先求重心,然后枚举,对于奇数个顶点的多边形,枚举一个点和一条边的中点所成的直线,对于偶数个顶点的多边形,对称轴可能过两条边的中点或者两个顶点(O(n)枚举,由于要满足过重心这个条件,可以使该层复杂度优化为常数),然后判断即可(O(n))!!这题解题报告是用后缀数组来做~_~ Holy Shit
代码(各种模板堆积):
#include <cstdio> #include <cstring> #include <cmath> using namespace std; #define EPS 1e-8 #define min(a,b) (a) + EPS < (b) ? (a) : (b) #define max(a,b) (a) > (b) + EPS ? (a) : (b) const int MAXN = 20010; struct Point{ double x,y; Point(){} Point(double _x,double _y):x(_x),y(_y){} void input(){ scanf("%lf%lf",&x,&y); } }; struct Line{ double a,b,c; }; Line l; Point ps[MAXN]; inline double difcross(Point a,Point b,Point c){ return (b.x-a.x)*(c.y-a.y) - (c.x-a.x)*(b.y-a.y); } inline double getarea(Point p[],int np){ double area = 0; p[np] = p[0]; for(int i = 0; i < np; ++i){ area += p[i].x*p[i+1].y - p[i].y*p[i+1].x; } return area/2.0; } inline Point getgc(Point p[],int np){ double area = getarea(p,np); Point tp; p[np] = p[0]; tp.x = tp.y =0; for(int i = 0; i < np; ++i) tp.x += (p[i].x+p[i+1].x)*(p[i].x*p[i+1].y-p[i].y*p[i+1].x); for(int i = 0; i < np; ++i) tp.y += (p[i].y+p[i+1].y)*(p[i].x*p[i+1].y-p[i].y*p[i+1].x); tp.x /= (6.0*area); tp.y /= (6.0*area); return tp; } inline void getline(Point x,Point y,Line &l){ l.a = y.y-x.y; l.b = x.x - y.x; l.c = y.x*x.y-x.x*y.y; } inline bool p2line(Point a,Line l){ if(fabs(l.a*a.x+l.b*a.y+l.c) < EPS)return true; return false; } inline Point Mid(Point a,Point b){ return Point((a.x+b.x)/2.0,(a.y+b.y)/2.0); } inline int abs(int a){ return a > 0 ? a : -a; } inline int got(int i,int j,int n){ if(j > 0)return (i+j)%n; else return got(i,n+j,n); } inline bool chuizhi(Line l,Line tl){ if(fabs(l.a*tl.a+l.b*tl.b) < EPS)return true; return false; } int main(){ int n; while (scanf("%d",&n) != EOF){ for(int i = 0; i < n; ++i)ps[i].input(); Point gc = getgc(ps,n); Line l,tl; bool duichen = 0; if(n&1){ for(int i = 0; i < n; ++i){ getline(ps[i],Mid(ps[got(i,n/2,n)],ps[got(i,-n/2,n)]),l); if(!p2line(gc,l))continue; bool ok = 1; for(int j = 1; j <= n/2; ++j){ getline(ps[got(i,j,n)],ps[got(i,-j,n)],tl); if(!p2line(Mid(ps[got(i,j,n)],ps[got(i,-j,n)]),l) || !chuizhi(l,tl)){ ok = 0; break; } } if(ok){ duichen = 1; break; } } } else{ for(int i = 0; i < n; ++i){ getline(ps[i],ps[got(i,n/2,n)],l); if(!p2line(gc,l))continue; bool ok = 1; for(int j = 1; j <= n/2-1; ++j){ getline(ps[got(i,j,n)],ps[got(i,-j,n)],tl); if(!p2line(Mid(ps[got(i,j,n)],ps[got(i,-j,n)]),l) || !chuizhi(l,tl)){ ok = 0; break; } } if(ok){ duichen = 1; goto abc; } } for(int i = 0; i < n; ++i){ int s = i,t = got(i,1,n); getline(Mid(ps[s],ps[t]),Mid(ps[got(s,-(n/2-1),n)],ps[got(t,n/2-1,n)]),l); if(!p2line(gc,l))continue; bool ok = 1; for(int j =1; j <= n/2-1; ++j){ getline(ps[got(s,-j,n)],ps[got(t,j,n)],tl); if(!p2line(Mid(ps[got(s,-j,n)],ps[got(t,j,n)]),l) || !chuizhi(l,tl)){ ok = 0; break; } } if(ok){ duichen = 1; break; } } } abc: if(duichen)puts("YES"); else puts("NO"); } return 0; }