HDU 4617 Weapon (三维计算几何)

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

题意：说白了就是求空间上两跳直线的距离 - 两个半径的最小值。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <list>
#include <deque>
#include <queue>
#include <iterator>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <cctype>
#include <ctime>
#pragma comment(linker, "/STACK:16777216")
using namespace std;

typedef __int64 LL;
const int N=50005;
const int M=55555555;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const double eps=1e-7;

bool zero(double x)
{
    if(fabs(x)<eps)
        return true;
    return false;
}

struct point3D
{
    double x,y,z;
    point3D(){};
    point3D(double a,double b,double c):x(a),y(b),z(c){}
    void input()
    {
        scanf("%lf%lf%lf",&x,&y,&z);
    }
    friend point3D operator -(const point3D &a,const point3D &b)
    {
        return point3D(a.x-b.x,a.y-b.y,a.z-b.z);
    }
    friend point3D operator +(const point3D &a,const point3D &b)
    {
        return point3D(a.x+b.x,a.y+b.y,a.z+b.z);
    }
}p[33][4];
double r[33];

struct line
{
    double a,b,c,d;
    point3D u,v;
}l[33];

double vlen(point3D a)
{
    return sqrt(a.x*a.x+a.y*a.y+a.z*a.z);
}

point3D xmult(point3D u,point3D v)
{
	point3D 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(point3D u,point3D v)
{
    return u.x*v.x+u.y*v.y+u.z*v.z;
}

point3D get_faline(point3D a,point3D b,point3D c)
{
    return xmult(b-a,c-a);
}


double xian_xian(line l1,line l2)
{
    point3D n=xmult(l1.u-l1.v,l2.u-l2.v);
	return fabs(dmult(l1.u-l2.u,n))/vlen(n);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=3;j++)
                p[i][j].input();
            r[i]=vlen(p[i][1]-p[i][2]);
        }
        for(int i=1;i<=n;i++)
        {
            l[i].u=p[i][1];
            l[i].v=p[i][1]+get_faline(p[i][1],p[i][2],p[i][3]);
        }
        double Min=INF;
        int flag=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                double tmp=xian_xian(l[i],l[j]);
                tmp-=(r[i]+r[j]);
                if(tmp<=0)
                {
                    flag=1;
                    i=j=300;
                    break;
                }
                if(tmp<Min)
                    Min=tmp;
            }
        }
        if(flag)
            printf("Lucky\n");
        else
            printf("%.2f\n",Min);
    }
    return 0;
}