hdu 5120 - Intersection(解题报告）

Intersection

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

Total Submission(s): 1036 Accepted Submission(s): 407

[/b]

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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 xi, yi (0 ≤ xi, yi
≤ 20) indicating the coordinates of the center of each ring.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2
2 3
0 0
0 0
2 3
0 0
5 0

[align=left]Sample Output[/align]

Case #1: 15.707963
Case #2: 2.250778

题意：
求两圆环相交的面积。
参考代码：

#include<stdio.h>
#include<math.h>
#define PI acos(-1.0)
double sum(double a1,double b1,double r1,double a2,double b2,double r2)
{
double A1,A2,s1,s2,s,d;
d=sqrt((a2-a1)*(a2-a1)+(b2-b1)*(b2-b1));
if(d>=r1+r2)
return 0.000;
else if(d<=fabs(r1-r2))
{
if(r1>r2)
return PI*r2*r2;
else
return PI*r1*r1;
}
else{

A1=2*acos((d*d+r1*r1-r2*r2)/(2*d*r1));
A2=2*acos((d*d+r2*r2-r1*r1)/(2*d*r2));
s1=0.5*r1*r1*sin(A1)+0.5*r2*r2*sin(A2);
s2=A1/2*r1*r1+A2/2*r2*r2;
s=s2-s1;
return s;
}
}
int main()
{
int t;
scanf("%d",&t);
for(int q=1;q<=t;q++)
{
double r1,r2,a1,a2,b1,b2;
scanf("%lf%lf%lf%lf%lf%lf",&r1,&r2,&a1,&b1,&a2,&b2);
printf("Case #%d: ",q);
printf("%.6lf\n",sum(a1,b1,r2,a2,b2,r2)-2*sum(a1,b1,r2,a2,b2,r1)+sum(a1,b1,r1,a2,b2,r1));
}
return 0;
}