计算几何RCpoj2187

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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[0]=R[n];// avoid to mod
  db ans=0.0;
  for(int u=0,v=1;u<n;u++){
    while(Cross(R[u+1]-R[u],R[v+1]-R[v])>0)v=(v+1)%n;
    ans=max(ans,Dis2(R[u],R[v]));
    ans=max(ans,Dis2(R[u+1],R[v+1]));
  }
  return ans;
}

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=1;i<=n;i++)if(R[i+1]==R[i])In[i+1]=1;else In[i]=0;
  int cnt=0;
  for(int i=1;i<=n;i++)if(!In[i])R[++cnt]=R[i];
  n=cnt;
}

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-10;
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=1;i<=n;i++)if(R[i+1]==R[i])In[i+1]=1;else In[i]=0;
  int cnt=0;
  for(int i=1;i<=n;i++)if(!In[i])R[++cnt]=R[i];
  n=cnt;
}

int Andrew(D*R,int&n,D*A){
  int m=0;
  sort(R+1,R+n+1);
  Unique(R,n);
  for(int i=1;i<=n;i++){
    while(m>=2 && Cross(A[m]-A[m-1],R[i]-A[m-1])<=0)m--;
    A[++m]=R[i];
  }
  int k=m;
  for(int i=n-1;i>=1;i--){
    while(m>k  && Cross(A[m]-A[m-1],R[i]-A[m-1])<=0)m--;
    A[++m]=R[i];
  }
  return n>1?m-1:m;
}

db RC(D*R,int n){    //Rotating calipers
  R[0]=R[n];             // avoid to mod
  db ans=0.0;
  for(int u=0,v=1;u<n;u++){
    while(Cross(R[u+1]-R[u],R[v+1]-R[v])>0)v=(v+1)%n;
    ans=max(ans,Dis2(R[u],R[v]));
    ans=max(ans,Dis2(R[u+1],R[v+1]));
  }
  return ans;
}

const int MAXN=(int)4e5+10;
D R[MAXN],T[MAXN];

int main(){
  int n;scanf("%d",&n);
  for(int i=1;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 0;
}