己推一下就知道了。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define ll long long
#define N 2010
using namespace std;
struct zz
{
	double x;
	double y;
}p[N];
bool cmp1(zz a,zz b)
{
	if(a.x==b.x)
		return a.y<b.y;
	return a.x<b.x;
}
struct ss
{
	double k;
	double b;
}pp[N*N];
bool cmp2(ss a,ss b)
{
	if(a.k==b.k)
		return a.b<b.b;
	return a.k<b.k;
}
ll C(int x,int k)
{
	int i,j;
	ll s=1;
	for(i=x,j=1;i>x-k;i--,j++)
		s*=i,s/=j;
	return s;
}
int main()
{
	int n,i,j,k;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=0;i<n;i++)
			scanf("%lf%lf",&p[i].x,&p[i].y);
		sort(p,p+n,cmp1);
		k=0;
		double x,y;
		for(i=0;i<n-1;i++)
		{
			for(j=i+1;j<n;j++)
			{
				x=(p[j].x-p[i].x);y=(p[j].y-p[i].y);
				if(x==0)
				{
					pp[k].k=1000.0;
					pp[k].b=p[j].x;
				}
				else
				{
					pp[k].k=y/x;
					pp[k].b=p[i].y-pp[k].k*p[i].x;
				}
				k++;
			}
		}
		sort(pp,pp+k,cmp2);
		int cnt=1,kk;
		ll num=C(n,3);
		for(i=0;i<k;i++)
		{
			if(fabs(pp[i].k-pp[i+1].k)<=1e-7&&fabs(pp[i].b-pp[i+1].b)<=1e-7)
				cnt++;
			else
			{
//				printf("%d\n",cnt);
				kk=(sqrt(1+8*cnt)+1)/2;
				num-=C(kk,3);
				cnt=1;
			}
		}
		printf("%lld\n",num);
	}
	return 0;
}