hdu4617 Weapon(立体几何题)
WeaponTime Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 279 Accepted Submission(s): 220
Problem 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 su易做图ce 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 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 su易做图ce 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
Source
2013 Multi-University Training Contest 2
没啥技巧吧，就是直接算
思路：对于每一个圆柱，算出它的中心轴方向向量同时保存圆心坐标，算出两个直线之间的距离d，如果距离大于两个圆柱的半径之和那就记录mind=d-rr[i]-rr[j],否则就输出Lucky。两直线平行时距离好算，不平行时有点麻烦，用向量的叉乘来计算。
#include<stdio.h>
#include<math.h>
double r[31][3],center[31][3],rr[31];
int main()
{
double x1,x2,x3,y1,y2,y3,z1,z2,z3,r1,r2,r3,mind,temp,temp1;
double a1,b1,c1,a2,b2,c2;
int t,n,m,i,j,k,flag;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%lf%lf%lf",&x1,&y1,&z1);
scanf("%lf%lf%lf",&x2,&y2,&z2);
scanf("%lf%lf%lf",&x3,&y3,&z3);
center[i][0]=x1;
center[i][1]=y1;
center[i][2]=z1;
rr[i]=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2));
a1=x2-x1;
b1=y2-y1;
c1=z2-z1;
a2=x3-x1;
b2=y3-y1;
c2=z3-z1;
r1=b1*c2-b2*c1;
r2=a2*c1-a1*c2;
r3=a1*b2-a2*b1;
r[i][0]=r1;
r[i][1]=r2;
r[i][2]=r3;
//printf("%.4f %.4f %.4f %.4f\n",rr[i],r1,r2,r3);
}
mind=1000000000;
flag=1;
for(i=0;i<n-1&&flag;i++)
{
for(j=i+1;j<n;j++)
{
a1=r[i][0];
b1=r[i][1];
c1=r[i][2];
a2=r[j][0];
b2=r[j][1];
c2=r[j][2];
r1=b1*c2-b2*c1;//垂直
r2=a2*c1-a1*c2;
r3=a1*b2-a2*b1;
//printf("%d %d %.4f %.4f %.4f\n",i,j,r1,r2,r3);
if(!(r1==0&&r2==0&&r3==0))//两直线不平行
{
temp=fabs(((center[i][0]-center[j][0])*r1+(center[i][1]-center[j][1])*r2+(center[i][2]-center[j][2])*r3)/sqrt(r1*r1+r2*r2+r3*r3));//利用
if(temp<=rr[i]+rr[j])
{
printf("Lucky\n");
flag=0;
break;
}
else if(mind>temp-(rr[i]+rr[j]))
mind=temp-rr[i]-rr[j];
//printf("%.2f\n",temp);
}
else //平行
{
a2=center[i][0]-center[j][0],b2=center[i][1]-center[j][1],c2=center[i][2]-center[j][2];
temp1=sqrt(((b1*c2-b2*c1)*(b1*c2-b2*c1)+(a2*c1-a1*c2)*(a2*c1-a1*c2)+(a1*b2-b1*a2)*(a1*b2-b1*a2))/(a1*a1+b1*b1+c1*c1));
if(temp1<=rr[i]+rr[j])
{
printf("Lucky\n");
flag=0;
break;
}
else if(mind>temp1-rr[i]-rr[j])
mind=temp1-rr[i]-rr[j];
}
}
}
if(flag)
printf("%.2f\n",mind);
}
return 0;
}
补充：软件开发 , C++ ,
