http://acm.pku.edu.cn/JudgeOnline/problem?id=2187凸包直徑 diameter of a convex polygon
#include <cstdlib>
#include<iostream>
#include<cmath>
using namespace std;
#define MAXN 50005
#define eps 1e-8
#define zero(x) (((x)>0?(x):-(x))<eps)

struct point{double x,y;}p5[MAXN],convex1[MAXN];;

//計算cross product (P1-P0)x(P2-P0)

double xmult(point p1,point p2,point p0){
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}


double dist(point p3,point p4){
return (p3.x-p4.x)*(p3.x-p4.x)+(p3.y-p4.y)*(p3.y-p4.y);
}

//graham算法順時針構造包含所有共線點的凸包,O(nlogn)
point p1,p2;

int graham_cp(const void* a,const void* b){
double ret=xmult(*((point*)a),*((point*)b),p1);
return zero(ret)?(xmult(*((point*)a),*((point*)b),p2)>0?1:-1):(ret>0?1:-1);
}

void _graham(int n,point* p,int& s,point* ch){
int i,k=0;
for (p1=p2=p[0],i=1;i<n;p2.x+=p[i].x,p2.y+=p[i].y,i++)
if (p1.y-p[i].y>eps||(zero(p1.y-p[i].y)&&p1.x>p[i].x))
p1=p[k=i];
p2.x/=n,p2.y/=n;
p[k]=p[0],p[0]=p1;
qsort(p+1,n-1,sizeof(point),graham_cp);
for (ch[0]=p[0],ch[1]=p[1],ch[2]=p[2],s=i=3;i<n;ch[s++]=p[i++])
for (;s>2&&xmult(ch[s-2],p[i],ch[s-1])<-eps;s--);
}

//構造凸包接口函數,傳入原始點集大小n,點集p(p原有順序被打亂!)
//返回凸包大小,凸包的點在convex中
//參數maxsize為1包含共線點,為0不包含共線點,缺省為1
//參數clockwise為1順時針構造,為0逆時針構造,缺省為1
//在輸入僅有若干共線點時算法不穩定,可能有此類情況請另行處理!
//不能去掉點集中重合的點

int graham(int n,point* p,point* convex,int maxsize=1,int dir=1){
point* temp=new point[n];
int s,i;
_graham(n,p,s,temp);
for (convex[0]=temp[0],n=1,i=(dir?1:(s-1));dir?(i<s):i;i+=(dir?1:-1))
if (maxsize||!zero(xmult(temp[i-1],temp[i],temp[(i+1)%s])))
convex[n++]=temp[i];
delete []temp;
return n;
}



int main()

{
int n,m;//m凸包頂點
double a,b;
//memset()
scanf("%d",&n);

for(int i=0;i<n;i++){
scanf("%lf %lf",&p5[i].x,&p5[i].y);
}
double maxn=0;
m=graham(n,p5,convex1,1,1);
for(int i=0;i<m;i++)

for(int j=0;j<m;j++){
double tmp=dist(convex1[i],convex1[j]);
if(maxn<tmp)maxn=tmp;
}
printf("%d\n",(int)maxn);
//system("pause");
return 0;
}

posted on 2009-03-03 23:24
爬 閱讀(1245)
評論(0) 編輯 收藏 引用 所屬分類:
computing geometry