hdu 4617 Weapon (几何)

Description

　Doctor D. are researching for a horrific weapon. The muzzle of the weapon is a circle. When it fires, rays form a cylinder that runs through the circle verticality in both side. If one cylinder of rays touch another, there will be an horrific explosion. Originally, all circles can rotate easily. But for some unknown reasons they can not rotate any more. If these weapon can also make an explosion, then Doctor D. is lucky that he can also test the power of the weapon. If not, he would try to make an explosion by other means. One way is to find a medium to connect two cylinder. But he need to know the minimum length of medium he will prepare. When the medium connect the surface of the two cylinder, it may make an explosion.

Input

The first line contains an integer T, indicating the number of testcases. For each testcase, the first line contains one integer N(1 < N < 30), the number of weapons. Each of the next 3N lines&#160; contains three float numbers. Every 3 lines represent one weapon. The first line represents the coordinates of center of the circle, and the second line and the third line represent two points in the circle which surrounds the center. It is supposed that these three points are not in one straight line. All float numbers are between -1000000 to 1000000.

Output

For each testcase, if there are two cylinder can touch each other, then output 'Lucky', otherwise output then minimum distance of any two cylinders, rounded to two decimals, where distance of two cylinders is the minimum distance of any two point in the surface of two cylinders.

Sample Input

 3
3
0 0 0
1 0 0
0 0 1
5 2 2
5 3 2
5 2 3
10 22 -2
11 22 -1
11 22 -3
3
0 0 0
1 0 1.5
1 0 -1.5
112 115 109
114 112 110
109 114 111
-110 -121 -130
-115 -129 -140
-104 -114 -119.801961
3
0 0 0
1 0 1.5
1 0 -1.5
112 115 109
114 112 110
109 114 111
-110 -121 -130
-120 -137 -150
-98 -107 -109.603922 

 

Sample Output

 Lucky
2.32
Lucky 

 

   题意是：D博士发明了一个新武器，要对它进行威力测试。可以威力测试的条件是圆形炮筒两端发出的射线构成无限长的圆柱，若有两个圆柱相交后相切，则可以测试，输出“Lucky”否则输出圆柱间的最短距离。

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
struct point
{
    double x,y,z;
    double a,b,c;
    double r;
}node[45];
double solve(point p1,point p2)
  {
      double a1,a2,b1,b2,c1,c2;
      double s1,s2,s3;
      double q1,q2,q3;
      double ans1,ans2,ans3;
      a1=p1.x,b1=p1.y,c1=p1.z;
      a2=p2.x,b2=p2.y,c2=p2.z;
      s1=b1*c2-b2*c1;
      s2=c1*a2-c2*a1;
      s3=a1*b2-a2*b1;
      q1=p2.a-p1.a;
      q2=p2.b-p1.b;
      q3=p2.c-p1.c;
      ans1=fabs(q1*s1+q2*s2+q3*s3);
      ans2=sqrt(s1*s1+s2*s2+s3*s3);
      ans3=ans1/ans2;
    return ans3;
  }

int main()
{
   int t,n,i,j;
   double x1,x2,x3,y1,y2,y3,z1,z2,z3;
   double a1,a2,b1,b2,c1,c2;
  scanf("%d",&t);
  while(t--)
  {
      scanf("%d",&n);
     for(i=0;i<n;i++)
      {
        cin>>x1>>y1>>z1>>x2>>y2>>z2>>x3>>y3>>z3;
         node[i].a=x1;node[i].b=y1;node[i].c=z1,
         a1=x2-x1;b1=y2-y1;c1=z2-z1;
         a2=x3-x1;b2=y3-y1;c2=z3-z1;
         node[i].r=sqrt(a1*a1+b1*b1+c1*c1);
         node[i].x=b1*c2-b2*c1;
         node[i].y=c1*a2-c2*a1;
         node[i].z=a1*b2-a2*b1;
      }
   bool flag=false;
   double ss=1e8;
   double tmp;
  for(i=0;i<n;i++)
  {
      for(j=i+1;j<n;j++)
      {
             tmp=solve(node[i],node[j]);
             tmp=tmp-node[i].r-node[j].r;
                if(tmp<=0)
                {
                    flag=true;
                    break;
                }
                if(tmp<ss)
                  ss=tmp;
      }
    if(flag) break;
  }
   if(flag)
      printf("Lucky\n");
   else
     printf("%.2lf\n",ss);
  }

   return 0;
    }