计算几何-RC-poj2187

This article is made by Jason-Cow.
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今天学习一下旋(xuan1)转(zhuan3)卡(qia3)壳(qiao4)

1.凸包

2.对踵点

定理：最远点对必然属于对踵点对集合

对踵点定义：

        如果过凸包上的两个点可以画一对平行直线，使凸包上的所有点都

夹在两条平行线之间或落在平行线上，那么这两个点叫做一对对踵点。

具体有两种情况：

1.两个平行线正好卡着两个点

2.两个平行线分别卡着一条边和一个点

Rotating calipers Algorithm 是基于情况2的

        考虑到，固定一条边，凸包上的点到线的距离构成一个单峰函数，

所以，有“单调性”(姑且叫做单调性)

直观的感受一下

 

 

post the Rujia Liu 's words :

/*
    当Area(p[u], p[u+1], p[v+1]) <= Area(p[u], p[u+1], p[v])时停止旋转
    即Cross(p[u+1]-p[u], p[v+1]-p[u]) - Cross(p[u+1]-p[u], p[v]-p[u]) <= 0
    根据Cross(A,B) - Cross(A,C) = Cross(A,B-C)
    化简得Cross(p[u+1]-p[u], p[v+1]-p[v]) <= 0
*/

旋转code

 db RC(D*R,int n){//Rotating calipers
   R[]=R[n];// avoid to mod
   db ans=0.0;
   for(int u=,v=;u<n;u++){
     while(Cross(R[u+]-R[u],R[v+]-R[v])>)v=(v+)%n;
     ans=max(ans,Dis2(R[u],R[v]));
     ans=max(ans,Dis2(R[u+],R[v+]));
   }
   return ans;
 }

RC

一个小技巧，手写unique（其实是不会用STL，PS:不去重可以过）

 bl operator==(D A,D B){return (fabs(A.x-B.x)<eps && fabs(A.y-B.y)<eps);}

 void Unique(D*R,int&n){
   bl*In=new bl[n];
   for(int i=;i<=n;i++)if(R[i+]==R[i])In[i+]=;else In[i]=;
   int cnt=;
   for(int i=;i<=n;i++)if(!In[i])R[++cnt]=R[i];
   n=cnt;
 }

Unique

ACcode

 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <cstdlib>
 #include <cstdio>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <map>
 #include <set>
 using namespace std;
 #define sqr(x) ((x)*(x))
 #define RG register
 #define op operator
 #define IL inline
 typedef double db;
 typedef bool bl;
 const db pi=acos(-1.0),eps=1e-;
 struct D{
   db x,y;
   D(db x=0.0,db y=0.0):x(x),y(y){}
 };
 typedef D V;
 bl operator<(D A,D B){return A.x<B.x||(A.x==B.x&&A.y<B.y);}
 bl operator==(D A,D B){return (fabs(A.x-B.x)<eps && fabs(A.y-B.y)<eps);}
 V operator+(V A,V B){return V(A.x+B.x,A.y+B.y);}
 V operator-(V A,V B){return V(A.x-B.x,A.y-B.y);}
 V operator*(V A,db N){return V(A.x*N,A.y*N);}
 V operator/(V A,db N){return V(A.x/N,A.y/N);}

 db Ang(db x){return(x*180.0/pi);}
 db Rad(db x){return(x*pi/180.0);}
 V Rotate(V A,db a){return V(A.x*cos(a)-A.y*sin(a),A.x*sin(a)+A.y*cos(a));}
 db Dis2(D A,D B){return sqr(A.x-B.x)+sqr(A.y-B.y);}
 db Dis(D A,D B){return sqrt(sqr(A.x-B.x)+sqr(A.y-B.y));}
 db Cross(V A,V B){return A.x*B.y-A.y*B.x;}
 db Dot(V A,V B){return A.x*B.x+A.y*B.y;}

 void Unique(D*R,int&n){
   bl*In=new bl[n];
   for(int i=;i<=n;i++)if(R[i+]==R[i])In[i+]=;else In[i]=;
   int cnt=;
   for(int i=;i<=n;i++)if(!In[i])R[++cnt]=R[i];
   n=cnt;
 }

 int Andrew(D*R,int&n,D*A){
   int m=;
   sort(R+,R+n+);
   Unique(R,n);
   for(int i=;i<=n;i++){
     while(m>= && Cross(A[m]-A[m-],R[i]-A[m-])<=)m--;
     A[++m]=R[i];
   }
   int k=m;
   for(int i=n-;i>=;i--){
     while(m>k  && Cross(A[m]-A[m-],R[i]-A[m-])<=)m--;
     A[++m]=R[i];
   }
   return n>?m-:m;
 }

 db RC(D*R,int n){    //Rotating calipers
   R[]=R[n];             // avoid to mod
   db ans=0.0;
   for(int u=,v=;u<n;u++){
     while(Cross(R[u+]-R[u],R[v+]-R[v])>)v=(v+)%n;
     ans=max(ans,Dis2(R[u],R[v]));
     ans=max(ans,Dis2(R[u+],R[v+]));
   }
   return ans;
 }

 const int MAXN=(int)4e5+;
 D R[MAXN],T[MAXN];

 int main(){
   int n;scanf("%d",&n);
   for(int i=;i<=n;i++)scanf("%lf%lf",&R[i].x,&R[i].y);
   int m=Andrew(R,n,T);
   printf("%.0lf\n",RC(T,m));
   return ;
 }