poj 2546 Circular Area（两个圆相交的面积）

Circular Area

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5716		Accepted: 2239

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

Source

Northeastern Europe 2000, Far-Eastern Subregion

题意：给第一个圆的圆心坐标和半径，再给第二个圆的圆心坐标和半径，求两个圆的相交面积。

几何题，刚开始接触，照着别人代码理解完成自己的代码。

两种情况一次分析就好，情况相同~

附上代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #define ll long long
 6 #define M 100005
 7 using namespace std;
 8 
 9 const double  pi = acos(-1.0);
10 
11 struct Round
12 {
13     double x,y;
14     double r;
15 } rr[2];
16 
17 double dis(Round a, Round b)
18 {
19     return sqrt((a.x - b.x)*(a.x - b.x)+(a.y - b.y)*(a.y - b.y)); ///求两点之间的距离
20 }
21 
22 double solve(Round a, Round b)
23 {
24     double d = dis(a, b);
25     if(d >= a.r + b.r)  ///相离的情况
26         return 0;
27     else if(d <= fabs(a.r - b.r)) ///内含的情况
28     {
29         double r = a.r < b.r?a.r : b.r;
30         return pi*r*r;
31     }
32     double ang1 = acos((a.r * a.r + d * d - b.r * b.r) / 2.0 / a.r / d); 
33     ///公式： cos(A)=(b^2+c^2-a^2)/2bc acos(cos(A))求出角A的弧度
34     double ang2 = acos((b.r * b.r + d * d - a.r * a.r) / 2.0 / b.r / d);
35     double ret = ang1 * a.r * a.r + ang2 * b.r * b.r - d * a.r * sin(ang1);
36     ///扇形面积s=弧度*r^2/2 三角形面积=a*b*sin(A)/2
37     return ret;
38 }
39 
40 int main()
41 {
42     while(~scanf("%lf%lf%lf%lf%lf%lf",&rr[0].x,&rr[0].y,&rr[0].r,&rr[1].x,&rr[1].y,&rr[1].r))
43     {
44         printf("%.3lf
",solve(rr[0], rr[1]));
45     }
46     return 0;
47 }