hdu 5120 Intersection(求相交环面积)

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

solution：

由容斥可得 两环相交面积=两大圆相交面积-2*一大圆与一小圆相交面积+两小圆相交面积

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
#define pi acos(-1.0)
struct node{
double x,y;
} c[10];
double area(double r1, double rr)
{
int i = 0, j = 1;
double d =sqrt((c[i].x - c[j].x)*(c[i].x - c[j].x) + (c[i].y - c[j].y)*(c[i].y - c[j].y));//圆心距
if (r1>rr){
double temp = r1;
r1 = rr;
rr = temp;
}//r1取小
if (r1 + rr <= d)
return 0;//相离
else if (rr - r1 >= d)
return pi*r1*r1;//内含
else {

double a1 = acos((r1*r1 + d*d - rr*rr) / (2.0*r1*d));
double a2 = acos((rr*rr + d*d - r1*r1) / (2.0*rr*d));
return (a1*r1*r1 + a2*rr*rr - r1*d*sin(a1));
}//相交
}
int main()
{
int t;
scanf("%d", &t);
double r, rr;
for (int k = 1; k <= t; k++)
{
scanf("%lf%lf", &r, &rr);
for (int i = 0; i < 2; i++)
scanf("%lf%lf", &c[i].x, &c[i].y);
double ans = area(rr, rr)+area(r, r) - area(r, rr)*2.0;
printf("Case #%d: %.6lf\n",k, ans);
}

}