BZOJ1100 : [POI2007]对称轴osi

将多边形转化为如下的环：

1到2的边,角2,2到3的边,角3,...,n-1到n的边,角n,n到1的边,角1

然后枚举对称轴，如果i是对称轴，那么[i-n,i+n]是一个回文串

用Manacher算法实现即可。

时间复杂度$O(n)$。

#include<cstdio>
#define N 100010
typedef long long ll;
int T,n,i,r,p,f[N<<2],ans;
struct P{
int x,y;
P(){x=y=0;}
P(int _x,int _y){x=_x,y=_y;}
inline P operator-(P b){return P(x-b.x,y-b.y);}
}a
;
inline ll sqr(ll x){return x*x;}
inline ll dis(P a,P b){return sqr(a.x-b.x)+sqr(a.y-b.y);}
inline ll cross(P a,P b){return (ll)a.x*b.y-(ll)a.y*b.x;}
struct Q{
ll x;int y;
Q(){x=y=0;}
Q(ll _x,int _y){x=_x,y=_y;}
inline bool operator==(Q b){return x==b.x&&y==b.y;}
}b[N<<2];
inline int min(int a,int b){return a<b?a:b;}
inline void read(int&a){
char c;bool f=0;a=0;
while(!((((c=getchar())>='0')&&(c<='9'))||(c=='-')));
if(c!='-')a=c-'0';else f=1;
while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';
if(f)a=-a;
}
int main(){
for(read(T);T--;printf("%d\n",ans>>1)){
read(n);
for(i=1;i<=n;i++)read(a[i].x),read(a[i].y);
a[n+1]=a[1],a[n+2]=a[2];
for(i=1;i<=n;i++){
b[i*2-1]=Q(dis(a[i],a[i+1]),1);
b[i*2]=Q(cross(a[i+1]-a[i],a[i+1]-a[i+2]),2);
}
for(i=1;i<=n*2;i++)b[i+n*2]=b[i];
b[n*4+1]=Q(0,3);
for(r=p=0,i=1;i<=n*4;i++){
for(f[i]=r>i?min(r-i,f[p*2-i]):1;b[i-f[i]]==b[i+f[i]];f[i]++);
if(i+f[i]>r)r=i+f[i],p=i;
}
for(ans=0,i=n+1;i<=n*3;i++)if(f[i]>n)ans++;
}
return 0;
}