hdu 5120 (求两圆相交的面积的公式)

Posted 2020-08-06 十目

S = A大B大 - A大B小 - A小B大 + A小B小。（A表示A环，大表示大圆，B同）。然后直接套模板，，，，

 

 1 #include <stdio.h>
 2 #include <algorithm>
 3 #include <string.h>
 4 #include <cmath>
 5 using namespace std;
 6 
 7 const double eps = 1e-8;
 8 const double PI = acos(-1.0);
 9 
10 int sgn(double x)
11 {
12     if(fabs(x) < eps) return 0;
13     if(x < 0) return - 1;
14     else return 1;
15 }
16 struct Point
17 {
18     double x, y;
19     Point(){}
20     Point(double _x, double _y)
21     {
22         x = _x; y = _y;
23     }
24     Point operator -( const Point &b) const
25     {
26         return Point(x - b. x, y - b. y);
27     }
28 
29     double operator ^ (const Point &b) const
30     {
31         return x*b. y - y*b. x;
32     }
33 
34     double operator * (const Point &b) const
35     {
36         return x*b. x + y*b. y;
37     }
38 
39     void transXY(double B)
40     {
41         double tx = x,ty = y;
42         x = tx* cos(B) - ty*sin(B);
43         y = tx* sin(B) + ty*cos(B);
44     }
45 };
46 
47 double dist( Point a, Point b)
48 {
49     return sqrt((a-b)*(a- b));
50 }
51 
52 double Ac(Point c1, double r1, Point c2, double r2)
53 {
54     double d = dist(c1,c2);
55     if(r1 + r2 < d + eps) return 0;
56     if(d < fabs(r1 - r2) + eps)
57     {
58         double r = min(r1,r2);
59         return PI*r*r;
60     }
61     double x = (d*d + r1*r1 - r2*r2)/(2*d);
62     double t1 = acos(x / r1);
63     double t2 = acos((d - x)/r2);
64     return r1*r1*t1 + r2*r2*t2 - d*r1*sin(t1);
65 }
66 
67 int main() {
68     int T; Point c1, c2;
69     double ans, r, R, x1, y1, x2, y2;
70     scanf("%d", &T);
71     for(int cas = 1; cas <= T; ++cas) {
72         scanf("%lf%lf%lf%lf%lf%lf", &r, &R, &x1, &y1, &x2, &y2);
73         c1.x = x1; c1.y = y1;
74         c2.x = x2; c2.y = y2;
75         ans = Ac(c1, R, c2, R) - Ac(c1, R, c2, r) - Ac(c1, r, c2, R)
76                 + Ac(c1, r, c2, r);
77         printf("Case #%d: %.6lf\n", cas, ans);
78     }
79     return 0;
80 }