bzoj2592: [Usaco2012 Feb]Symmetry

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;
}