Geometry

uva1473 这题说的是 在空间中给了n个点 然后用体积最小的圆锥将这些点包含在内可以在表面上, 将这些点 映射到xoz平面上然后,然后枚举每个上凸包的边和每个点的极值进行判断求得最小的体积 我们会发现最小的体积 要不就紧贴一个点要不然就会贴住两个点

#include <iostream>
#include <cstdio>
#include <string.h>
#include <cmath>
#include <algorithm>
using namespace std;
const double INF =1.79769e+308;
const double eps = 0.000000000001;
int dcmp(double a)
{
     if(fabs(a)<=eps) return ;
     return a>?:-;
}
struct point {
   double x,y;
   point(double a=, double b=){
       x=a; y=b;
   }
   point operator -(point A){
     return point(x-A.x, y-A.y);
   }
   bool operator <(point A)const{
      return dcmp(x-A.x)<||(dcmp(x-A.x)==&&dcmp(y-A.y)<=);
   }
}P[],ch[];
int n;
double Cross(point A, point B)
{
     return A.x*B.y-A.y*B.x;
}
int Conxtull(int &G)
{
    sort(P,P+n);
    int m=;
    for(int i =; i<n ; ++i )
    {
        while(m>&&dcmp( Cross( ( ch[m-] - ch[m-] ), ( P[i] - ch[m-] ) ) )<=)m--;
        ch[ m++ ] = P[i];
    }
    int k = m;
    for(int i = n-; i>=; --i )
        {
             while( k<m&&dcmp( Cross( ch[m-]-ch[m-], P[i]-ch[m-] ) )<= ) m--;
              ch[m++] = P[i] ;
        }
        if(n>)m--;
        G=m;
        return k-;
}
point getpoint(double k,point F)
{
      point ans;
      ans.x = F.x+(F.y/k);
      ans.y = F.y+k*F.x;
      return ans;
}
double Volun(double radio, double hight)
{
      return  acos(-)*radio*radio*hight/3.0;
}
double Volume,ansR,ansH;
void solve(double k,point T)
{
       point e = getpoint(k,T);
       double V =Volun(e.x,e.y);
       if(dcmp(Volume-V)>){
          ansR=e.x; ansH=e.y;
          Volume=V;
       }
}
int main()
{
      freopen("data.in","r",stdin);
      freopen("data.out","w",stdout);
     while(scanf("%d",&n)==)
        {
            n++;
             double Minx=0.0;
             P[]=point(,);
               for(int i = ; i<n; ++i)
                {
                     double x,y,z;
                     scanf("%lf%lf%lf",&x,&y,&z);
                     P[i].x=sqrt(x*x+y*y);
                     P[i].y=z;
                     Minx = max(Minx,P[i].x);
                }
                int m;
                int k =Conxtull(m);
                ch[m]=ch[];
                Volume=INF;
                double K=INF;
                int we;
                for(int i=k; i<m; ++i )
                {
                     if(i==m-||dcmp(ch[i].x-ch[i+].x)<=||dcmp(ch[i].y-ch[i+].y)>=)
                     {
                            we=i;
                            break;
                     }
                     double R =(ch[i+].y-ch[i].y)/(ch[i].x-ch[i+].x);
                     double t =ch[i].y*;
                     t/=ch[i].x;
                     int f1 =dcmp(t-R);
                     int f2=dcmp(t-K);
                     if(f1>=&&f2<=)
                       solve(t,ch[i]);
                      solve(R,ch[i]);
                      K=R;
                }
                double t = 2.0*ch[we].y/ch[we].x;
                if(dcmp(t-K)<)
                    solve(t,ch[we]);
                printf("%.3lf %.3lf\n",ansH,ansR);
        }

     return ;
}

uva 12165 这题说的是  用梅涅劳斯 计算图中三角形的对应的比例列出一堆后 开始拆分那些边然后化简就会达到所要的公式

#include <iostream>
#include <string.h>
#include <cmath>
#include <cstdio>
using namespace std;
struct point{
    double  x,y;
    point(double a=,double b=){
       x=a; y =b;
    }
    point operator +(point A){
       return point(x+A.x,y+A.y);
    }
    point operator -(point A){
         return point(x-A.x,y-A.y);
    }
    point operator *(double  A){
          return point(x*A,y*A);
    }
};
double Cross(point A,point B){
   return A.x*B.y-A.y*B.x;
}
double Dot(point A,point B){
   return A.x*B.x+A.y*B.y;
}
double Length(point A){
   return sqrt(Dot(A,A));
}
int main()
{

      double m1,m2,m3,m4,m5,m6;
      point P,Q,R;
      int cas;
      scanf("%d",&cas);
      for(int cc=; cc<=cas; ++cc){

           scanf("%lf%lf%lf%lf%lf%lf",&P.x,&P.y,&Q.x,&Q.y,&R.x,&R.y);
           scanf("%lf%lf%lf%lf%lf%lf",&m1,&m2,&m3,&m4,&m5,&m6);
           double c =Length(P-Q),a =Length(R-Q), b =Length(P-R);
           double m = m3*m5/(m6*(m3+m4));
           double n = m4*m2/((m3+m4)*m1);
           double BP = (c+m*c)/(n-m);
           m = (m5*m1)/((m5+m6)*m2);
           n = m6*m4/((m5+m6)*m3);
           double CQ = (m*a+a)/(n-m);
           m = m1*m3/((m1+m2)*m4);
           n =m2*m6/((m1+m2)*m5);
           double AR = (m*b+b)/(n-m);
           point PR = (R-P)*(/Length(R-P));
           point A = R +(PR*AR);
           point QP = (P-Q)*(/Length(P-Q));
           point B = P+(QP*BP);
           point RQ = (Q-R)*(/Length(Q-R));
           point C = Q+(RQ*CQ);
           printf("%.8lf %.8lf %.8lf %.8lf %.8lf %.8lf\n",A.x,A.y,B.x,B.y,C.x,C.y);
     }

      return ;
}

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