poj1228 Grandpa's Estate【凸包稳定性判断】

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1e5+100;
const double eps = 1e-10;
const double pi = acos(-1.0);
struct point
{
    double x,y;
    point() {}
    point(double _x,double _y)
    {
        x = _x;
        y = _y;
    }
    bool operator < (const point &b) const
    {
        if(y==b.y)
            return x<b.x;
        return y<b.y;
    }
    bool operator == (const point &b)const
    {
        return x==b.x && y==b.y;
    }
}a[maxn],ans[maxn];
double x_mul(point p0,point p1,point p2)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
double dis(point p1,point p2)
{
    return sqrt((p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));
}
bool cmp(point t1,point t2)
{
    double tmp = x_mul(a[0],t1,t2);
    if(tmp==0)
        return dis(a[0],t1)<dis(a[0],t2);
    else
        return tmp>0;
}
int graham(int n)
{
    sort(a,a+n);
    sort(a+1,a+n,cmp);
    ans[0] = a[0];
    ans[1] = a[1];
    ans[2] = a[2];
    int top = 2;
    for(int i=3;i<n;i++)
    {
        while(top>=2&&x_mul(ans[top-1],ans[top],a[i])<0)
            top--;
        ans[++top] = a[i];
    }
    return top;
}
bool slove(int n)
{
    int top = graham(n);
    if(top<3)
        return false;
    for(int i=1;i<top;i++)
    {
        if(x_mul(ans[i-1],ans[i+1],ans[i])!=0&&
           x_mul(ans[i],ans[i+2],ans[i+1])!=0)
            return false;
    }
    return true;
}
int main(void)
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%lf %lf",&a[i].x,&a[i].y);
        if(slove(n))
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}