UESTC 1854

题目意思  就是说 有一个起点一个终点  和一些虫洞，从一个虫洞进去然后出来，是不需要消耗时间的，注意点就是，虫洞是一条线段，你可以从线段的任意位置进去，从任意位置出来； 所以从一个虫洞到另一个虫洞的距离变成了空间的直线距离；

线段到线段的最短距离  用三分的方法

                        #include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<cmath>
#define eps 1e-10
#define inf 0x1f1f1f1f
using namespace std;

int dcmp( double a ){ if( abs(a) < eps ) return 0; if(a > 0) return 1;return -1;}
struct point{
   double x,y,z;
   point(){}
   point( double x,double y,double z ):x(x),y(y),z(z){}
};
typedef point Vector ;
point operator + ( point a,point b ){
    return point( a.x+b.x , a.y+b.y , a.z+b.z );
}
point operator - ( point a,point b ){
    return point( a.x-b.x , a.y-b.y, a.z-b.z );
}
point operator * ( point a,double p ){
    return point( a.x*p , a.y*p , a.z*p );
}
point operator / ( point a,double p ){
    return point( a.x/p , a.y/p , a.z/p );
}
bool operator == ( point a,point b ){
    if( dcmp(a.x-b.x)  == 0&& dcmp(a.y-b.y) == 0 && dcmp(a.z-b.z) == 0 )return 1;
    return 0;
}
double Dot( point a,point b ){ return a.x*b.x + a.y*b.y + a.z*b.z; }
double Length( point a ){ return sqrt(Dot(a,a));}
double Angle(  point a,point b ){ return acos(Dot(a,b))/Length(a)/Length(b); }
//  点到  平面p0 - n 的距离；n 必须为单位向量 n 是平面法向量
double p_po_n( const point &p,const point &po,const point n ){
    return abs(Dot( p-po,n ));
}
//  点在 平面p0 - n 的投影 ； n 必须为单位向量 n 是平面法向量
point p_po_n_jec( const point &p,const point &p0,const point &n ){
    return p-n*Dot( p-p0,n );
}
//  直线p1 p2 在平面 p0 - n 的交点，假设交点存在；唯一
point p_p_p( point p1,point p2,point p0,point n ){
    point  v = p2 - p1;
    double t = ( Dot(n,p0-p1)/Dot(n,p2-p1) );
    return p1 + v*t;
}
point cross( point a,point b ){
    return point( a.y*b.z - a.z*b.y, a.z*b.x - a.x*b.z,a.x*b.y - a.y*b.x );
}
double area( point a,point b,point c ){ return Length(cross(b-a,c-a)); }
double dis_to_line( point p, point a,point b ){
    point v1 = b-a, v2 = p-a;
    return Length( cross(v1,v2)/Length(v1) );
}
// p 到 ab 线段的距离
double dis_to_segm( point p,point a,point b ){
    if( a == b )return Length(p-a);
    point v1 = b-a, v2 = p-a, v3 = p-b;
        if( dcmp( Dot(v1,v2) ) < 0 ) return Length(v2);
   else if( dcmp( Dot(v1,v3) ) > 0 ) return Length(v3);
   else return Length(cross(v1,v2))/Length(v1);
}
// 四面体的体积
double volume( point a,point b,point c,point d ){
    return Dot( d-a,cross(b-a,c-a) );
}
// 线段到线段的距离
double dis_l_to_l( point a,point b,point lt,point rt )
{
    while( abs(rt.x - lt.x) > eps || abs(rt.y - lt.y) > eps || abs( rt.z - lt.z) > eps )
    {
        point temp; temp = lt + ( rt - lt )/3.0;
        point now;   now = lt + ( rt - lt )/3.0*2.0;
        if( dis_to_segm(temp,a,b) < dis_to_segm(now,a,b) )
             rt = now;
        else lt = temp;
    }
    return dis_to_segm( rt,a,b );
}
point A[100],B[100];
double map[100][100];
double dis[100]; int que[200];bool vis[100];
void spfa( int N )
{
    int tail,hed; tail = hed = 0;
    for( int i = 0; i < 100; i++ )dis[i] = (1<<30);
    memset( vis,0,sizeof(vis) );
    dis[0] = 0; que[tail++] = 0; vis[0] = true;
    while( tail > hed )
    {
        int temp = que[hed++]; vis[temp] = false;
        for( int i = 0; i <= N; i++ )
        if(  dis[temp] + map[temp][i] < dis[i] )
        {
             dis[i] = dis[temp] + map[temp][i];
            if( !vis[i] )
            {
                que[tail++] = i;
                vis[i] = true;
            }
        }
    }
}
int main( )
{
   int T,N; scanf("%d",&T);
   while( T-- )
   {
        scanf("%d",&N);
        scanf("%lf%lf%lf",&A[0].x,&A[0].y,&A[0].z);
        scanf("%lf%lf%lf",&A[N+1].x,&A[N+1].y,&A[N+1].z);
        for( int i = 1; i <= N; i++ )
        {
            scanf("%lf%lf%lf",&A[i].x,&A[i].y,&A[i].z);
            scanf("%lf%lf%lf",&B[i].x,&B[i].y,&B[i].z);
        }
        for( int i = 0; i <= N+1; i++ )
        for( int j = 0; j <= N+1; j++ )
        map[i][j] = (1<<30);
        map[0][N+1] = map[N+1][0] = Length( A[0]-A[N+1] );
        for( int i = 1; i <= N; i++ ){
            map[i][0] = map[0][i] = dis_to_segm( A[0],A[i],B[i] );
            map[i][N+1] = map[N+1][i] = dis_to_segm( A[N+1],A[i],B[i] );
        }
        for( int i = 1; i <= N; i++ )
        for( int j = i+1; j <= N; j++ ){
            map[i][j] = dis_l_to_l( A[i],B[i],A[j],B[j] );
            map[j][i] = map[i][j];
        }
        spfa( N + 1 );
        printf("%.15lf
",dis[N+1]);
   }
   return 0;
}