HDU 5120 Intersection(圆的面积交)

题目大意：给你两个圆环，让你求出来圆环的面积交，需要用到圆的面积交，然后容斥一下，就可以得到圆环的面积交。画一下图就会很清晰。

Intersection

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 526    Accepted Submission(s): 226

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 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

22 30 00 02 30 05 0

 

Sample Output

Case #1: 15.707963Case #2: 2.250778

 

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <math.h>#include <time.h>#include <stack>#include <map>#include <set>#define eps 1e-8///#define LL long long#define LL __int64#define INF 0x3f3f3f#define PI acos(-1)#define mod 1000000007using namespace std;struct Point{    double x, y;};double Distance(Point a, Point b){    return sqrt(((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y))*1.0);}double area_of_overlap(Point c1, double r1, Point c2, double r2){    double a = Distance(c1, c2);    double b = r1;    double c = r2;    if((a >= b+c)) return 0.0;    if(a < abs(b-c))    {        double r = min(r1, r2);        return PI*r*r;    }    double cta1 = acos((a*a+b*b-c*c)/2.0/(a*b)),           cta2 = acos((a*a+c*c-b*b)/2.0/(a*c));    double s1 = r1*r1*cta1                -r1*r1*sin(cta1)*(a*a+b*b-c*c)/2/(a*b);    double s2 = r2*r2*cta2                -r2*r2*sin(cta2)*(a*a+c*c-b*b)/2/(a*c);    return s1+s2;}int main(){    int T;    int Case = 1;    scanf("%d",&T);    while(T--)    {        double r1, r2;        Point a, b;        scanf("%lf %lf",&r1, &r2);        scanf("%lf %lf", &a.x, &a.y);        scanf("%lf %lf", &b.x, &b.y);        printf("Case #%d: ",Case++);        if(a.x == b.x && a.y == b.y)        {            printf("%.6lf\n",PI*(r2*r2-r1*r1));            continue;        }        double sum = 0;        sum += area_of_overlap(a, r2, b, r2);        sum -= area_of_overlap(a, r1, b, r2);        sum -= area_of_overlap(a, r2, b, r1);        sum += area_of_overlap(a, r1, b, r1);        printf("%.6lf\n",sum);    }    return 0;}/*51 40 0 0 6*/