HDU 4617

复习了一下大一的东西 都忘记了

题意是  n 个无限长的圆柱

给你 n  然后 圆心   圆上2点    n个

求 能不能相交  相交 lucky 

否则 输出最短距离

先求  2 平面 法向量  求出 这2个法向量 的法向量 n  然后取直线上2点 ab    d  =  fabs(n*ab)/|n|;

难受 

#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<queue>

using namespace std;

#define MAXN 510
#define inf  1000000000
struct point
{
    double x,y,z;
}z[MAXN][3],fa[MAXN];
double r[MAXN];
point de(point a,point b)
{
    point c;
    c.x=a.x-b.x;
    c.y=a.y-b.y;
    c.z=a.z-b.z;
    return c;
}
point chaji(point a,point b)
{
    point c;
    c.x=a.y*b.z-a.z*b.y;
    c.y=-(a.x*b.z-a.z*b.x);
    c.z=a.x*b.y-a.y*b.x;
    return c;
}
double dis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));
}
double dianji(point a,point b)
{
    return a.x*b.x+a.y*b.y+a.z*b.z;
}
double len(point a)
{
    return sqrt(a.x*a.x+a.y*a.y+a.z*a.z);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<3;j++)
                scanf("%lf%lf%lf",&z[i][j].x,&z[i][j].y,&z[i][j].z);
        }

        for(int i=1;i<=n;i++)
        {
            point a,b;
            a=de(z[i][1],z[i][0]);
            b=de(z[i][2],z[i][0]);
            point c;
            fa[i]=chaji(a,b);
            r[i]=dis(z[i][1],z[i][0]);
        }
        double ans =inf;
        int ok=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i==j)
                    continue;
                point a =chaji(fa[i],fa[j]);
                point b =de(z[i][0],z[j][0]);

                double d = fabs(dianji(a,b))/len(a);
                if(d<=r[i]+r[j])
                {
                    ok=1;
                }
                ans = min(ans,d-r[i]-r[j]);
            }
        }

        if(ok==1)
            printf("Lucky\n");
        else
            printf("%.2lf\n",ans);

    }
    return 0;
}