BZOJ 1100 POI2007 对称轴osi 计算几何+KMP算法

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-6
using 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;
	}
}