两圆相交求面积 hdu5120

转载

 

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)/(2*r1*d));
double ang2=acos((r2*r2+d*d-r1*r1)/(2*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 0;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)/(2*r1*d));double ang2=acos((r2*r2+d*d-r1*r1)/(2*r2*d));return ang1*r1*r1 + ang2*r2*r2 - r1*d*sin(ang1);
}int main()
{point a, b;a.x=2, a.y=2;b.x=7, b.y=2;double result = AREA(a, 3, b, 5);printf("%lf\n", result);return 0;
}

 

 

 

Intersection

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 3443    Accepted Submission(s): 1302

Problem Description

Matt is a big fan of logo design. Recently he falls in love with logo made up by rings. The following figures are some famous examples you may know.


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.

 

Input

The first line contains only one integer T (T ≤ 10⁵), which indicates the number of test cases. For each test case, the first line contains two integers r, R (0 ≤ r < R ≤ 10).

Each of the following two lines contains two integers x_i, y_i (0 ≤ x_i, y_i ≤ 20) indicating the coordinates of the center of each ring.

 

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y is the area of intersection rounded to 6 decimal places.

 

Sample Input

2

2 3

0 0

0 0

2 3

0 0

5 0

 

Sample Output

Case #1: 15.707963

Case #2: 2.250778

#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 0;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=2*acos(x/R);double seta2=2*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=1,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++,s1-2*s2+s3);}
}