HDU5120 Intersection 【求圆的面积交】

最新推荐文章于 2021-08-19 22:25:16 发布

长风Qiu

最新推荐文章于 2021-08-19 22:25:16 发布

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文章标签： HDU5120

本文链接：https://blog.youkuaiyun.com/chang_mu/article/details/41598405

版权

本文介绍了一种计算两个相同大小的环形在二维平面上交集面积的方法。通过定义环形的基本参数，利用数学公式计算不同配置下两环形相交部分的面积，并提供了一个具体的C++实现示例。

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Intersection

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

Total Submission(s): 41 Accepted Submission(s): 22

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 ≤ 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

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）

S = A大B大 - A大B小 - A小B大 + A小B小。（A表示A环，大表示大圆，B同）。然后直接套模板。

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <cmath>
using namespace std;

const double eps = 1e-8;
const double PI = acos(-1.0);

int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    if(x < 0) return - 1;
    else return 1;
}
struct Point
{
    double x, y;
    Point(){}
    Point(double _x, double _y)
    {
        x = _x; y = _y;
    }
    Point operator -( const Point &b) const
    {
        return Point(x - b. x, y - b. y);
    }

    double operator ^ (const Point &b) const
    {
        return x*b. y - y*b. x;
    }

    double operator * (const Point &b) const
    {
        return x*b. x + y*b. y;
    }

    void transXY(double B)
    {
        double tx = x,ty = y;
        x = tx* cos(B) - ty*sin(B);
        y = tx* sin(B) + ty*cos(B);
    }
};

double dist( Point a, Point b)
{
    return sqrt((a-b)*(a- b));
}

double Ac(Point c1, double r1, Point c2, double r2)
{
    double d = dist(c1,c2);
    if(r1 + r2 < d + eps) return 0;
    if(d < fabs(r1 - r2) + eps)
    {
        double r = min(r1,r2);
        return PI*r*r;
    }
    double x = (d*d + r1*r1 - r2*r2)/(2*d);
    double t1 = acos(x / r1);
    double t2 = acos((d - x)/r2);
    return r1*r1*t1 + r2*r2*t2 - d*r1*sin(t1);
}

int main() {
    int T; Point c1, c2;
    double ans, r, R, x1, y1, x2, y2;
    scanf("%d", &T);
    for(int cas = 1; cas <= T; ++cas) {
        scanf("%lf%lf%lf%lf%lf%lf", &r, &R, &x1, &y1, &x2, &y2);
        c1.x = x1; c1.y = y1;
        c2.x = x2; c2.y = y2;
        ans = Ac(c1, R, c2, R) - Ac(c1, R, c2, r) - Ac(c1, r, c2, R)
                + Ac(c1, r, c2, r);
        printf("Case #%d: %.6lf\n", cas, ans);
    }
    return 0;
}