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;
}
}
BZOJ 1100 POI2007 对称轴osi 计算几何+KMP算法
原文地址：http://blog.csdn.net/popoqqq/article/details/41847901
