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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 97 */
 98 
 99 
100 
101 #include<cstdio>
102 #include<iostream>
103 #include<cstring>
104 #include<algorithm>
105 #include<queue>
106 #include<set>
107 #include<map>
108 #include<stack>
109 #include<time.h>
110 #include<cstdlib>
111 #include<cmath>
112 #include<list>
113 using namespace std;
114 #define MAXN 10000006
115 #define eps 0
116 #define For(i,a,b) for(int i=a;i<=b;i++)
117 #define Fore(i,a,b) for(int i=a;i>=b;i--)
118 #define lson l,mid,rt<<1
119 #define rson mid+1,r,rt<<1|1
120 #define mkp make_pair
121 #define pb push_back
122 #define cr clear()
123 #define sz size()
124 #define met(a,b) memset(a,b,sizeof(a))
125 #define iossy ios::sync_with_stdio(false)
126 #define fr freopen
127 #define pi acos(-1.0)
128 #define Vector Point
129 #define fir first
130 #define sec second
131 #define it_s_too_hard main
132 const long long inf=1LL<<62;
133 const int iinf=1<<30;
134 const double dinf=1e17;
135 const int Mod=1e9+7;
136 typedef long long ll;
137 typedef long double ld;
138 struct Point{
139     ll x,y;
140     double ang;
141     Point(ll x=0,ll y=0):x(x),y(y) {}
142     Point operator - (const Point &a)const { return Point(x-a.x,y-a.y);}
143 };
144 ll Cross(Vector a,Vector b){ return a.x*b.y-a.y*b.x;}
145 int n; 
146 Point o;
147 Vector v1,v2;
148 bool cmp(Point a,Point b){
149     return a.ang<b.ang;
150 }
151 Point p[2050],tp[5050];
152 void I_can_t_solve_it(){
153     cin>>n;
154     ll ans=0,cnt1,l,r,ct;
155     For(i,1,n) cin>>p[i].x>>p[i].y;
156     For(i,1,n){
157         For(j,1,n) {
158             if(j==i) continue;
159             if(j>i) tp[j-1]=p[j]-p[i],tp[j-1].ang=atan2(tp[j-1].y,tp[j-1].x);
160             else tp[j]=p[j]-p[i],tp[j].ang=atan2(tp[j].y,tp[j].x);
161         }
162         ct=n-1;cnt1=1;
163         sort(tp+1,tp+ct+1,cmp);
164         For(j,1,ct) tp[j+ct]=tp[j];
165         For(j,1,ct){
166             if(j>cnt1) cnt1=j;
167             if(cnt1>ct*2) break;
168             while(cnt1-j+2<=ct && cnt1<ct*2 && Cross(tp[j],tp[cnt1+1])>0) cnt1++;
169             l=cnt1-j;r=n-1-l-1;
170             ans+=l*(l-1)*r*(r-1)/4;
171         }
172     }
173     cout<<ans/2<<endl;
174     
175 }
176 int it_s_too_hard(){
177     int t=1;
178     iossy;
179     For(i,1,t){
180         I_can_t_solve_it();
181     }
182     return 0;
183 }
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