hdu5120（相交环面积）

Intersection

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

Total Submission(s): 41    Accepted Submission(s): 22

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

题意：给你两个环，求相交部分面积。

分析：两环相交也就是图中黑色部分，然后用大圆相交面积 - 大小圆相交面积 +小圆相交面积。直接套模板。

代码：

#include<bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
// 圆的结构
struct circle {
       double x; // 圆心横坐标
       double y; // 圆心纵坐标
       double r; // 半径
};
// 计算圆心距
double dist(circle a, circle b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
// 计算两相交圆的面积
double area(circle a, circle b) {


       if((dist(a, b)+min(a.r,b.r))<=max(a.r,b.r)) // 内含或重合
        {
            if(a.r<b.r)
                 return PI*a.r*a.r;
            else
                 return PI*b.r*b.r;
        }
        else if(dist(a, b)>=(a.r+b.r)) // 相离或相切
            return 0.0;
        else // 相交
        {
            double length=dist(a, b);
            // 利用三角形余弦定理求圆心角
            double d1=2*acos((a.r*a.r+length*length-b.r*b.r)/(2*a.r*length));
            double d2=2*acos((b.r*b.r+length*length-a.r*a.r)/(2*b.r*length));
            // 利用圆心角求得扇形面积再减去三角形面积后两部分相加就是相交部分面积
            double area1=a.r*a.r*d1/2-a.r*a.r*sin(d1)/2;
            double area2=b.r*b.r*d2/2-b.r*b.r*sin(d2)/2;
            double area=area1+area2;
            return area;
        }
}
int main()
{
    int T;
    circle a, b, A, B;
    double r, R;
    scanf("%d", &T) ;
    while(T--)
    {
        scanf("%lf%lf%lf%lf%lf%lf", &r,&R,&a.x,&a.y,&b.x,&b.y);
        a.r = b.r = r;
        A.r = B.r = R;
        A.x = a.x;
        A.y = a.y;
        B.x = b.x;
        B.y = b.y;
        double sec = area(A, B) - area(A, b) - area(B, a) + area(a, b);
        static int t=1;
        printf("Case #%d: %.6f\n", t++, sec);
    }
    return 0;
}