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 3

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);

     }

 }

巴特西

Intersection(计算几何)

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