HDU 4741 Save Labman No.004 2013 ACM/ICPC 杭州网络赛

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=4741

 

题意：给你两条异面直线，然你求着两条直线的最短距离，并求出这条中垂线与两直线的交点。

需要注意的是，不知道为什么用double就WA了，但是改为long double就AC了。

 

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>
using namespace std;

typedef __int64 LL;
const int N=1100;
const int INF=0x3f3f3f3f;
const long double PI=acos(-1.0);
const long double eps=1e-7;

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

struct point3D
{
    long double x,y,z;
    point3D(){};
    point3D(long double a,long double b,long double c):x(a),y(b),z(c){}
    void input()
    {
        double a,b,c;
        scanf("%lf%lf%lf",&a,&b,&c);
        x=a,    y=b,    z=c;
    }
    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);
    }
};

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

long double vlen(point3D a)//向量长度
{
    return sqrt(a.x*a.x+a.y*a.y+a.z*a.z);
}

long double dis(point3D a,point3D b)//两点距离
{
    long double x=a.x-b.x;
    long double y=a.y-b.y;
    long double z=a.z-b.z;
    return sqrt(x*x+y*y+z*z);
}

point3D xmult(point3D u,point3D v)//叉积,法向量
{
    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;
}

long double dmult(point3D u,point3D v)//点积
{
    return u.x*v.x+u.y*v.y+u.z*v.z;
}

point3D get_faline(point3D a,point3D b,point3D c)//平面的法向量
{
    return xmult(b-a,c-a);
}

bool dian_inline(point3D a,point3D b,point3D c)//判断三点共线
{
    return vlen(xmult(b-a,c-a))<eps;
}

bool dian_inmian(point3D a,point3D b,point3D c,point3D d)//四点公面
{
    return zero(dmult(get_faline(a,b,c),d-a));
}

long double xian_xian(line l1,line l2)//直线到直线的距离
{
    point3D n=xmult(l1.u-l1.v,l2.u-l2.v);//法向量
    return fabs(dmult(l1.u-l2.u,n))/vlen(n);
}

point3D a,b,c,d;

long double F1(point3D a,point3D b)
{
    return ((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y)+(b.z-a.z)*(b.z-a.z));
}

long double F2()
{
    return ((b.x-a.x)*(d.x-c.x)+(b.y-a.y)*(d.y-c.y)+(b.z-a.z)*(d.z-c.z));
}

long double F3ab(point3D a,point3D b)
{
    return ((b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y)+(b.z-a.z)*(c.z-a.z));
}

long double F3cd(point3D c,point3D d)
{
    return ((d.x-c.x)*(c.x-a.x)+(d.y-c.y)*(c.y-a.y)+(d.z-c.z)*(c.z-a.z));
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        a.input();  b.input();  c.input();  d.input();
        line l1,l2;
        l1.u=a;   l1.v=b;
        l2.u=c;   l2.v=d;
        printf("%.6lf\n",(double)xian_xian(l1,l2));
        long double x[6];
        long double xh1,xh2;
        xh1=F3ab(a,b)*F1(c,d)-F3cd(c,d)*F2();
        xh2=F1(a,b)*F1(c,d)-F2()*F2();
        x[0]=(b.x-a.x)*xh1/xh2+a.x;
        x[1]=(b.y-a.y)*xh1/xh2+a.y;
        x[2]=(b.z-a.z)*xh1/xh2+a.z;

        long double xx1,xx2,xxx;
        xx1=F3cd(c,d)*F1(a,b)-F3ab(a,b)*F2();
        xx2=F2()*F2()-F1(a,b)*F1(c,d);
        xxx=xx1/xx2;
        x[3]=(d.x-c.x)*xxx+c.x;
        x[4]=(d.y-c.y)*xxx+c.y;
        x[5]=(d.z-c.z)*xxx+c.z;

        for(int i=0;i<5;i++)
            printf("%.6lf ",(double)x[i]);
        printf("%.6lf\n",(double)x[5]);
    }
    return 0;
}