[CodeForces]CodeForces - 1025F Disjoint Triangles

　　题意：

　　　　给出平面上n个点，问能在其中选出6个点，组成两个三角形，使得其互不相交

　　　　问有多少种选法

　　大致思路

　　　　 考虑枚举一条直线，将所有得点分为左右两部分，其中有两个点在直线上，

　　　　以这两个点为顶点，分别统计左边有多少个三角形，右边有多少个三角形即可。

　　　　枚举直线时，枚举一个点，然后对其他点极角排序，
　　　　枚举另一个点，可快速算出该直线左右两边的点的数量(二分，叉积），组合一下即可。

　　　　

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 */

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<map>
 #include<stack>
 #include<time.h>
 #include<cstdlib>
 #include<cmath>
 #include<list>
 using namespace std;
 #define MAXN 10000006
 #define eps 0
 #define For(i,a,b) for(int i=a;i<=b;i++)
 #define Fore(i,a,b) for(int i=a;i>=b;i--)
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define mkp make_pair
 #define pb push_back
 #define cr clear()
 #define sz size()
 #define met(a,b) memset(a,b,sizeof(a))
 #define iossy ios::sync_with_stdio(false)
 #define fr freopen
 #define pi acos(-1.0)
 #define Vector Point
 #define fir first
 #define sec second
 #define it_s_too_hard main
 const long long inf=1LL<<;
 const int iinf=<<;
 const double dinf=1e17;
 const int Mod=1e9+;
 typedef long long ll;
 typedef long double ld;
 struct Point{
     ll x,y;
     double ang;
     Point(ll x=,ll y=):x(x),y(y) {}
     Point operator - (const Point &a)const { return Point(x-a.x,y-a.y);}
 };
 ll Cross(Vector a,Vector b){ return a.x*b.y-a.y*b.x;}
 int n;
 Point o;
 Vector v1,v2;
 bool cmp(Point a,Point b){
     return a.ang<b.ang;
 }
 Point p[],tp[];
 void I_can_t_solve_it(){
     cin>>n;
     ll ans=,cnt1,l,r,ct;
     For(i,,n) cin>>p[i].x>>p[i].y;
     For(i,,n){
         For(j,,n) {
             if(j==i) continue;
             if(j>i) tp[j-]=p[j]-p[i],tp[j-].ang=atan2(tp[j-].y,tp[j-].x);
             else tp[j]=p[j]-p[i],tp[j].ang=atan2(tp[j].y,tp[j].x);
         }
         ct=n-;cnt1=;
         sort(tp+,tp+ct+,cmp);
         For(j,,ct) tp[j+ct]=tp[j];
         For(j,,ct){
             if(j>cnt1) cnt1=j;
             if(cnt1>ct*) break;
             while(cnt1-j+<=ct && cnt1<ct* && Cross(tp[j],tp[cnt1+])>) cnt1++;
             l=cnt1-j;r=n--l-;
             ans+=l*(l-)*r*(r-)/;
         }
     }
     cout<<ans/<<endl;

 }
 int it_s_too_hard(){
     int t=;
     iossy;
     For(i,,t){
         I_can_t_solve_it();
     }
     return ;
 }