HDU 5120 两圆环覆盖的面积(几何)

Intersection

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

Total Submission(s): 1340 Accepted Submission(s): 516



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2
2 3
0 0
0 0
2 3
0 0
5 0

[align=left]Sample Output[/align]

Case #1: 15.707963
Case #2: 2.250778

[align=left]Source[/align]
2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

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此题题意是让求两个圆心坐标不同但内径和外径相同的的两个圆环,相交覆盖的面积

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define eps 1e-9
#define PI acos(-1)
using namespace std;
struct Point
{
double x,y;
Point(){};
Point(double a,double b)
{
x=a,y=b;
}
Point operator - (const Point a)const
{
return Point(x-a.x,y-a.y);
}
double operator *(const Point a)const
{
return x*a.x+y*a.y;
}
};
double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
}
double Area(Point a,double r1,Point b,double r2)
{
double k=dist(a,b);
if(k+eps>=r1+r2) //相切或相离
return 0;
if(k<=fabs(r1-r2)+eps)  //大圆包含小圆 ,eps应该放在<后
{
double Min=min(r1,r2);
return PI*Min*Min;
}
double x=(k*k+r1*r1-r2*r2)/(2.0*k);
double j1=acos(x/r1);  //直角三角形
double j2=acos((k-x)/r2);
return (j1*r1*r1+j2*r2*r2-k*r1*sin(j1));
}
int main()
{
int T;
while(~scanf("%d",&T))
{
int Case=0;
while(T--)
{
double r,R;
Point C1,C2;
Case++;
scanf("%lf%lf",&r,&R);
scanf("%lf%lf%lf%lf",&C1.x,&C1.y,&C2.x,&C2.y);
double area=Area(C1,R,C2,R)+Area(C1,r,C2,r)-2*Area(C1,r,C2,R);
printf("Case #%d: ",Case);
printf("%.6lf\n",area);
}
}
}