Intersection

Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)

Total Submission(s): 602 Accepted Submission(s): 247

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 xi, yi (0 ≤ xi, yi ≤ 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 2 3 0 0 0 0 2 3 0 0 5 0

Sample Output

Case #1: 15.707963 Case #2: 2.250778

题意

给你两个完全一样的圆环，圆环，圆环

因为很重要，所以要说三遍

然后求他们相交的面积

题解

那就直接套模板吧！！！

套板大作战，相交面积=大圆交-2*大小交+小交

代码

#define inf 0x7fffffff

#define exp 1e-10

#define PI 3.141592654

using namespace std;

typedef long long ll;

struct Point

{

    double x,y;

    Point (double x=0,double y=0):x(x),y(y){}

};

double dist(Point a,Point b)

{

    double x=(a.x-b.x)*(a.x-b.x);

    double y=(a.y-b.y)*(a.y-b.y);

    return sqrt(x+y);

}

double Area_of_overlap(Point c1,double r1,Point c2,double r2)

{

    double d=dist(c1,c2);

    if (r1+r2<d+exp) return 0;

    if (d<fabs(r1-r2)+exp)

    {

        double r=min(r1,r2);

        return PI*r*r;

    }

    double x=(d*d+r1*r1-r2*r2)/(2*d);

    double t1=acos(x/r1);

    double t2=acos((d-x)/r2);

    return r1*r1*t1+r2*r2*t2-d*r1*sin(t1);

}

int main()

{

    int t,ncase=1;

    double r,R;

    Point a,b;

    scanf("%d",&t);

    while (t--)

    {

        scanf("%lf%lf",&r,&R);

        scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);

        double bb_area=Area_of_overlap(a,R,b,R);

        double bs_area=Area_of_overlap(a,R,b,r);

        double ss_area=Area_of_overlap(a,r,b,r);

        printf("Case #%d: %.6lf\n",ncase++,bb_area-2.0*bs_area+ss_area);

    }

    return 0;

}