就是给你一堆点，看这些点能否构成一个 稳定的凸包。
凸包每条边上有3个及以上的点就可以了。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <vector>
typedef double db;
const db eps = 1e-;
const db pi = acos(-);
using namespace std;
int sign(db k){
if (k>eps) return ; else if (k<-eps) return -; return ;
}
int cmp(db k1,db k2){return sign(k1-k2);}
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=;}// k3 在 [k1,k2] 内
struct point{
db x,y;
point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
point operator * (db k1) const{return (point){x*k1,y*k1};}
point operator / (db k1) const{return (point){x/k1,y/k1};}
bool operator <(const point &k1)const {
int c=cmp(x,k1.x);
if(c)return c==-;
return cmp(y,k1.y)==-;
}
};
int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
vector<point> convexHull(vector<point>ps){
int n = ps.size();if(n<=)return ps;
sort(ps.begin(),ps.end());
vector<point> qs(n*);int k=;
for(int i=;i<n;qs[k++]=ps[i++])
while (k>&&cross(qs[k-]-qs[k-],ps[i]-qs[k-])<=)--k;
for(int i=n-,t=k;i>=;qs[k++]=ps[i--])
while (k>t&&cross(qs[k-]-qs[k-],ps[i]-qs[k-])<=)--k;
qs.resize(k-);
return qs;
}
vector<point> v;
int t,n;
point p[];
point tmp;
int main(){
scanf("%d",&t);
while (t--){
scanf("%d",&n);
for(int i=;i<=n;i++){
scanf("%lf%lf",&tmp.x,&tmp.y);
v.push_back(tmp);
p[i]=tmp;
}
v=convexHull(v);
int m = v.size();
bool f=;
for(int i=;i<m;i++){
int cnt=;
for(int j=;j<=n;j++){
if(inmid(v[i],v[(i+)%m],p[j])){
cnt++;
}
}
if(cnt<){
f=;
break;
}
}
if(v.size()<=)f=;
if(f)printf("YES\n");
else printf("NO\n");
v.clear();
}
}