hdu 5120 Intersection 两环面积交

Intersection

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 2499    Accepted Submission(s): 929

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 x_i, y_i (0 ≤ x_i, y_i ≤ 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

22 30 00 02 30 05 0

 

Sample Output

Case #1: 15.707963Case #2: 2.250778

 

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

 

解法：答案=两外圆面积交-外1内2面积交-外2内1面积交+两内圆面积交。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)#define ysk(x)  (1<<(x))#define sqr(x)  ((x)*(x))typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;//const int maxn=    ;const double PI=acos(-1.0);const double eps=1e-10;int dcmp(double x){    if(fabs(x)<eps)  return 0;    else return x<0?-1:1;}struct Point{    double x,y;    Point(double x=0,double y=0):x(x),y(y) {};    bool operator ==(const Point B)const {return dcmp(x-B.x)==0&&dcmp(y-B.y)==0;}    bool operator<(const Point& b)const    {        return dcmp(x-b.x)<0|| dcmp(x-b.x)==0 &&dcmp(y-b.y)<0;    }};typedef Point Vector;Vector operator -(Vector A,Vector B) {return Vector(A.x-B.x,A.y-B.y); }double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}Vector operator +(Vector A,Vector B) {return Vector(A.x+B.x,A.y+B.y); }Vector operator *(Vector A,double p) {return Vector(A.x*p,A.y*p); }Vector operator /(Vector A,double p) {return Vector(A.x/p,A.y/p); }Vector operator -(Vector A)  {return  Vector(-A.x,-A.y);}double Length(Vector A){return sqrt(Dot(A,A));}double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));}struct Circle{    Point c;    double r;    Circle(){}    Circle(Point c,double r):c(c),r(r){}    Point point(double a)    {        return Point(c.x+cos(a)*r,c.y+sin(a)*r);    }};double Helen(Point A,Point B,Point C)//海伦公式{     Vector AB=B-A;double c= Length(AB);     Vector AC=C-A;double b= Length(AC);     Vector BC=C-B;double a= Length(BC);     double p=(a+b+c)/2;     return sqrt( p*(p-a)*(p-b)*(p-c) );}double angle(Vector v){    return atan2(v.y,v.x);}double dis(Point A,Point B){    return sqrt(sqr(A.x-B.x)+sqr(A.y-B.y));}double getS(Point c1, double r1, Point c2, double r2){    double d = dis(c1,c2);    if(dcmp(r1 + r2 - d )<=0 ) return 0;//相离 外切    if(dcmp(d - fabs(r1 - r2) )<=0  )//内含 内切    {        double r = min(r1,r2);        return PI*r*r;    }    double x = (d*d + r1*r1 - r2*r2)/(2*d);//x=rad1*r1    double rad1 = acos(x / r1);    double rad2 = acos((d - x)/r2);    return r1*r1*rad1 + r2*r2*rad2 - d*r1*sin(rad1);}Circle a[2],b[2];double solve(){    double S1=getS(a[0].c,a[0].r,a[1].c,a[1].r);    double S2=getS(a[0].c,a[0].r,b[1].c,b[1].r);    double S3=getS(a[1].c,a[1].r,b[0].c,b[0].r);    double S4=getS(b[0].c,b[0].r,b[1].c,b[1].r);//    cout<<S1<<" "<<S2<<" "<<S3<<" "<<S4<<endl;    double ans=S1-S2-S3+S4;    return ans;}int main(){//   std::ios::sync_with_stdio(false);   int T,kase=0;cin>>T;   while(T--)   {       scanf("%lf%lf",&b[0].r,&a[0].r);       b[1].r=b[0].r;       a[1].r=a[0].r;//       cin>>b[0].c.x>>b[0].c.y;       scanf("%lf%lf",&b[0].c.x,&b[0].c.y);       a[0].c=b[0].c;//       cin>>b[1].c.x>>b[1].c.y;       scanf("%lf%lf",&b[1].c.x,&b[1].c.y);       a[1].c=b[1].c;       printf("Case #%d: %.6f\n",++kase,solve());   }   return 0;}