题意:就是给你一个圆,和你一个矩形,求面积并,且 保证是一种情况:三角剖分后 一个点在圆内 两个在圆外
题解:可以直接上圆与凸多边形交的板子,也可以由这题实际情况,面积等于扇形减两个三角形
#include<bits/stdc++.h>
using namespace std;
int main()
{
int T;
double dx,dy,yx,ux,uy,rx,ry,R;
double s1,s2,s3,x1,y1,x2,y2,cosA;
scanf("%d",&T);
for (int i=;i<=T;i++)
{
scanf("%lf%lf%lf",&rx,&ry,&R);
scanf("%lf%lf%lf%lf",&dx,&dy,&ux,&uy);
dx-=rx;ux-=rx;dy-=ry;uy-=ry;
x1=dx;
y1=-sqrt(R*R-x1*x1);
y2=uy;
x2=sqrt(R*R-y2*y2);
cosA=acos((x1*x2+y1*y2)/sqrt((x1*x1+y1*y1)*(x2*x2+y2*y2)));
s1=R*R*cosA/;
s2=x1*(y2-y1)/;
s3=y2*(x1-x2)/;
printf("Case %d: %.5lf\n",i,s1-s2-s3);
}
}
#include<bits/stdc++.h>
#define inf 1000000000000
#define M 100009
#define eps 1e-12
#define PI acos(-1.0)
using namespace std;
struct Point
{
double x,y;
Point(){}
Point(double xx,double yy){x=xx;y=yy;}
Point operator -(Point s){return Point(x-s.x,y-s.y);}
Point operator +(Point s){return Point(x+s.x,y+s.y);}
double operator *(Point s){return x*s.x+y*s.y;}
double operator ^(Point s){return x*s.y-y*s.x;}
}p[M];
double max(double a,double b){return a>b?a:b;}
double min(double a,double b){return a<b?a:b;}
double len(Point a){return sqrt(a*a);}
double dis(Point a,Point b){return len(b-a);}//两点之间的距离
double cross(Point a,Point b,Point c)//叉乘
{
return (b-a)^(c-a);
}
double dot(Point a,Point b,Point c)//点乘
{
return (b-a)*(c-a);
}
int judge(Point a,Point b,Point c)//判断c是否在ab线段上（前提是c在直线ab上）
{
if (c.x>=min(a.x,b.x)
&&c.x<=max(a.x,b.x)
&&c.y>=min(a.y,b.y)
&&c.y<=max(a.y,b.y)) return ;
return ;
}
double area(Point b,Point c,double r)
{
Point a(0.0,0.0);
if(dis(b,c)<eps) return 0.0;
double h=fabs(cross(a,b,c))/dis(b,c);
if(dis(a,b)>r-eps&&dis(a,c)>r-eps)//两个端点都在圆的外面则分为两种情况
{
double angle=acos(dot(a,b,c)/dis(a,b)/dis(a,c));
if(h>r-eps) return 0.5*r*r*angle;else
if(dot(b,a,c)>&&dot(c,a,b)>)
{
double angle1=*acos(h/r);
return 0.5*r*r*fabs(angle-angle1)+0.5*r*r*sin(angle1);
}else return 0.5*r*r*angle;
}else
if(dis(a,b)<r+eps&&dis(a,c)<r+eps) return 0.5*fabs(cross(a,b,c));//两个端点都在圆内的情况
else//一个端点在圆上一个端点在圆内的情况
{
if(dis(a,b)>dis(a,c)) swap(b,c);//默认b在圆内
if(fabs(dis(a,b))<eps) return 0.0;//ab距离为0直接返回0
if(dot(b,a,c)<eps)
{
double angle1=acos(h/dis(a,b));
double angle2=acos(h/r)-angle1;
double angle3=acos(h/dis(a,c))-acos(h/r);
return 0.5*dis(a,b)*r*sin(angle2)+0.5*r*r*angle3;
}else
{
double angle1=acos(h/dis(a,b));
double angle2=acos(h/r);
double angle3=acos(h/dis(a,c))-angle2;
return 0.5*r*dis(a,b)*sin(angle1+angle2)+0.5*r*r*angle3;
}
}
}
int main()
{
int T,n=;
double rx,ry,R;
scanf("%d",&T);
for (int ii=;ii<=T;ii++)
{
scanf("%lf%lf%lf",&rx,&ry,&R);
scanf("%lf%lf%lf%lf",&p[].x,&p[].y,&p[].x,&p[].y);
p[].x=p[].x;p[].y=p[].y;
p[].x=p[].x;p[].y=p[].y;
p[]=p[];
Point O(rx,ry);
for (int i=;i<=n+;i++) p[i]=p[i]-O;
O=Point(,);
double sum=;
for (int i=;i<=n;i++)
{
int j=i+;
double s=area(p[i],p[j],R);
if (cross(O,p[i],p[j])>) sum+=s;else sum-=s;
}
printf("Case %d: %.5lf\n",ii,fabs(sum));
}
return ;
}