计算几何（容斥原理，圆交）：HDU 5120 Intersection

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 xi, yi (0 ≤ xi, yi ≤ 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

　　容斥原理很明显，圆交的求法是我自己想的，很简单。

1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <cmath>
5 using namespace std;
6 int T,cas,r,R,px,py,qx,qy;
7 const double Pi=acos(-1.0);
8 double sqr(double x){return x*x;}
9 double Q(double r1,double r2){
10     double d=sqrt(sqr(px-qx)+sqr(py-qy));
11     if(r1+r2<=d)return 0;
12     if(r1-r2>=d)return Pi*sqr(r2);
13     double a1=acos((sqr(r1)+sqr(d)-sqr(r2))/(2*r1*d));
14     double a2=acos((sqr(r2)+sqr(d)-sqr(r1))/(2*r2*d));
15     double s1=sqr(r1)*a1;
16     double s2=sqr(r2)*a2;
17     return s1+s2-r1*sin(a1)*d;
18 }
19 int main(){
20     scanf("%d",&T);
21     while(T--){
22         scanf("%d%d",&r,&R);
23         scanf("%d%d%d%d",&px,&py,&qx,&qy);
24         printf("Case #%d: %.6lf\n",++cas,Q(R,R)-2*Q(R,r)+Q(r,r));
25     }
26     return 0;
27 }