HDOJ 5120-Intersection【几何数学】

Intersection

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

Total Submission(s): 1857    Accepted Submission(s): 708

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 ≤ 10 ⁵), 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
   
   
   
   

 

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

解题思路：

求出两个圆环重叠之后的阴影面积，也有可能是两个圆。但是两圆环的重叠面积可以简化成  

S=大大+小小-2*大小。

最后记住这个模板，可以计算整数坐标。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const double pi=acos(-1.0);
double f(int x1,int x2,int y1,int y2,int r1,int r2)
{
	int t;
	if(r1<r2)
	{
		t=r1;
		r1=r2;
		r2=t;
	}
	double s,l1,l2,s1,s2,s3;
	double rr;
	double d=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
	rr=(double)(r1+r2);
	//printf("%lf %lf\n",d,rr);
	if(d>=rr)
	{
		return 0;
	}
	else if(d<=(r1-r2))
	{
		s=pi*r2*r2;
	}
	else 
	{
		l1=acos((r1*r1+d*d-r2*r2)/(2*r1*d));
		l2=acos((r2*r2+d*d-r1*r1)/(2*r2*d));
		s1=l1*r1*r1;
		s2=l2*r2*r2;
		s3=r1*sin(l1)*d;
		//printf("%lf %lf %lf \n",s1,s2,s3);
		s=s1+s2-s3;
	}
	return s;
}
int main()
{
	double s1,s2,s3;
	int t;
	scanf("%d",&t);
	int xx=1;
	while(t--)
	{
		int r1,r2;
		scanf("%d%d",&r1,&r2);
		int x1,y1,x2,y2;
		scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
		s1=f(x1,x2,y1,y2,r2,r2);
		s2=f(x1,x2,y1,y2,r2,r1);
		s3=f(x1,x2,y1,y2,r1,r1);
		double s=s1+s3-2*s2;
		//printf("%lf,%lf,%lf\n",s1,s2,s3);
		printf("Case #%d: ",xx++);
		//sz
		printf("%.6lf\n",s);
	}
	return 0;
}