poj2187 旋转卡(qia)壳(ke)

题意：求凸包的直径

关于对踵点对、旋转卡壳算法的介绍可以参考这里：

http://www.cnblogs.com/Booble/archive/2011/04/03/2004865.html

http://www.cppblog.com/staryjy/archive/2009/11/19/101412.html

http://blog.csdn.net/ACMaker

这里使用了lrj的complex<double>大法来表示复数。

注意别忘了复数乘法的定义：(a+bi)*(c+di)=(ac-bd)+(bc+ad)i

 #include <iostream>
 #include <complex>
 #include <algorithm>
 #include <cstdio>
 using namespace std;
 typedef complex<double> Point;      //Point A:complex   x+yi
 typedef Point Vector;
 const double eps=1e-;

 int dcmp(double x)      //return  0:x==0  -1:x<0  1:x>0
 {
     if (fabs(x)<eps)    return ;
         else return x<?-:;
 }

 bool operator == (const Point &A,const Point &B)
 {
     double ax=real(A),ay=imag(A),bx=real(B),by=imag(B);
     return dcmp(ax-bx)== && dcmp(ay-by)==;
 }
 /*
 bool operator <  (const Point &A,const Point &B)
 {
     double ax=real(A),ay=imag(A),bx=real(B),by=imag(B);
     return ((ax<bx)||(ax==bx && ay<by));
 }
 */
 bool cmp(const Point &A,const Point &B)
 {
     double ax=real(A),ay=imag(A),bx=real(B),by=imag(B);
     //return ((ax<bx)||(ax==bx && ay<by));
     int dx=dcmp(ax-bx),dy=dcmp(ay-by);   //return  0:ax==bx  -1:ax<bx  1:ax>bx
     return ((dx==-)||((dx==)&&(dy==-)));
 }

 double Dot(Vector A,Vector B)        //Ax*Bx+Ay*By
 {
     return real(conj(A)*B);
 }

 double Cross(Vector A,Vector B)     //Ax*By-Ay*Bx
 {
     return imag(conj(A)*B);
 }

 Vector Rotate(Vector A,double rad)
 {
     return A*exp(Point(,rad));
 }

 double Dist(Point A,Point B)        //distance^2
 {
     double ax=real(A),ay=imag(A),bx=real(B),by=imag(B);
     //cout<<ax<<" "<<ay<<" "<<bx<<" "<<by<<" "<<(ax-bx)*(ax-bx)+(ay-by)*(ay-by)<<endl;
     return ((ax-bx)*(ax-bx)+(ay-by)*(ay-by));
 }

 double PolygonArea(Point *p,int n)
 {
     double area=;
     for (int i=;i<n-;i++)
         area+=Cross(p[i]-p[],p[i+]-p[]);
     return area/;
 }

 int convexhull(Point *p,int n,Point *ch)
 {
     sort(p,p+n,cmp);
     int m=;
     for (int i=;i<n;i++)
     {
         while (m> && Cross(ch[m-]-ch[m-],p[i]-ch[m-])<=)
             m--;
         ch[m++]=p[i];
     }
     int k=m;
     for (int i=n-;i>=;i--)
     {
         while (m>k && Cross(ch[m-]-ch[m-],p[i]-ch[m-])<=)
             m--;
         ch[m++]=p[i];
     }
     if (n>)    m--;
     return m;
 }

 double rotating_calipers(Point *ch,int num)
 {
     int q=;
     double ans=;
     ch[num]=ch[];
     for(int p=;p<num;p++)
     {
         while(Cross(ch[p+]-ch[p],ch[q+]-ch[p])>Cross(ch[p+]-ch[p],ch[q]-ch[p]))
             q=(q+)%num;
         ans=max(ans,max(Dist(ch[p],ch[q]),Dist(ch[p+],ch[q+])));
     }
     return ans;
 }

 Point p[],ch[];
 int n,x,y;

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("ou.txt","w",stdout);

     while (cin>>n)
     {
         for (int i=;i<n;i++)
         {
             cin>>x>>y;
             p[i]=Point(x,y);
         }

         int num=convexhull(p,n,ch);
         int ans=rotating_calipers(ch,num);

         //cout<<num<<" "<<ans<<endl;

         cout<<ans<<endl;
     }
     return ;
 }

发现一个下载USACO Contest数据的好地方：http://iskren.info/tasks/USACO/