题目链接：

题意：求多边形最小外接矩形。

思路：模板。

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #define min(x,y) ((x)<(y)?(x):(y))using namespace std;struct point{    double x,y;    point(){}    point(double _x,double _y)    {        x=_x;        y=_y;    }    void get()    {        scanf("%lf%lf",&x,&y);    }};const double EPS=1e-8;const int MAX=1005;point p[MAX],q[MAX],temp;int n,m;int DB(double x){    if(x>EPS) return 1;    if(x<-EPS) return -1;    return 0;}double Dis(point a,point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}//判断p在向量ab的哪一侧//右侧：返回正值//左侧：返回负值//在向量ab上返回0double cross(point a,point b,point p){    return (b.x-a.x)*(p.y-a.y)-(b.y-a.y)*(p.x-a.x);}int cmp(point a,point b){    double x=Dis(a,temp),y=Dis(b,temp);    int flag=DB(cross(temp,a,b));    if(flag) return flag==1;    return DB(x-y)<=0;}void Graham(point p[],int n,point q[],int &m){    point t;    int i,k=0,a,b;    for(i=1;i
          
           1&&DB(cross(q[m-2],q[m-1],p[i]))<=0) m--;        q[m++]=p[i];    }    m--;}//向量点乘//ab*acdouble dot(point a,point b,point c){    return (c.x-a.x)*(b.x-a.x)+(c.y-a.y)*(b.y-a.y);}double calMinRect(point pt[],int n){    if(n<3) return 0;    int i,p=1,q=1,r;    double ans=1e10,a,b,c;    for(i=0;i
           
            0) break; r=(r+1)%n; } a=cross(pt[i],pt[i+1],pt[p]); b=dot(pt[i],pt[i+1],pt[q])-dot(pt[i],pt[i+1],pt[r]); c=dot(pt[i],pt[i+1],pt[i+1]); ans=min(ans,a*b/c); } return ans;}int main(){ while(scanf("%d",&n),n) { int i; for(i=0;i