http://acm.hdu.edu.cn/showproblem.php?pid=4617

三维几何简单题 多谢高尚博学长留下的模板

代码:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

typedef long long ll;
typedef pair<double,double>ppd;
const double PI = acos(-1.);
const int MAXL = (1 << 18)+100;
const double eps = (1e-9);
const int N=50;
struct point3{double x,y,z;};
struct line3{point3 a,b;};
double r[N];
line3 L[N];
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0?-1:1;
}
point3 xmult(point3 u,point3 v)
{
    point3 ret;
    ret.x=u.y*v.z-v.y*u.z;
    ret.y=u.z*v.x-u.x*v.z;
    ret.z=u.x*v.y-u.y*v.x;
    return ret;
}
double dmult(point3 u,point3 v)
{
    return u.x*v.x+u.y*v.y+u.z*v.z;
}
point3 subt(point3 u,point3 v)
{
    point3 ret;
    ret.x=u.x-v.x;
    ret.y=u.y-v.y;
    ret.z=u.z-v.z;
    return ret;
}
double vlen(point3 p)
{
    return sqrt(p.x*p.x+p.y*p.y+p.z*p.z);
}
double dist(point3 u,point3 v)
{
    return (double)(sqrt((u.x-v.x)*(u.x-v.x)+(u.y-v.y)*(u.y-v.y)+(u.z-v.z)*(u.z-v.z)));
}
double ptoline(point3 p,line3 l)
{
    l.b.x=l.a.x+l.b.x;
    l.b.y=l.a.y+l.b.y;
    l.b.z=l.a.z+l.b.z;
    return vlen(xmult(subt(p,l.a),subt(l.b,l.a)))/dist(l.a,l.b);
}
double linetoline(line3 u,line3 v)
{
    point3 n=xmult(u.b,v.b);
    if(dcmp(vlen(n))==0)
    return ptoline(u.a,v);
    return fabs(dmult(subt(v.a,u.a),n))/vlen(n);
}
point3 pvec(point3 s1,point3 s2,point3 s3)
{
    return xmult(subt(s1,s2),subt(s2,s3));
}

int main()
{
    //freopen("data.in","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;++i)
        {
            point3 p[3];
            for(int j=0;j<3;++j)
            scanf("%lf %lf %lf",&p[j].x,&p[j].y,&p[j].z);
            L[i].a=p[0];
            L[i].b=pvec(p[0],p[1],p[2]);
            r[i]=dist(p[0],p[1]);
        }
        double k=1e9;
        for(int i=0;i<n;++i)
        for(int j=i+1;j<n;++j)
        {
            //cout<<linetoline(L[i],L[j])<<endl;
            k=min(k,linetoline(L[i],L[j])-r[i]-r[j]);
        }
        if(dcmp(k)<=0)
        printf("Lucky\n");
        else
        printf("%.2lf\n",k);

    }
    return 0;
}

 

posted on 2013-07-29 11:54  夜->  阅读(154)  评论(0编辑  收藏  举报