poj 1228

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();
     }
 }