题目链接：
POJ 1228 Grandpa’s Estate
题意：
给出n个点，问由这n个点能不能位唯一确定一个凸包？
分析：
看到有网友将这称为稳定凸包。
其实就是确定这n个点所形成的凸包每条边上是不是至少有三个点。
当顶点数小于6的时候肯定是不能构成凸包的，这个特判下就好了。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <cmath>
using namespace std;
const double EPS=1e-10;
const int MAX_N=1010;
const double PI=acos(-1.0);int n;struct Point{double x,y;Point () {}Point (double x,double y) : x(x),y(y) {}Point operator + (const Point& rhs) const {return Point(x+rhs.x,y+rhs.y);}Point operator - (const Point& rhs) const {return Point(x-rhs.x,y-rhs.y);}Point operator * (const double d) const {return Point(x*d,y*d);}double cross(const Point& rhs) const {return (x*rhs.y-y*rhs.x);}
}point[MAX_N],res[MAX_N];bool cmp_x(Point a,Point b)
{if(a.x==b.x) return a.y<b.y;else return a.x<b.x;
}int Andrew()
{sort(point,point+n,cmp_x);int k=0;for(int i=0;i<n;i++){while(k>1 && (res[k-1]-res[k-2]).cross(point[i]-res[k-1])<=0) k--;res[k++]=point[i];}int m=k;for(int i=n-2;i>=0;i--){while(k>m && (res[k-1]-res[k-2]).cross(point[i]-res[k-1])<=0) k--;res[k++]=point[i];}if(k>1) k--;return k;
}void solve()
{int total=Andrew();//printf("total=%d\n",total);/*将边上的点也存进凸包顶点里，在判断的时候只需要判断凸包顶点中是否一定存在相邻向量叉积为0即可上面Andrew（）函数中需要将(res[k-1]-res[k-2]).cross(point[i]-point[k-1])<=0改为(res[k-1]-res[k-2]).cross(point[i]-point[k-1])int flag=1;for(int i=1;i<total;i++){if((res[i-1]-res[i]).cross(res[(i+1)%total]-res[i])!=0&&(res[i]-res[(i+1)%total]).cross(res[(i+2)%total]-res[(i+1)%total])!=0){printf("NO\n");flag=0;break;}}if(flag){printf("YES\n");}*/int flag=1;for(int i=0;i<total;i++){ //遍历凸包边Point a=res[i],b=res[(i+1)%total];//该边端点是a和bint cnt=0;for(int j=0;j<n;j++){ //检查该边上点的个数//printf("i=%d j=%d %.2f\n",i,j,fabs((point[j]-a).cross(b-point[j])));if((point[j]-a).cross(b-point[j])==0){
// 在这条边上的点cnt++;if(cnt>=3) break;}}if(cnt<3){ //少于三个点//printf("i=%d\n",i);printf("NO\n");flag=0;break;}}if(flag) printf("YES\n");}int main()
{//freopen("Din.txt","r",stdin);int T;scanf("%d",&T);while(T--){scanf("%d",&n);for(int i=0;i<n;i++){scanf("%lf%lf",&point[i].x,&point[i].y);}//要构成稳定凸包至少需要5个顶点(构成三角形)if(n<6){printf("NO\n");continue;}solve();}return 0;
}