zoj 1597 Circular Area (求两园相交面积)

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# include <algorithm># include <stdio.h># include <string.h># include <iostream># include <math.h>using namespace std;# define PI acos(-1.0)const double eps = 1e-8;struct  node{    double x;    double y;    double r;};struct node a1[10];double dis(double x1,double y1,double x2,double y2){    return sqrt( (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2) );}double area()///不相离不相含求相交面积{    double a = dis(a1[1].x,a1[1].y,a1[0].x,a1[0].y), b = a1[0].r, c = a1[1].r;    double cta1 = acos((a * a + b * b - c * c) / 2 / (a * b)),           cta2 = acos((a * a + c * c - b * b) / 2 / (a * c));    double s1 = a1[0].r*a1[0].r*cta1 - a1[0].r*a1[0].r*sin(cta1)*(a * a + b * b - c * c) / 2 / (a * b);    double s2 = a1[1].r*a1[1].r*cta2 - a1[1].r*a1[1].r*sin(cta2)*(a * a + c * c - b * b) / 2 / (a * c);    return s1 + s2;}int main(){    double x1,x2,y1,y2,xx1,xx2,yy1,yy2;    double d1,d2,d3,d4,d5,d6,d7,d8;    while(~scanf("%lf %lf %lf %lf %lf %lf",&a1[0].x,&a1[0].y,&a1[0].r,&a1[1].x,&a1[1].y,&a1[1].r))    {        if(dis(a1[0].x,a1[0].y,a1[1].x,a1[1].y)<=fabs(a1[0].r-a1[1].r))///相包含        {            if(a1[1].r>a1[0].r)                  printf("%.3lf\n",PI*a1[0].r*a1[0].r);            else                printf("%.3lf\n",PI*a1[1].r*a1[1].r);        }        else if(dis(a1[0].x,a1[0].y,a1[1].x,a1[1].y)>=a1[0].r+a1[1].r)///相离            printf("0.000\n");        else ///相交        printf("%.3lf\n",area());    }    return 0;}
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