hdu5120Intersection+圆环的面积交

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 ≤ 10^5), 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

Source
2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

Recommend
liuyiding

给两个一样大小的圆环。位置不一样。求两个圆环的面积交
其实就是容斥加圆的面积交。外圆的面积交-大圆与内圆的面积交（2个）+内圆的面积交
这里被自己的模板坑了。以前的模板都是保证相交的，这里的圆的面积交可以是相离、包含的，要自己再判一下。（罚时两发，orz）。。

#include<bits/stdc++.h>#define pi acos(-1.0)using namespace std;#define eps 1e-8struct point {    double x,y;    point(){};    point(double x,double y):x(x),y(y){};};double dis(const point &a,const point &b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}struct Circle {    point p;    double r;    Circle(){};    Circle(point p,double r):p(p),r(r){};};double Circle_area(Circle x,Circle y){    double a=dis(x.p,y.p),b=x.r,c=y.r;    //相离、相切    if(a>=b+c) return 0.0;      //某个圆是点时    if(b<eps||c<eps) return 0.0;     //包含    if(b<c) swap(b,c);    if(a+c<=b) return pi*c*c;    //相交。    double cta1=acos((a*a+b*b-c*c)/2/(a*b)),           cta2=acos((a*a+c*c-b*b)/2/(a*c));    double s1=b*b*cta1-b*b*sin(cta1)*(a*a+b*b-c*c)/2/(a*b);    double s2=c*c*cta2-c*c*sin(cta2)*(a*a+c*c-b*b)/2/(a*c);    return fabs(s1+s2);}int main(){    int t;    //cout<<Circle_area(Circle(point(0,0),2),Circle(point(0,1),2));    scanf("%d",&t);    for(int cas=1;cas<=t;cas++){        double r1,r2;        double x1,y1,x2,y2;        scanf("%lf %lf",&r2,&r1);        scanf("%lf %lf",&x1,&y1);        scanf("%lf %lf",&x2,&y2);        Circle a1=Circle(point(x1,y1),r1);        Circle a2=Circle(point(x1,y1),r2);        Circle a3=Circle(point(x2,y2),r1);        Circle a4=Circle(point(x2,y2),r2);        double ans=Circle_area(a1,a3);        ans-=Circle_area(a1,a4);        ans-=Circle_area(a2,a3);        ans+=Circle_area(a2,a4);        printf("Case #%d: %.6f\n",cas,ans);    }    return 0;}