多校5 1004 HDU5784 统计锐角三角形数目

http://acm.hdu.edu.cn/showproblem.php?pid=5784

题意：n个点，找多少个锐角三角形数目

思路：极角排序+two pointers

当前选择的点集要倍增一倍，点集过大时，极角排序后，后面的点有可能和前面的点形成钝角

ans=总的三角形数目 - 三点共线的情况-直角和钝角

 // #pragma comment(linker, "/STACK:102c000000,102c000000")
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <sstream>
 #include <string>
 #include <algorithm>
 #include <list>
 #include <map>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <cstdlib>
 // #include <conio.h>
 using namespace std;
 #define clc(a,b) memset(a,b,sizeof(a))
 #define inf 0x3f3f3f3f
 #define lson l,mid,rt<<1
 // #define rson mid+1,r,rt<<1|1
 const int N = ;
 const int M = 1e6+;
 const int MOD = 1e9+;
 #define LL long long
 #define LB long double
 // #define mi() (l+r)>>1
 double const pi = acos(-);
 const double eps = 1e-;
 void fre(){freopen("in.txt","r",stdin);}
 inline int read(){int x=,f=;char ch=getchar();while(ch>''||ch<'') {if(ch=='-') f=-;ch=getchar();}while(ch>=''&&ch<='') { x=x*+ch-'';ch=getchar();}return x*f;}

 struct Point{
     LL x,y;
     Point(){}
     Point(LL _x,LL _y):x(_x),y(_y){}
     Point operator + (const Point &t)const{
         return Point(x+t.x,y+t.y);
     }
     Point operator - (const Point &t)const{
         return Point(x-t.x,y-t.y);
     }
     LL operator * (const Point &t)const{
         return x*t.y-y*t.x;
     }
     LL operator ^ (const Point &t)const{
         return x*t.x+y*t.y;
     }
     bool operator < (const Point &b)const{
         if (y * 1LL * b.y <= ) {
         if (y >  || b.y > ) return y < b.y;
         if (y ==  && b.y == ) return x < b.x;
         }
         return (*this)*b > ;
     }
 }p[N],v[N<<];

 int main(){
     int n;
     while(~scanf("%d",&n)){
         for(int i=;i<n;i++) scanf("%I64d%I64d",&p[i].x,&p[i].y);
         LL ans=1LL*n*(n-)*(n-)/,tem=;
         for(int k=;k<n;k++){
             int cnt=;
             for(int i=;i<n;i++){
                 if(k==i)continue;
                 v[cnt++]=p[i]-p[k];
             }
             sort(v,v+cnt);
             for(int i=;i<cnt;i++) v[i+cnt]=v[i];
             int cxt=;
             for(int i=;i<cnt;i++){
                 if(v[i-]*v[i]==&&(v[i-]^v[i])>) cxt++;
                 else cxt=;
                 tem+=cxt;
             }
             for(int i=,p1=,p2=;i<cnt;i++){
                 while(p1<=i||(p1<i+cnt&&v[p1]*v[i]<&&(v[p1]^v[i])>)) p1++;
                 while(p2<=i||(p2<i+cnt&&v[p2]*v[i]<)) p2++;
                 ans-=p2-p1;
             }
         }
         printf("%I64d\n",ans-tem/);
     }
     return ;
 }