Intersection(计算几何)

Intersection

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 3018    Accepted Submission(s): 1135

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

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）

//题意:问两个同样大的圆环相交的面积是多大

画图后，可以发现，用容斥定理很简单，area = 两个大圆的相交面积 - 2 * 大圆和小圆相交面积 + 两个小圆相交面积

求面积要用余弦定理，然后就简单了

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <stdlib.h>
 #include <math.h>
 #include <algorithm>
 #include <map>
 #include <stack>
 #include <queue>
 #include <set>
 #include <vector>
 using namespace std;
 #define LL long long
 #define PI acos(-1.0)
 #define lowbit(x) (x&(-x))
 #define INF 0x7f7f7f7f      // 21 E
 #define MEM 0x7f            // memset 都变为 INF
 #define MOD 4999            // 模
 #define eps 1e-9            // 精度
 #define MX  1000005         // 数据范围
 
 int read() {    //输入外挂
     int res = , flag = ;
     char ch;
     if((ch = getchar()) == '-') flag = ;
     else if(ch >= '' && ch <= '') res = ch - '';
     while((ch = getchar()) >= '' && ch <= '') res = res *  + (ch - '');
     return flag ? -res : res;
 }
 // code... ...
 
 double area(double x1,double y1,double r1,double x2,double y2,double r2)
 {
     double d=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
     if(d>=(r1+r2)*(r1+r2)) return ;
     if(d<=(r1-r2)*(r1-r2)) return r1<r2 ? PI*r1*r1 : PI*r2*r2;
     d=sqrt(d);
     double a1=acos((r1*r1+d*d-r2*r2)/(2.0*r1*d));
     double a2=acos((r2*r2+d*d-r1*r1)/(2.0*r2*d));
     double s1=a1*r1*r1; //扇形s1
     double s2=a2*r2*r2;
     double sinx = sqrt(-cos(a1)*cos(a1));
     double t = sinx * d * r1;    //三角形
     return s1+s2-t;
 }
 
 int main()
 {
     int T;
     cin>>T;
     for (int cnt=;cnt<=T;cnt++)
     {
         double x1,y1,x2,y2,r1,r2;
         scanf("%lf%lf",&r1,&r2);
         scanf("%lf%lf",&x1,&y1);
         scanf("%lf%lf",&x2,&y2);
         double ans = area(x1,y1,r2,x2,y2,r2);
         ans -= *area(x1,y1,r1,x2,y2,r2);
         ans += area(x1,y1,r1,x2,y2,r1);
         printf("Case #%d: %.6f\n",cnt,ans);
     }
 }