凸包稳定性判断：每条边上是否至少有三点 POJ 1228

//凸包稳定性判断：每条边上是否至少有三点
// POJ 1228

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <math.h>
using namespace std;
#define LL long long
typedef pair<int,int> pii;
const double inf = 0x3f3f3f3f;
const LL MOD =100000000LL;
const int N =110;
#define clc(a,b) memset(a,b,sizeof(a))
const double eps = 1e-8;
void fre() {freopen("in.txt","r",stdin);}
void freout() {freopen("out.txt","w",stdout);}
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;}

int sgn(double x){
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}

struct Point{
    double x,y;
    Point(){}
    Point(double _x,double _y){
        x = _x;y = _y;
    }
    Point operator -(const Point &b)const{
        return Point(x - b.x,y - b.y);
    }
    double operator ^(const Point &b)const{
        return x*b.y - y*b.x;
    }
    double operator *(const Point &b)const{
        return x*b.x + y*b.y;
    }
    friend bool operator<(const Point &a,const Point &b){
        if(fabs(a.y-b.y)<eps) return a.x<b.x;
        return a.y<b.y;
    }
    friend double dis2(Point a){
        return a.x*a.x+a.y*a.y;
    }
};

double dis(Point a,Point b){
    return sqrt(dis2(a-b));
}

int mult(Point a,Point b,Point o){
    return sgn((a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y));
}
//返回top个点
int graham(Point p[],int n,Point q[]){
     int top=1;
     sort(p,p+n);
     if(n==0) return 0;
     q[0]=p[0];
     if(n==1) return 1;
     q[1]=p[1];
     if(n==2) return 2;
     q[2]=p[2];
     for(int i=2;i<n;i++){
         while(top&&(mult(p[i],q[top],q[top-1])>0)) top--;
         q[++top]=p[i];
     }
     int len=top;
     q[++top]=p[n-2];
     for(int i=n-3;i>=0;i--){
         while(top!=len&&(mult(p[i],q[top],q[top-1])>0)) top--;
         q[++top]=p[i];
     }
     return top;
}

// 判断凸包边上是否至少有三点
bool judge(Point p[],int m){
    p[m]=p[0];
    p[m+1]=p[1];
    for(int i=1;i<m;i++){
       if(mult(p[i-1],p[i+1],p[i])!=0&&mult(p[i],p[i+2],p[i+1])!=0)
        return false;
    }
    return true;
}


Point p[1011];
Point ch[1011];
int main(){
    // fre();
    int T;
    scanf("%d",&T);
    while(T--){
       int n;
       scanf("%d",&n);
       for(int i=0;i<n;i++){
           double x,y;
           scanf("%lf%lf",&x,&y);
           p[i]=Point(x,y);
       }
       int m=graham(p,n,ch);
       if(n<6) {
        puts("NO");
        continue;
       }
       if(judge(ch,m)) puts("YES");
       else puts("NO");
    }
    return 0;
}