保证存在点q和ABC不共线。 Input 第一行是样例个数。
每一个样例前三行是ABC坐标。
最后一行是q到平面ABC三点的距离。
Output 对于每一个样例，输出一个两位小数。
Sample Input 17.000000 49.000000 73.00000058.000000 30.000000 72.00000044.000000 78.000000 23.000000199.231022 148.680866 163.300337 Sample Output
0.00
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<map>#include<set>using namespace std;const double eps=1e-8;double a1,b1,c1,a2,b2,c2,a3,b3,c3;double x,y,z;double P( double a,double b,double c,double d,double e ){ return a*(b*c-d*e); }double V(double OA,double OB,double OC,double AB,double CA,double BC){ OA*=OA;OB*=OB;OC*=OC;AB*=AB;CA*=CA;BC*=BC; double ans=0; ans+=P( OA,OB,OC,(OB+OC-BC)/2.,(OB+OC-BC)/2. ); ans-=P( (OA+OB-AB)/2.,(OA+OB-AB)/2.,OC,(OA+OC-CA)/2.,(OB+OC-BC)/2. ); ans+=P( (OA+OC-CA)/2.,(OA+OB-AB)/2.,(OB+OC-BC)/2.,OB,(OA+OC-CA)/2.); if(ans<0) return 0; else return sqrt(ans);}double dis(double a,double b,double c){ return sqrt(a*a+b*b+c*c);}double S(double a,double b,double c){ double s=(a+b+c)/2; double area=sqrt(s*(s-a)*(s-b)*(s-c)); return area;}int main(){ int T; int i,j,k; scanf("%d",&T); while(T--){ scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf",&a1,&b1,&c1,&a2,&b2,&c2,&a3,&b3,&c3); scanf("%lf%lf%lf",&x,&y,&z); double x4=dis(a1-a2,b1-b2,c1-c2); double x5=dis(a1-a3,b1-b3,c1-c3); double x6=dis(a2-a3,b2-b3,c2-c3); double ans=V(x,y,z,x4,x5,x6)/(S(x4,x5,x6)*2); printf("%.2fn",ans); } return 0;}
转载于:https://www.cnblogs.com/Basasuya/p/8433775.html