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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
1 #include <iostream> 2 #include <complex> 3 #include <algorithm> 4 #include <cstdio> 5 using namespace std; 6 typedef complex<double> Point; //Point A:complex x+yi 7 typedef Point Vector; 8 const double eps=1e-10; 9 10 int dcmp(double x) //return 0:x==0 -1:x<0 1:x>0 11 { 12 if (fabs(x)<eps) return 0; 13 else return x<0?-1:1; 14 } 15 16 bool operator == (const Point &A,const Point &B) 17 { 18 double ax=real(A),ay=imag(A),bx=real(B),by=imag(B); 19 return dcmp(ax-bx)==0 && dcmp(ay-by)==0; 20 } 21 /* 22 bool operator < (const Point &A,const Point &B) 23 { 24 double ax=real(A),ay=imag(A),bx=real(B),by=imag(B); 25 return ((ax<bx)||(ax==bx && ay<by)); 26 } 27 */ 28 bool cmp(const Point &A,const Point &B) 29 { 30 double ax=real(A),ay=imag(A),bx=real(B),by=imag(B); 31 //return ((ax<bx)||(ax==bx && ay<by)); 32 int dx=dcmp(ax-bx),dy=dcmp(ay-by); //return 0:ax==bx -1:ax<bx 1:ax>bx 33 return ((dx==-1)||((dx==0)&&(dy==-1))); 34 } 35 36 37 double Dot(Vector A,Vector B) //Ax*Bx+Ay*By 38 { 39 return real(conj(A)*B); 40 } 41 42 double Cross(Vector A,Vector B) //Ax*By-Ay*Bx 43 { 44 return imag(conj(A)*B); 45 } 46 47 Vector Rotate(Vector A,double rad) 48 { 49 return A*exp(Point(0,rad)); 50 } 51 52 double Dist(Point A,Point B) //distance^2 53 { 54 double ax=real(A),ay=imag(A),bx=real(B),by=imag(B); 55 //cout<<ax<<" "<<ay<<" "<<bx<<" "<<by<<" "<<(ax-bx)*(ax-bx)+(ay-by)*(ay-by)<<endl; 56 return ((ax-bx)*(ax-bx)+(ay-by)*(ay-by)); 57 } 58 59 double PolygonArea(Point *p,int n) 60 { 61 double area=0; 62 for (int i=1;i<n-1;i++) 63 area+=Cross(p[i]-p[0],p[i+1]-p[0]); 64 return area/2; 65 } 66 67 int convexhull(Point *p,int n,Point *ch) 68 { 69 sort(p,p+n,cmp); 70 int m=0; 71 for (int i=0;i<n;i++) 72 { 73 while (m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) 74 m--; 75 ch[m++]=p[i]; 76 } 77 int k=m; 78 for (int i=n-2;i>=0;i--) 79 { 80 while (m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) 81 m--; 82 ch[m++]=p[i]; 83 } 84 if (n>1) m--; 85 return m; 86 } 87 88 double rotating_calipers(Point *ch,int num) 89 { 90 int q=1; 91 double ans=0; 92 ch[num]=ch[0]; 93 for(int p=0;p<num;p++) 94 { 95 while(Cross(ch[p+1]-ch[p],ch[q+1]-ch[p])>Cross(ch[p+1]-ch[p],ch[q]-ch[p])) 96 q=(q+1)%num; 97 ans=max(ans,max(Dist(ch[p],ch[q]),Dist(ch[p+1],ch[q+1]))); 98 } 99 return ans;100 }101 102 Point p[51000],ch[51000];103 int n,x,y;104 105 int main()106 {107 //freopen("in.txt","r",stdin);108 //freopen("ou.txt","w",stdout);109 110 while (cin>>n)111 {112 for (int i=0;i<n;i++)113 {114 cin>>x>>y;115 p[i]=Point(x,y);116 }117 118 int num=convexhull(p,n,ch);119 int ans=rotating_calipers(ch,num);120 121 //cout<<num<<" "<<ans<<endl;122 123 cout<<ans<<endl;124 }125 return 0;126 }
发现一个下载USACO Contest数据的好地方:http://iskren.info/tasks/USACO/
poj2187 旋转卡(qia)壳(ke)
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