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;
}