题目链接：http://poj.org/problem?id=2546
Description
Your task is to write a program, which, given two circles, calculates the area of their intersection with the accuracy of three digits after decimal point.
Input
In the single line of input file there are space-separated real numbers x1 y1 r1 x2 y2 r2. They represent center coordinates and radii of two circles.
Output
The output file must contain single real number - the area.
Sample Input
20.0 30.0 15.0 40.0 30.0 30.0
Sample Output
608.366
大意是给出两圆心，及圆的半径，求两个圆相交部分的面积，求解关键是对分好case进行处理
Accepted：
1 #include <iostream> 2 #include <math.h> 3 #include <iomanip> 4 #define Pi 3.1415926 5 6 using namespace std; 7 typedef struct Point 8 { 9 float x, y; 10 }Point; 11 typedef struct Circle 12 { 13 Point p; 14 float r; 15 }Circle; 16 double Dis(Point p1, Point p2); 17 double CArea( Circle c1, Circle c2); 18 int main() 19 { 20 float x1, y1, r1, x2, y2, r2; 21 cin >> x1 >> y1 >> r1 >> x2 >> y2 >> r2; 22 double s = 0; 23 Circle c1, c2; 24 Point p1, p2; 25 if(r1 <= r2) 26 { 27 p1.x = x1, p1.y = y1, c1.p = p1, c1.r = r1; 28 p2.x = x2, p2.y = y2, c2.p = p2, c2.r = r2; 29 } 30 else 31 { 32 p1.x = x2, p1.y = y2, c1.p = p1, c1.r = r2; 33 p2.x = x1, p2.y = y1, c2.p = p2, c2.r = r1; 34 } 35 36 s = CArea(c1, c2); 37 cout << setiosflags(ios::fixed)<<setprecision(3)<< s << endl; 38 return 0; 39 } 40 double CArea( Circle c1, Circle c2) 41 { 42 double s = 0; 43 double x = 0, y = 0, h = 0; 44 double alpha = 0, beta = 0; 45 double d = Dis(c1.p, c2.p); 46 if(d >= (c1.r + c2.r)) 47 { 48 s = 0; 49 //cout << "case 1" << endl; 50 } 51 else if(c2.r >= (c1.r + d)) 52 { 53 s = Pi*c1.r*c1.r; 54 //cout << "case 2" << endl; 55 } 56 else if(d >= sqrt(pow(c2.r, 2) - pow(c1.r, 2)) && d < c2.r + c1.r) 57 { 58 x = (pow(c2.r, 2) - pow(c1.r, 2) + pow(d, 2)) / (2*d); 59 h = sqrt(pow(c2.r, 2) - pow(x, 2)); 60 alpha = asin(h/c1.r); 61 beta = asin(h/c2.r); 62 s = alpha*pow(c1.r, 2) + beta*pow(c2.r, 2) - d*h; 63 //cout << "case 3" << endl; 64 } 65 else if(d <= sqrt(pow(c2.r, 2) - pow(c1.r, 2)) && d >= c2.r - c1.r) 66 { 67 68 y = (pow(c2.r, 2) - pow(c1.r, 2) - pow(d, 2)) /(2*d); 69 h = sqrt(pow(c1.r, 2) - pow(y, 2)); 70 alpha = asin(h/c1.r); 71 beta = asin(h/c2.r); 72 s = (Pi - alpha)*pow(c1.r, 2) + h*y + beta*pow(c2.r, 2) - h*(d+y); 73 //cout << "case 4" << endl; 74 } 75 return s; 76 } 77 double Dis(Point p1, Point p2) 78 { 79 return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y)); 80 }