POJ 1444 计算几何趣题

求长方体上两点表面距离。

用到了比较神奇的递归，很有趣。学习之。

将1点旋转至XOY平面，再用递归四面展开，限制步数，得到最近的距离。

/*********************
 * Creater:Sevenster *
 * Time:2012.07.31 13:28  *
 * PID:POJ 1444      *
 *********************/
#include<iostream>
#include<cmath>
using namespace std;

double length2( double x1, double y1, double x2, double y2 )
{
    return (x2-x1)*(x2-x1)+(y2-y1)*(y2-y1);
}

double ans;
void getAns( int i, int j, double x, double y, double x2, double y2, double z2, double l, double w, double h )
{
    if( z2==0 && ans>length2( x, y, x2, y2 ) )
        ans=length2( x, y, x2, y2 );
    //else
    {
        if( i>= 0 && i< 2 )
            getAns( i+ 1, j, x+ h, y, h- z2, y2, x2, h, w, l);
        if( i<= 0 && i> -2 )
            getAns( i- 1, j, x- l, y, z2, y2, l- x2, h, w, l);
        if( j>= 0 && j< 2 )
            getAns( i, j+ 1, x, y- w, x2, z2, w- y2, l, h, w);
        if( j<= 0 && j> -2 )
            getAns( i, j- 1, x, y+ h, x2, h- z2, y2, l, h, w);
    }
}

int main()
{
    double l, w, h;
    double INF = 1e20;
    double x1, y1, z1;
    double x2, y2, z2;

    while( scanf( "%lf%lf%lf",&l, &w, &h )!= EOF )
    {
        scanf( "%lf%lf%lf", &x1, &y1, &z1 );
        scanf( "%lf%lf%lf", &x2, &y2, &z2 );
        if( z1!=0 && z1!=h )
        {
            if( x1!=0 && x1!=l )
            {
                swap( y1,z1 );
                swap( y2,z2 );
                swap( w,h );
            }
            else
            {
                swap( x1,z1 );
                swap( x2,z2 );
                swap( l,h );
            }
        }
        if( z1==h )
        {
            z1=0;
            z2=h-z2;
        }
        ans=INF;
        getAns(0, 0, x1, y1, x2, y2, z2, l, w, h );
        printf("%.0lf\n",ans);
    }
    return 0;
}