Intersection(计算几何)-白红宇

Intersection(计算几何)

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发布时间：2019-06-20

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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

//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

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
             8 #include 
             
               9 #include 
              
               10 #include 
               
                11 #include 
                
                 12 using namespace std;13 #define LL long long14 #define PI acos(-1.0)15 #define lowbit(x) (x&(-x))16 #define INF 0x7f7f7f7f // 21 E17 #define MEM 0x7f // memset 都变为 INF18 #define MOD 4999 // 模19 #define eps 1e-9 // 精度20 #define MX 1000005 // 数据范围21 22 int read() { //输入外挂23 int res = 0, flag = 0;24 char ch;25 if((ch = getchar()) == '-') flag = 1;26 else if(ch >= '0' && ch <= '9') res = ch - '0';27 while((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + (ch - '0');28 return flag ? -res : res;29 }30 // code... ...31 32 double area(double x1,double y1,double r1,double x2,double y2,double r2)33 {34 double d=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);35 if(d>=(r1+r2)*(r1+r2)) return 0;36 if(d<=(r1-r2)*(r1-r2)) return r1
                 
                  >T;51 for (int cnt=1;cnt<=T;cnt++)52 {53 double x1,y1,x2,y2,r1,r2;54 scanf("%lf%lf",&r1,&r2);55 scanf("%lf%lf",&x1,&y1);56 scanf("%lf%lf",&x2,&y2);57 double ans = area(x1,y1,r2,x2,y2,r2);58 ans -= 2*area(x1,y1,r1,x2,y2,r2);59 ans += area(x1,y1,r1,x2,y2,r1);60 printf("Case #%d: %.6f\n",cnt,ans);61 }62 }

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转载于:https://www.cnblogs.com/haoabcd2010/p/6776400.html

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