hdu 5120 Intersection (圆环面积相交->圆面积相交)

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

2 3

0 0

2 3

0 0

5 0

Sample Output

Case #1: 15.707963

Case #2: 2.250778

两个圆环相交,然而我只有圆相交的板子

首先拿其中一个大圆A与另一个大圆B和小圆b算交面积,两者相减,求的是A与圆环Bb相交的面积 area1

然后拿小圆a另一个大圆B和小圆b算交面积,两者相减,求的是a与圆环Bb相交的面积,area2

我们输出area1-area2就行了

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 const double PI = acos(-1.0);
 5 const double eps = 1e-8;
 6 int dblcmp (double k)
 7 {
 8     if (fabs(k)<eps) return 0;
 9     return k>0?1:-1;
10 }
11 struct Point
12 {
13     double x,y;
14 };
15 double dis (Point a,Point b)
16 {
17     return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
18 }
19 double area_of_overlap (Point c1,double r1,Point c2,double r2)//圆相交模板
20 {
21     double d = dis(c1,c2);
22     if (r1+r2<d+eps) return 0;
23     if (d<fabs(r1-r2)+eps){
24         double r = min(r1,r2);
25         return PI*r*r;
26     }
27     double x = (d*d+r1*r1-r2*r2)/(2*d);
28     double t1 = acos(x/r1);
29     double t2 = acos((d-x)/r2);
30     return r1*r1*t1+r2*r2*t2-d*r1*sin(t1);
31 }
32 int t;
33 int casee = 0;
34 int main()
35 {
36     //freopen("de.txt","r",stdin);
37     scanf("%d",&t);
38     while (t--){
39         Point p1,p2;
40         double r1,r2;
41         scanf("%lf%lf",&r1,&r2);
42         scanf("%lf%lf",&p1.x,&p1.y);
43         scanf("%lf%lf",&p2.x,&p2.y);
44         if (dblcmp(r1-r2)>0) swap(r1,r2);
45         double ans = area_of_overlap (p1,r2,p2,r2) -area_of_overlap (p1,r2,p2,r1)
46                     -(area_of_overlap (p1,r1,p2,r2) - area_of_overlap(p1,r1,p2,r1) );
47         /*double ans = area (p1,r2,p2,r2) -area (p1,r2,p2,r1)
48         -(area(p1,r1,p2,r2) - area(p1,r1,p2,r1) );*/
49         printf("Case #%d: %.6f
",++casee,ans);
50     }
51     return 0;
52 }