FOJ 2148（求凸四边形数量）

#include <iostream>
#include <stdio.h>
using namespace std;
struct point
{
int x,y;
} points[40];
int abs(int k){if(k<0) return -k;return k;}
int mult(point a,point b,point c)
{
a.x-=c.x;
a.y-=c.y;
b.x-=c.x;
b.y-=c.y;
return abs(a.x*b.y-a.y*b.x);
}
bool OK(point a,point b,point c,point d)
{
if(mult(a,b,c)+mult(a,c,d)+mult(a,b,d)<=mult(b,c,d))
return false;
if(mult(b,a,c)+mult(b,c,d)+mult(b,a,d)<=mult(a,c,d))
return false;
if(mult(c,a,b)+mult(c,b,d)+mult(c,a,d)<=mult(a,b,d))
return false;
if(mult(d,a,b)+mult(d,b,c)+mult(d,a,c)<=mult(a,b,c))
return false;
return true;
}
int n;
int main()
{
int t;cin>>t;int an=1;
while(t--)
{
cin>>n;
for(int i=0;i<n;i++)
scanf("%d%d",&points[i].x,&points[i].y);
int ans=0;
for(int i=0;i<n;i++)
for(int j=i+1;j<n;j++)
for(int k=j+1;k<n;k++)
for(int h=k+1;h<n;h++)
{
if(OK(points[i],points[j],points[k],points[h]))
ans++;
}
printf("Case %d: %d\n",an,ans);an++;
}
return 0;
}