HDU 5120 Intersection（计算几何+容斥）

思路：直接容斥搞，然后就是两圆相交面积的模板题了

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=100010;
const int INF=0x3f3f3f3f;
int cas=1,T;
#define PI acos(-1.0)


double area(double x1, double y1, double r1, double x2, double y2, double r2)
{
	double d=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
	if(d>=(r1+r2)*(r1+r2)) return 0;
	if(d<=(r1-r2)*(r1-r2)) return r1<r2 ? PI * r1 * r1 : PI * r2 *r2;

	d=sqrt(d);
	double a1=acos((r1*r1+d*d-r2*r2)/(2.0*r1*d));
	double a2=acos((r2*r2+d*d-r1*r1)/(2.0*r2*d));
	double s1=a1*r1*r1;
	double s2=a2*r2*r2;
	double t=(r1+r2+d)/2.0;
	t=2.0*sqrt(t*(t-r1)*(t-r2)*(t-d));
	return s1+s2-t;
}
int main()
{
	//freopen("1.in","w",stdout);
	//freopen("1.in","r",stdin);
	//freopen("1.out","w",stdout);
	scanf("%d",&T);
	int ca=1;
	//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
	while(T--)
	{
		int r, R, x1,y1,x2,y2;	
		scanf("%d%d", &r, &R);
		scanf("%d%d", &x1, &y1);
		scanf("%d%d", &x2, &y2);
		double ans = area(x1,y1,R,x2,y2,R)-area(x1,y1,r,x2,y2,R)-area(x1,y1,R,x2,y2,r)+area(x1,y1,r,x2,y2,r);
		printf("Case #%d: %.6lf\n", ca++, ans);
	}
	return 0;
}

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