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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[N]; 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; }
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原文地址:http://www.cnblogs.com/clrs97/p/4610049.html