ACM--两圆交集的面积--HDOJ 1798--Tell me the area

HDOJ题目地址：传送门

Tell me the area

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2451    Accepted Submission(s): 747

Problem Description

    There are two circles in the plane (shown in the below picture), there is a common area between the two circles. The problem is easy that you just tell me the common area.

 

Input

There are many cases. In each case, there are two lines. Each line has three numbers: the coordinates (X and Y) of the centre of a circle, and the radius of the circle.

 

Output

For each case, you just print the common area which is rounded to three digits after the decimal point. For more details, just look at the sample.

 

Sample Input

   
   
   
   

    
    
    
    0 0 2
2 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0.108
   
   
   
   
   
   
   
   


   
   
   
   

解题思路

高中知识，先判断两个圈的状态（相交，相切，相离）。如果相交，用余弦定理：cosA=(b*b+c*c-a*a)/2*b*b分别求出两个弧的角度，用两个圆弧面积之和剪掉两个三角形面积

    #include<stdio.h>
    #include<math.h>
    #define PI acos(-1.0)
    int main()
    {
        double a1,b1,r1,a2,b2,r2,d;
        double A1,A2,s1,s2,s;
        while(scanf("%lf%lf%lf%lf%lf%lf",&a1,&b1,&r1,&a2,&b2,&r2)!=EOF)
        {
            d=sqrt((a2-a1)*(a2-a1)+(b2-b1)*(b2-b1));
            if(d>=r1+r2)    printf("0.000\n");
            else if(d<=fabs(r1-r2)&&d>=0)
            {
                if(r1>r2)
                        printf("%0.3lf\n",PI*r2*r2);
                else printf("%0.3lf\n",PI*r1*r1);
            }
            else {
                A1=2*acos((d*d+r1*r1-r2*r2)/(2*d*r1));//余弦定理
                A2=2*acos((d*d+r2*r2-r1*r1)/(2*d*r2));
                s1=0.5*r1*r1*sin(A1)+0.5*r2*r2*sin(A2);//三角形面积 s=a*b*sinC/2;
                s2=A1/2*r1*r1+A2/2*r2*r2;//圆弧面积 s=A*l*l/2;
                s=s2-s1;
                printf("%.03lf\n",s);
            }
        }
    }

参考博客：传送门