多校2016 1004 HDU5784 统计锐角三角形数目

 1 // #pragma comment(linker, "/STACK:102c000000,102c000000")
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <sstream>
 6 #include <string>
 7 #include <algorithm>
 8 #include <list>
 9 #include <map>
10 #include <vector>
11 #include <queue>
12 #include <stack>
13 #include <cmath>
14 #include <cstdlib>
15 // #include <conio.h>
16 using namespace std;
17 #define clc(a,b) memset(a,b,sizeof(a))
18 #define inf 0x3f3f3f3f
19 #define lson l,mid,rt<<1
20 // #define rson mid+1,r,rt<<1|1
21 const int N = 2010;
22 const int M = 1e6+10;
23 const int MOD = 1e9+7;
24 #define LL long long
25 #define LB long double
26 // #define mi() (l+r)>>1
27 double const pi = acos(-1);
28 const double eps = 1e-8;
29 void fre(){freopen("in.txt","r",stdin);}
30 inline int read(){int x=0,f=1;char ch=getchar();while(ch>‘9‘||ch<‘0‘) {if(ch==‘-‘) f=-1;ch=getchar();}while(ch>=‘0‘&&ch<=‘9‘) { x=x*10+ch-‘0‘;ch=getchar();}return x*f;}
31 
32 struct Point{
33     LL x,y;
34     Point(){}
35     Point(LL _x,LL _y):x(_x),y(_y){}
36     Point operator + (const Point &t)const{
37         return Point(x+t.x,y+t.y);
38     }
39     Point operator - (const Point &t)const{
40         return Point(x-t.x,y-t.y);
41     }
42     LL operator * (const Point &t)const{
43         return x*t.y-y*t.x;
44     }
45     LL operator ^ (const Point &t)const{
46         return x*t.x+y*t.y;
47     }
48     bool operator < (const Point &b)const{
49         if (y * 1LL * b.y <= 0) {
50         if (y > 0 || b.y > 0) return y < b.y;
51         if (y == 0 && b.y == 0) return x < b.x;
52         }
53         return (*this)*b > 0;
54     }
55 }p[N],v[N<<1];
56 
57 int main(){
58     int n;
59     while(~scanf("%d",&n)){
60         for(int i=0;i<n;i++) scanf("%I64d%I64d",&p[i].x,&p[i].y);
61         LL ans=1LL*n*(n-1)*(n-2)/6,tem=0;
62         for(int k=0;k<n;k++){
63             int cnt=0;
64             for(int i=0;i<n;i++){
65                 if(k==i)continue;
66                 v[cnt++]=p[i]-p[k];
67             }
68             sort(v,v+cnt);
69             for(int i=0;i<cnt;i++) v[i+cnt]=v[i];
70             int cxt=0;
71             for(int i=1;i<cnt;i++){
72                 if(v[i-1]*v[i]==0&&(v[i-1]^v[i])>0) cxt++;
73                 else cxt=0;
74                 tem+=cxt; 
75             }
76             for(int i=0,p1=0,p2=0;i<cnt;i++){
77                 while(p1<=i||(p1<i+cnt&&v[p1]*v[i]<0&&(v[p1]^v[i])>0)) p1++;
78                 while(p2<=i||(p2<i+cnt&&v[p2]*v[i]<0)) p2++;
79                 ans-=p2-p1;
80             }
81         }
82         printf("%I64d\n",ans-tem/2);
83     }
84     return 0;
85 }