Description
After taking a modern art class, Farmer John has become interested in finding geometric patterns in everything around his farm. He carefully plots the locations of his N cows (2 <= N <= 1000), each one occupying a distinct point in the 2D plane, and he wonders how many different lines of symmetry exist for this set of points. A line of symmetry, of course, is a line across which the points on both sides are mirror images of each-other. Please help FJ answer this most pressing geometric question.
上过现代艺术课后,FJ开始感兴趣于在他农场中的几何图样。他计划将奶牛放置在二维平面上的N个互不相同的点(1<=N<=1000),他希望找出这个点集有多少条对称轴。他急切地需要你帮忙解决这个几何问题。
Input
* Line 1: The single integer N.
* Lines 2..1+N: Line i+1 contains two space-separated integers representing the x and y coordinates of the ith cow (-10,000 <= x,y <= 10,000).
Output
* Line 1: The number of different lines of symmetry of the point set.
圆上的整点很少,所以可以找出点集的重心并把点按到重心的距离排序,到重心相同距离的点在同一圆上,计算这些点的对称轴(最多四条),最后取交集即可,时间复杂度是O(n2logn)但因为整点且范围小所以不会达到最坏情况
#include<bits/stdc++.h> typedef long double ld; const ld pi=std::acos(-1),_0=1e-7l; int n,ans=0,xs=0,ys=0,lp=0,ap=0,ad=0; struct pos{int x,y;long long d;ld a;}ps[1007]; ld ls[2007],as[2007]; bool operator<(pos a,pos b){ return a.d!=b.d?a.d<b.d:a.a<b.a; } ld dis(ld x,ld y){ return std::sqrt(x*x+y*y); } int fix(int a,int l,int r){ if(a<l)return a+r-l; if(a>=r)return a-r+l; return a; } bool feq(ld a,ld b){ return fabs(a-b)<_0; } bool chk(ld x){ while(x>pi*2-_0)x-=pi*2; while(x<_0-pi*2)x+=pi*2; return std::fabs(x)<_0; } int main(){ scanf("%d",&n); for(int i=0;i<n;i++)scanf("%d%d",&ps[i].x,&ps[i].y); for(int i=0;i<n;i++){ xs+=ps[i].x; ys+=ps[i].y; } for(int i=0;i<n;i++){ (ps[i].x*=n)-=xs; (ps[i].y*=n)-=ys; ps[i].d=1ll*ps[i].x*ps[i].x+1ll*ps[i].y*ps[i].y; ps[i].a=atan2(ps[i].y,ps[i].x); } std::sort(ps,ps+n); int p0=0,p1; while(ps[p0].x==0&&ps[p0].y==0)++p0; for(p1=p0;p0<n;p0=p1){ while(p1<n&&ps[p1].d==ps[p0].d)++p1; lp=0; for(int p2=p0;p2<p1;p2++){ bool ab=1; ld m=ps[p2].a*2; for(int l=fix(p2-1,p0,p1),r=fix(p2+1,p0,p1);;l=fix(l-1,p0,p1),r=fix(r+1,p0,p1)){ if(!chk(ps[l].a+ps[r].a-m)){ ab=0; break; } if(l==r)break; } if(ab==1)ls[lp++]=ps[p2].a; if(p1-p0&1)continue; ab=1; m=(ps[p2].a+ps[fix(p2+1,p0,p1)].a); for(int l=fix(p2,p0,p1),r=fix(p2+1,p0,p1),t=p1-p0>>1;t;--t,l=fix(l-1,p0,p1),r=fix(r+1,p0,p1)){ if(!chk(ps[l].a+ps[r].a-m)){ ab=0; break; } if(l==r)break; } if(ab)ls[lp++]=m/2.; } for(int i=0;i<lp;i++){ ld x=ls[i]; while(x<-_0)x+=pi; while(x>pi-_0)x-=pi; ls[i]=x; } std::sort(ls,ls+lp); lp=std::unique(ls,ls+lp,feq)-ls; if(ad){ int lp1=0,ap1=0; for(int i=0;i<ap;i++){ while(lp1<lp&&ls[lp1]<as[i]-_0)++lp1; if(lp1==lp)break; if(feq(as[i],ls[lp1]))as[ap1++]=as[i]; } ap=ap1; }else{ ad=1; for(int i=0;i<lp;i++)as[ap++]=ls[i]; } } printf("%d",ap); return 0; }