HDU3902 判断简单多边形是否对称

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;}