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poj2187 旋转卡(qia)壳(ke)

时间:2015-02-01 19:07:15      阅读:115      评论:0      收藏:0      [点我收藏+]

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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

 

技术分享
  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 }
View Code

 

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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原文地址:http://www.cnblogs.com/pdev/p/4265803.html

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