Intersection
Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 2699 Accepted Submission(s): 1011
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 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
Source 2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）
解析：求两圆相交面积。画个图简单分析可知，圆环相交面积 = 两大圆相交面积-2*（大圆小圆相交面积）+小圆相交面积。注意两圆外离、外切和内含、内切的情况。
``` #include
const double PI = acos(-1.0);
struct Point
{
double x, y, r;
} A, B, a, b;
double dis(const Point &a, const Point &b)
{
return sqrt((a.x-b.x)(a.x-b.x)+(a.y-b.y)(a.y-b.y));
}
double interarea(Point a,Point b)
{
double d = dis(a, b);
if(d >= a.r+b.r) //外离或外切
return 0;
double r = min(a.r, b.r);
if(d <= fabs(a.r-b.r)) //内含或内切
return PIrr;
double angle1 = acos((a.ra.r+dd-b.rb.r)/2.0/a.r/d);
double angle2 = acos((b.rb.r+dd-a.ra.r)/2.0/b.r/d);
double ret = 0;
ret -= da.rsin(angle1);
ret += angle1a.ra.r+angle2b.rb.r;
return ret;
}
int main()
{
int t, cn = 0;
scanf("%d", &t);
while(t--){
double r,R,x1,y1,x2,y2;
scanf("%lf%lf%lf%lf%lf%lf",&r,&R,&x1,&y1,&x2,&y2);
A.r = R, a.r =r, A.x = a.x = x1, A.y = a.y = y1;
B.r = R, b.r =r, B.x = b.x = x2, B.y = b.y = y2;
double S = interarea(A,B);
double s1 = interarea(a,B);
// double s2=interarea(A,b);
double s3 = interarea(a,b);
// double res=S-(s1+s2-s3);
double res = S-2*s1+s3;
printf("Case #%d: %.6f
", ++cn, res);
}
return 0;
}