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  • POJ 2546 & ZOJ 1597 Circular Area(求两圆相交的面积 模板)

    题目链接:

    POJ:http://poj.org/problem?

    id=2546

    ZOJ:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=597

    Description

    Your task is to write a program, which, given two circles, calculates the area of their intersection with the accuracy of three digits after decimal point.

    Input

    In the single line of input file there are space-separated real numbers x1 y1 r1 x2 y2 r2. They represent center coordinates and radii of two circles.

    Output

    The output file must contain single real number - the area.

    Sample Input

    20.0 30.0 15.0 40.0 30.0 30.0

    Sample Output

    608.366
    

    Source

    Northeastern Europe 2000, Far-Eastern Subregion

    题意:

    求两圆相交的面积!

    直接上模板

    代码例如以下:

    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <iostream>
    #include <algorithm>
    //#include <complex>
    //#include <iomanip>
    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);
        }
        //叉积
        double operator ^ (const Point &b) const
        {
            return x*b. y - y*b. x;
        }
        //点积
        double operator * (const Point &b) const
        {
            return x*b. x + y*b. y;
        }
        //绕原点旋转角度B(弧度值),后x,y的变化
        void transXY(double B)
        {
            double tx = x,ty = y;
            x = tx* cos(B) - ty*sin(B);
            y = tx* sin(B) + ty*cos(B);
        }
    };
    //*两点间距离
    double dist( Point a, Point b)
    {
        return sqrt((a-b)*(a- b));
    }
    //两个圆的公共部分面积
    double Area_of_overlap(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()
    {
        double x1, y1, r1, x2, y2, r2;
        while(~scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&r1,&x2,&y2,&r2))
        {
            Point c1, c2;
            c1.x = x1;
            c1.y = y1;
            c2.x = x2;
            c2.y = y2;
            double ans = Area_of_overlap(c1,r1,c2,r2);
            printf("%.3lf
    ",ans);
            //cout<<setiosflags(ios::fixed)<<setprecision(3)<<ans<<endl;
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/5407546.html
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