ACM2014亚洲区北京赛区1009 Intersection（数学几何）

最新推荐文章于 2018-01-22 16:16:39 发布

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本文介绍了一种计算二维平面上两个相同大小圆环交集面积的方法，并提供了一个具体的C语言实现示例。该算法适用于当两个圆环的中心距离、内径与外径已知的情况。

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Intersection

Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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

Sample Output

Case #1: 15.707963
Case #2: 2.250778

ac代码

#include<stdio.h>
#include<math.h>
double fun(double x1,double y1,double r1,double x2,double y2,double r2)
{
	double s1,s2,p,q,h1,h2,rs,s;
	double pi=2*asin(1.0);
	rs=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
		if(rs>=r1+r2||!r1||!r2)//情况要考虑完全注
			s=0;
		else
			if(rs<=fabs(r1-r2))
			{
				if(r1>=r2)
					s=pi*r2*r2;
				else
					s=pi*r1*r1;
			}
			else
			{
				p=acos((rs*rs+r1*r1-r2*r2)/(2*rs*r1));
				q=acos((rs*rs+r2*r2-r1*r1)/(2*rs*r2));
				h1=p*r1*r1;
				h2=q*r2*r2;
				s1=r1*r1*cos(p)*sin(p);
				s2=r2*r2*cos(q)*sin(q);
				s=h1+h2-(s1+s2);
			}
			return s;
}
int main()
{
	double x1,x2,y1,y2,r1,r2;
	
	int t,c=0;
	scanf("%d",&t);
	while(t--)
	{
		
		double rs,s,p,q,h1,h2,s2,ans,max,min,s1,s3,s4;
		scanf("%lf%lf%lf%lf%lf%lf",&r1,&r2,&x1,&y1,&x2,&y2);
		s1=fun(x1,y1,r1,x2,y2,r1);
		s2=fun(x1,y1,r2,x2,y2,r2);
		s3=fun(x1,y1,r1,x2,y2,r2);
		ans=s1-2*s3+s2;
		printf("Case #%d: ",++c);
		printf("%.6lf\n",ans);
			
	}
}