POJ 2986 A Triangle and a Circle

POJ 2986 A Triangle and a Circle

题目链接：POJ 2986

/*POJ 2986 三角形和圆的公共面积题意：给出三角形的三个点的坐标和圆的坐标、半径斯特瓦尔特定理设已知△ABC及其底边上B、C两点间的一点D，则有AB^2·DC+AC^2·BD-AD²·BC=BC·DC·BD。*/#include<cstdio>#include<cmath>#define eps 1e-8#define PI acos(-1.0)struct point {double x,y;};double xmult(point p1,point p2,point p0){return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);}double dmult(point p1,point p2,point p0){return (p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y);}double tirarea(double a,double b,double c){double l=(a+b+c)/2;return sqrt(l*(l-a)*(l-b)*(l-c));}double distance(point p1,point p2){return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}double root(double a,double b,double c)//{if(a==0)return -c/b;return (-b+sqrt(b*b-4*a*c))/(2*a);}double slove(point pa,point pb,point po,double r)//圆与三角形求交{double a,b,c,x,y;double area=xmult(pa,pb,po)/2;a=distance(po,pb);b=distance(po,pa);c=distance(pa,pb);if(a<=r&&b<=r)//1{return area;}else if(a<r&&b>=r)//2{//斯特瓦特公式与海伦公式推出x=(dmult(pa,po,pb)+sqrt(c*c*r*r-xmult(pa,po,pb)*xmult(pa,po,pb)))/c;return asin(area*(c-x)*2/c/b/r)*r*r/2+area*x/c;}else if(a>=r&&b<r)//3{y=(dmult(pb,po,pa)+sqrt(c*c*r*r-xmult(pb,po,pa)*xmult(pb,po,pa)))/c;return asin(area*(c-y)*2/c/a/r)*r*r/2+area*y/c;}elseif(fabs(2*area) >=r*c||dmult(pb,po,pa)<=0||dmult(pa,po,pb)<=0)//4{if(dmult(pa,pb,po)<0){if(xmult(pa,pb,po)<0){return (-PI-asin(area*2/a/b))*r*r/2;}else return (PI-asin(area*2/a/b))*r*r/2;}else return asin(area*2/a/b)*r*r/2;}else //5{x=(dmult(pa,po,pb)+sqrt(c*c*r*r-xmult(pa,po,pb)*xmult(pa,po,pb)))/c;y=(dmult(pb,po,pa)+sqrt(c*c*r*r-xmult(pb,po,pa)*xmult(pb,po,pa)))/c;return (asin(area*(1-x/c)*2/r/b)+asin(area*(1-y/c)*2/r/a))*r*r/2+area*((y+x)/c-1);}}int main(){point p[4],cen;double r;while(scanf("%lf%lf %lf%lf %lf%lf %lf%lf%lf",&p[0].x,&p[0].y,&p[1].x,&p[1].y,&p[2].x,&p[2].y,&cen.x,&cen.y,&r)!=EOF){if(xmult(p[0],p[1],p[2])==0){printf("0.00\n");continue;}p[3]=p[0];double res=0;for(int i=0;i<3;i++)res+=slove(p[i],p[i+1],cen,r);printf("%.2lf\n",fabs(res));}return 0;}