hdu-5120-Intersection

http://acm.hdu.edu.cn/showproblem.php?pid=5120

Intersection

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 231    Accepted Submission(s): 101

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

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

 

Sample Output

Case #1: 15.707963 Case #2: 2.250778

 

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

解题思路：代码中解释的很详细

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define pi acos(-1.0)

/*
已知：圆环心距，两圆环的内圆半径和外圆半径。
S = S（大圆1交大圆2） + S（小圆1交小圆2）- S（大圆1交小圆2） - S（小圆1交大圆2） 
*/
struct P{
    double x;
    double y;
    double r;
}p[4];;

//求两圆相交面积
double area(int i, double r1, int j, double r2){
    double d = sqrt((p[i].x - p[j].x) * (p[i].x - p[j].x) + (p[i].y - p[j].y) * (p[i].y - p[j].y));//圆心距
    if(r1 > r2){
        double temp = r1;
        r1 = r2;
        r2 = temp;
    }//r1取小
    if(r1 + r2 <= d)
        return 0;//相离
    else if(r2 - r1 >= d)
        return pi * r1 * r1;//内含
    else {
        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));
        return (a1 * r1 * r1 + a2 * r2 * r2 - r1 * d * sin(a1));
    }//相交
}

int main(){
    int t;
    double r1, r2;
    double x1, y1, x2, y2;
    double s1, s2, s3, s4;
    double ans;
    int Case = 1;
    scanf("%d", &t);
    while(t--){
        scanf("%lf %lf", &r1, &r2);
        scanf("%lf %lf", &x1, &y1);
        scanf("%lf %lf", &x2, &y2);
        //小圆1
        p[0].x = x1, p[0].y = y1, p[0].r = r1;
        //大圆1
        p[1].x = x1, p[1].y = y1, p[1].r = r2;
        //小圆2
        p[2].x = x2, p[2].y = y2, p[2].r = r1;
        //大圆2
        p[3].x = x2, p[3].y = y2, p[3].r = r2;

        //大1小2
        s1 = area(1, r2, 2, r1);
        //小1大2
        s2 = area(0, r1, 3, r2);
        //大1大2
        s3 = area(1, r2, 3, r2);
        //小1小2
        s4 = area(0, r1, 2, r1);
        ans = s3 + s4 - s1 - s2;
        printf("Case #%d: %.6lf ", Case++, ans);
    }
    return 0;

}