【BZOJ】【POI2007】【对称轴osi】【题解】【计算几何】

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=1100

vfk是厉害http://vfleaking.blog.163.com/blog/static/17480763420123183831110/

我觉得这种做法随便卡……

Code:

#include<bits/stdc++.h>
using namespace std;
const double eps=1e-7;
const double pi=acos(-1);
#define prev(i) ((i)-1>=0?(i)-1:(n<<1)-1)
#define next(i) ((i)+1<(n<<1)?(i)+1:0)
const int maxn=1e5+5;
struct Point{
	double x,y;
	Point(double _x=0,double _y=0):
		x(_x),y(_y){}
}p[maxn<<1];
typedef Point Vector;
Point operator+(Point a,Point b){return Point(a.x+b.x,a.y+b.y);}
Point operator-(Point a,Point b){return Point(a.x-b.x,a.y-b.y);}
Point operator*(Point a,double p){return Point(a.x*p,a.y*p);}
int dcmp(double x){if(fabs(x)<eps)return 0;return x>0?1:-1;} 
bool operator==(Vector A,Vector B){return !dcmp(A.x-B.x)&&!dcmp(A.y-B.y);}  
bool operator!=(Vector A,Vector B){return !(A==B);}  
double sqr(double x){return x*x;}
double Length(Vector a){return sqrt(sqr(a.x)+sqr(a.y));}
double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}
double Angle(Vector a,Vector b){return acos(Dot(a,b)/Length(a)/Length(b));}
Vector Normal(Vector a){double Len=Length(a);return Vector(-a.y/Len,a.x/Len);}  
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}  
struct Line{  
    Point p;Vector v;  
    Line(Point _p=Point(0,0),Vector _v=Vector(0,0)):  
        p(_p),v(_v){}  
}L[2];
double Distance(Point P,Line L){  
    Point A=L.p,B=L.p+L.v;  
    Vector u=P-A,v=B-A;  
    return fabs(Cross(u,v))/Length(v);  
}   
int _,n;
bool can(int a,int b){
	int l=prev(a);
	int r=next(a);
	Line L=Line(p[a],p[b]-p[a]);
	Vector v=Normal(L.v);
	while(l!=r){
		double dl=Distance(p[l],L);
		if(p[l]+v*2*dl!=p[r]&&p[l]-v*2*dl!=p[r])return false;
		l=prev(l);
		r=next(r);
	}return true;
}
int main(){
	scanf("%d",&_);
	while(_--){
		scanf("%d",&n);int len=0;
		for(int i=0;i<n;i++)scanf("%lf%lf",&p[i<<1].x,&p[i<<1].y);
		for(int i=0;i<n;i++)p[i<<1|1]=(p[prev(i<<1|1)]+p[next(i<<1|1)])*0.5;
		for(int i=0;i<n;i++){
			if(can(i,(i+n)%(n<<1)))
				L[len++]=(Line){p[i],p[(i+n)%(n<<1)]-p[i]};
			if(len==2)break;			
		}
		double rad=Angle(L[0].v,L[1].v);
		if(len==0)puts("0");
		else if(len==1)puts("1");
		else printf("%d\n",int(pi/rad+.5));
	}
	return 0;
}