POJ 2546 Circular Area（计算几何 两圆相交面积）

题目描述：

Circular Area

Time Limit: 1000MS Memory Limit: 65536K

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

计算几何初接触，本来想先做个水题找找手感结果被无限卡制，最后没办法只好查题解，原来是g++在卡精度……用c++交就很顺利的A过了。

整理了一下自己写的比较顺手的方式。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define MAX 100010
#define MM 105
#define INF 0x3f3f3f3f
#define PI acos(-1.0)

using namespace std;

struct point
{
double x, y;

double len(point t)
    {
return sqrt((x - t.x) * (x - t.x) + (y - t.y) * (y - t.y));
    }
};

struct circle
{
    point p;
double r;

double dis(circle c)
    {
return p.len(c.p);
    }
};

double cosine(double a, double b, double c)
{
return (a * a + b * b - c * c) / (2.0 * a * b);
}

double Straingle(double a, double b, double angle)
{
return a * b * sin(angle);
}

double Scircle(double r, double angle)
{
return angle * r * r;
}

double getans(circle a, circle b)
{
double d = a.dis(b);
if(d >= a.r + b.r)
    {
return 0;
    }
else if(d <= fabs(a.r - b.r))
    {
double r = a.r < b.r ? a.r : b.r;
return PI * r * r;
    }
else
    {
double ang1 = acos(cosine(a.r, d, b.r));
double ang2 = acos(cosine(b.r, d, a.r));
double ans = Scircle(a.r, ang1) + Scircle(b.r, ang2) - Straingle(a.r, d, ang1);
return ans;
    }
}

int main()
{
    circle c1, c2;
while(~scanf("%lf %lf %lf %lf %lf %lf", &c1.p.x, &c1.p.y, &c1.r, &c2.p.x, &c2.p.y, &c2.r))
    {
printf("%.3lf\n", getans(c1, c2));
    }
return 0;
}

运行结果：