POJ3675 Telescope 圆和多边形的交

Posted 2020-09-11 逐雪

POJ3675

用三角剖分可以轻松搞定，数据也小 随便AC。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const double eps=1e-10,PI=acos(-1.0);

int cmp(double k)
{
 return k<-eps ? -1:k>eps ? 1:0;
}

const double pi=acos(-1.0);

inline double sqr(double x)
{
 return x*x;
}






struct point
{
 double x,y;
 point (){}
 point (double a,double b):x(a),y(b){}
 bool input()
 	{
 	 return scanf("%lf%lf",&x,&y)!=EOF;
	}
 friend point operator +(const point &a,const point &b)
 	{
 	 return point(a.x+b.x,a.y+b.y);
	}	
 friend point operator -(const point &a,const point &b)
 	{
 	 return point(a.x-b.x,a.y-b.y);
	}
 friend bool operator ==(const point &a,const point &b)
 	{
 	 return cmp(a.x-b.x)==0&&cmp(a.y-b.y)==0;
	}
 friend point operator *(const point &a,const double &b)
 	{
 	 return point(a.x*b,a.y*b);
	}
 friend point operator*(const double &a,const point &b)
 	{
 	 return point(a*b.x,a*b.y);
	}
 friend point operator /(const point &a,const double &b)
 	{
 	 return point(a.x/b,a.y/b);
	}
 double norm()
 	{
 	 return sqr(x)+sqr(y);
	}
};

double mysqrt(double x)
{
 return sqrt(x);
}
double dot(const point &a,const point &b)
{
 return a.x*b.x+a.y*b.y;
}
double cross(const point &a,const point &b)
{
 return a.x*b.y-a.y*b.x;
}
double abs(const point &o)
{
 return sqrt(dot(o,o));
}
void circle_cross_line(point a,point b,point o,double r,point ret[],int &num)
{
 double x0=o.x,y0=o.y;
 double x1=a.x,y1=a.y;
 double x2=b.x,y2=b.y;
 double dx=x2-x1,dy=y2-y1;
 double A=dx*dx+dy*dy;
 double B=2*dx*(x1-x0)+2*dy*(y1-y0);
 double C=sqr(x1-x0)+sqr(y1-y0)-sqr(r);
 double delta=B*B-4*A*C;
 num=0;
 if(cmp(delta)>=0)
 	{
 	 double t1=(-B-mysqrt(delta))/(2*A);
 	 double t2=(-B+mysqrt(delta))/(2*A);
 	 if(cmp(t1-1)<=0&&cmp(t1)>=0)
 	 	{
 	 	 ret[num++]=point(x1+t1*dx,y1+t1*dy);
		}
	 if(cmp(t2-1)<=0&&cmp(t2)>=0)
 	 	{
 	 	 ret[num++]=point(x1+t2*dx,y1+t2*dy);
		}
	}
}


point crosspt(const point &a,const point &b,const point &p,const point &q)
{
 double a1=cross(b-a,p-a);
 double a2=cross(b-a,q-a);
 return (p*a2-q*a1)/(a2-a1);
}

 
double sector_area(const point &a,const point &b,const double r)
{
 double theta=atan2(a.y,a.x)-atan2(b.y,b.x);
 while(cmp(theta)<=0)theta+=2*PI;
 while(cmp(theta-2*PI)>0)theta-=2*PI;
 theta=min(theta,2*PI-theta);
 return r*r*theta/2.;
}

double calc_c(const point &a,const point &b,const double r)
{
 point p[2];
 int num=0;
 bool ina=cmp(abs(a)-r)<0;
 bool inb=cmp(abs(b)-r)<0;
 if(ina)
 	{
 	 if(inb)return fabs(cross(a,b))/2.;
 	 	else
 	 		{
 	 		 circle_cross_line(a,b,point(0,0),r,p,num);
 	 		 return fabs(sector_area(b,p[0],r))+fabs(cross(a,p[0]))/2.;
			}
	}
		else
			{
			 if(inb)
			 	{
			 	 circle_cross_line(a,b,point(0,0),r,p,num);
 	 		 			return fabs(sector_area(p[0],a,r))+fabs(cross(p[0],b))/2.;	
				}
				else
					{
					 circle_cross_line(a,b,point(0,0),r,p,num);
					 if(num==2)
					 	{
					 	 return fabs(sector_area(a,p[0],r))+fabs(sector_area(p[1],b,r))+fabs(cross(p[0],p[1]))/2.;
					 	 
						}else
							{
							 return fabs(sector_area(a,b,r));
							}
					}
			}
}

double area(int n,point res[],double r)
{
 double ret=0.;
 for(int i=0;i<n;i++)
 	{
 	 int sgn=cmp(cross(res[i],res[(i+1)%n]));
	 if(sgn!=0)
	 	{
	 	 ret+=sgn*calc_c(res[i],res[(i+1)%n],r);
		} 	
	}
 return ret;
} 
point pp[100];
int main()
{freopen("t.txt","r",stdin);
 double r;
 int n;
 while(scanf("%lf",&r)!=EOF)
 {

 scanf("%d",&n);
 for(int i=0;i<n;i++)
 	pp[i].input();
 pp[n]=pp[0];
 printf("%.2f\n",fabs(area(n,pp,r))+eps);
}
 return 0;
}