BZOJ 1100 POI2007 对称轴osi 计算几何+KMP算法

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题目大意：给定一个多边形，求对称轴数量

我X 这究竟是怎么想到KMP的……

首先将边字符化即找到这个多边形的中心然后用与中心构成的三角形的边-角-边的方式表示这条边

将边顺时针扫一遍然后倍增至长度为2n-1 再逆时针扫一遍逆时针扫的那遍在顺时针那遍中出现的次数就是对称轴数目

用KMP算法就能搞出来证明自己YY吧

出题人卡精度丧心病狂。。。

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define M 100100#define EPS 1e-6using namespace std;struct point{double x,y;point(){}point(double _,double __):x(_),y(__){}void Read(){scanf("%lf%lf",&x,&y);}void operator += (const point &Y){x+=Y.x;y+=Y.y;}point operator - (const point &Y) const{return point(x-Y.x,y-Y.y);}double operator * (const point &Y) const{return x*Y.y-Y.x*y;}point operator / (double a) const{return point(x/a,y/a);}}points[M],centre;struct line{double d1,d2,cross;line(){}line(double _,double __,double ___):d1(_),d2(__),cross(___/_/__){}bool operator == (const line &Y) const;}a[M],b[M<<1];int n;bool line :: operator == (const line &Y) const{if(fabs(d1-Y.d1)>EPS)return false;if(fabs(d2-Y.d2)>EPS)return false;if(fabs(cross-Y.cross)>EPS)return false;return true;}double Distance(const point &p1,const point &p2){return sqrt( (p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y) );}int KMP(int len){static int next[M];int i,fix=0,re=0;for(i=2;i<=n;i++){while( fix && !(a[fix+1]==a[i]) )fix=next[fix];if( a[fix+1]==a[i] ) ++fix;next[i]=fix;}fix=0;for(i=1;i<=len;i++){while( fix && !(a[fix+1]==b[i]) )fix=next[fix];if( a[fix+1]==b[i] ) ++fix;if(fix==n)++re,fix=next[fix];}return re;}int main(){//freopen("osi.in","r",stdin);//freopen("osi.out","w",stdout);int T;for(cin>>T;T;T--){int i;cin>>n;centre=point(0.0,0.0);for(i=1;i<=n;i++)points[i].Read(),centre+=points[i]/static_cast<double>(n);for(i=1;i<=n;i++)b[i]=b[i+n]=line(Distance(points[i],centre),Distance(points[i%n+1],centre), (points[i]-centre)*(points[i%n+1]-centre) );for(i=n;i;i--)a[n-i+1]=line(b[i].d2,b[i].d1,0),a[n-i+1].cross=b[i].cross;cout<<KMP(n+n-1)<<endl;}}

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