hdoj 5120 Intersection 【计算圆环相交面积】

Intersection

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

Total Submission(s): 973    Accepted Submission(s): 372

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 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
   
   
   
   

 

题意，给你两个一样的圆环1和圆环2，让那个求出圆环相交面积。

思路：两个外圆相交面积 - 外圆1内圆2相交面积 - 内圆1外圆2相交面积 + 两个内圆相交面积。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define PI acos(-1.0)
#define eps 1e-10
using namespace std;
struct Point//点
{
    double x, y;
};
double dist(Point c1, Point c2)//两点间距离
{
    return sqrt((c1.x - c2.x) * (c1.x - c2.x) + (c1.y - c2.y) * (c1.y - c2.y));
}
double area(Point c1, double r1, Point c2, double r2)
{
    double d = dist(c1, c2);
    if(r1 + r2 < d + eps) return 0;
    if(d < fabs(r1 - r2) + eps)
    {
        double r = min(r1, r2);
        return PI * r * r;
    }
    double x = (d * d + r1 * r1 - r2 * r2) / (2 * d);
    double t1 = acos(x / r1);
    double t2 = acos((d - x) / r2);
    return r1 * r1 * t1 + r2 * r2 * t2 - d * r1 *sin(t1);
}
int main()
{
    int t, k = 1;
    double r, R;//内圆半径 外圆半径
    scanf("%d", &t);
    while(t--)
    {
        Point c1, c2;
        scanf("%lf%lf", &r, &R);
        scanf("%lf%lf", &c1.x, &c1.y);
        scanf("%lf%lf", &c2.x, &c2.y);
        double R_R = area(c1, R, c2, R);//大大相交面积
        double R_r = area(c1, R, c2, r);//大小相交面积
        double r_R = area(c1, r, c2, R);//小大相交面积
        double r_r = area(c1, r, c2, r);//小小相交面积
        printf("Case #%d: %.6lf\n", k++, R_R - R_r - r_R + r_r);
    }
    return 0;
}