两圆相交求面积 hdu5120
两圆相交分如下集中情况:相离、相切、相交、包含。
设两圆圆心分别是O1和O2,半径分别是r1和r2,设d为两圆心距离。又因为两圆有大有小,我们设较小的圆是O1。
相离相切的面积为零,代码如下:
- double d = sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
- if (d >= r1+r2)
- return 0;
double d = sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
if (d >= r1+r2)
return 0;
包含的面积就是小圆的面积了,代码如下:
- if(r2 - r1 >= d)
- return pi*r1*r1;
if(r2 - r1 >= d)
return pi*r1*r1;
接下来看看相交的情况。
相交面积可以这样算:扇形O1AB - △O1AB + 扇形O2AB - △O2AB,这两个三角形组成了一个四边形,可以用两倍的△O1AO2求得,
所以答案就是两个扇形-两倍的△O1AO2
因为
所以
那么
同理
接下来是四边形面积:
代码如下:
double ang1=acos((r1*r1+d*d-r2*r2)/(*r1*d));
double ang2=acos((r2*r2+d*d-r1*r1)/(*r2*d));
return ang1*r1*r1 + ang2*r2*r2 - r1*d*sin(ang1);
#include<iostream>
#include<cmath>
using namespace std; #define pi acos(-1.0) typedef struct node
{
int x;
int y;
}point; double AREA(point a, double r1, point b, double r2)
{
double d = sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
if (d >= r1+r2)
return ;
if (r1>r2)
{
double tmp = r1;
r1 = r2;
r2 = tmp;
}
if(r2 - r1 >= d)
return pi*r1*r1;
double ang1=acos((r1*r1+d*d-r2*r2)/(*r1*d));
double ang2=acos((r2*r2+d*d-r1*r1)/(*r2*d));
return ang1*r1*r1 + ang2*r2*r2 - r1*d*sin(ang1);
} int main()
{
point a, b;
a.x=, a.y=;
b.x=, b.y=;
double result = AREA(a, , b, );
printf("%lf\n", result);
return ;
}
Intersection
Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 3443 Accepted Submission(s): 1302
A ring is a 2-D figure bounded by two circles sharing the common center. The radius for these circles are denoted by r and R (r < R). For more details, refer to the gray part in the illustration below.
Matt just designed a new logo consisting of two rings with the same size in the 2-D plane. For his interests, Matt would like to know the area of the intersection of these two rings.
Each of the following two lines contains two integers xi, yi (0 ≤ xi, yi ≤ 20) indicating the coordinates of the center of each ring.
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
double x1,ya,x2,y2,dis,s1,s2,s3,R,r;
double sov(double R,double r){
if(dis>=r+R) return ;
if(dis<=R-r) return acos(-1.0)*r*r;
double x=(R*R-r*r+dis*dis)/2.0/dis;
double y=(r*r-R*R+dis*dis)/2.0/dis;
double seta1=*acos(x/R);
double seta2=*acos(y/r);
double ans=seta1*R*R/2.0+seta2*r*r/2.0;
double h=sqrt(R*R-x*x);
return ans-dis*h;
}
int main(){
int tas=,T;
for(scanf("%d",&T);T--;){
scanf("%lf%lf",&r,&R);
scanf("%lf%lf%lf%lf",&x1,&ya,&x2,&y2);
dis=sqrt((x1-x2)*(x1-x2)+(ya-y2)*(ya-y2));
s1=sov(R,R),s2=sov(R,r),s3=sov(r,r);
printf("Case #%d: %.6f\n",tas++,s1-*s2+s3);
}
}
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