首页 > 代码库 > POJ 2546 & ZOJ 1597 Circular Area（求两圆相交的面积 模板）
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
代码如下：
#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\n",ans);
//cout<<setiosflags(ios::fixed)<<setprecision(3)<<ans<<endl;
}
return 0;
}
POJ 2546 & ZOJ 1597 Circular Area（求两圆相交的面积 模板）
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