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 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）

 

 

//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 <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <stdlib.h>
 5 #include <math.h>
 6 #include <algorithm>
 7 #include <map>
 8 #include <stack>
 9 #include <queue>
10 #include <set>
11 #include <vector>
12 using namespace std;
13 #define LL long long
14 #define PI acos(-1.0)
15 #define lowbit(x) (x&(-x))
16 #define INF 0x7f7f7f7f      // 21 E
17 #define MEM 0x7f            // memset 都变为 INF
18 #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<r2 ? PI*r1*r1 : PI*r2*r2;
37     d=sqrt(d);
38     double a1=acos((r1*r1+d*d-r2*r2)/(2.0*r1*d));
39     double a2=acos((r2*r2+d*d-r1*r1)/(2.0*r2*d));
40     double s1=a1*r1*r1; //扇形s1
41     double s2=a2*r2*r2;
42     double sinx = sqrt(1-cos(a1)*cos(a1));
43     double t = sinx * d * r1;    //三角形
44     return s1+s2-t;
45 }
46 
47 int main()
48 {
49     int T;
50     cin>>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 }
View Code

 

 

 

 

 

 

posted @ 2017-04-27 20:32  happy_codes  阅读(275)  评论(0)  编辑  收藏  举报