• HDU 4617 Weapon 三维计算几何


    题意:给你一些无限长的圆柱,知道圆柱轴心直线(根据他给的三个点确定的平面求法向量即可)与半径,判断是否有圆柱相交。如果没有,输出柱面最小距离。

    一共只有30个圆柱,直接暴力一下就行。

    判相交/相切:空间直线距离是否小于等于两圆柱半径和
    若无相交,求最小:空间直线距离-两圆柱半径和

      1 #include <cstdio>
      2 #include <cstdlib>
      3 #include <cstring>
      4 #include <algorithm>
      5 #include <cmath>
      6 
      7 using namespace std;
      8 
      9 const double EPS = 1e-9;
     10 const int MAXN = 40;
     11 
     12 struct Point3  //空间点
     13 {
     14     double x, y, z;
     15 };
     16 
     17 struct Line3   //空间直线
     18 {
     19     Point3 a, b;
     20 };
     21 
     22 struct plane3   //空间平面
     23 {
     24     Point3 a, b, c;
     25     plane3(){}
     26     plane3( Point3 a, Point3 b, Point3 c ):
     27     a(a), b(b), c(c) { }
     28 };
     29 
     30 struct Cylinder
     31 {
     32     Point3 center;   //圆柱轴线经过的一点
     33     Point3 a, b;     //与中心点在同一平面的两点
     34     Line3 FaXian;    //轴线
     35     double R;        //圆柱半径
     36 };
     37 
     38 Cylinder CC[MAXN];
     39 
     40 Point3 Read_Point()
     41 {
     42     Point3 p;
     43     scanf("%lf%lf%lf", &p.x, &p.y, &p.z );
     44     return p;
     45 }
     46 
     47 double dcmp( double a )
     48 {
     49     if ( fabs( a ) < EPS ) return 0;
     50     return a < 0 ? -1 : 1;
     51 }
     52 
     53 //三维叉积
     54 Point3 Cross3( Point3 u, Point3 v )
     55 {
     56     Point3 ret;
     57     ret.x = u.y * v.z - v.y * u.z;
     58     ret.y = u.z * v.x - u.x * v.z;
     59     ret.z = u.x * v.y - u.y * v.x;
     60     return ret;
     61 }
     62 
     63 //三维点积
     64 double Dot3( Point3 u, Point3 v )
     65 {
     66     return u.x * v.x + u.y * v.y + u.z * v.z;
     67 }
     68 
     69 //矢量差
     70 Point3 Subt( Point3 u, Point3 v )
     71 {
     72     Point3 ret;
     73     ret.x = u.x - v.x;
     74     ret.y = u.y - v.y;
     75     ret.z = u.z - v.z;
     76     return ret;
     77 }
     78 
     79 //取平面法向量
     80 Point3 NormalVector( plane3 s )
     81 {
     82     return Cross3( Subt( s.a, s.b ), Subt( s.b, s.c ) );
     83 }
     84 Point3 NormalVector( Point3 a, Point3 b, Point3 c )
     85 {
     86     return Cross3( Subt( a, b ), Subt( b, c ) );
     87 }
     88 
     89 //两点距离
     90 double TwoPointDistance( Point3 p1, Point3 p2 )
     91 {
     92     return sqrt( (p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y) + (p1.z - p2.z)*(p1.z - p2.z) );
     93 }
     94 
     95 //向量的模
     96 double VectorLenth( Point3 p )
     97 {
     98     return sqrt( p.x*p.x + p.y*p.y + p.z*p.z );
     99 }
    100 
    101 //空间直线距离
    102 double LineToLine( Line3 u, Line3 v )
    103 {
    104     Point3 tmp = Cross3( Subt( u.a, u.b ), Subt( v.a, v.b ) );
    105     return fabs( Dot3( Subt(u.a, v.a), tmp ) ) / VectorLenth(tmp);
    106 }
    107 
    108 int N;
    109 
    110 int main()
    111 {
    112     int T;
    113     scanf( "%d", &T );
    114     while ( T-- )
    115     {
    116         scanf( "%d", &N );
    117         for ( int i = 0; i < N; ++i )
    118         {
    119             CC[i].center = Read_Point();
    120             CC[i].a = Read_Point();
    121             CC[i].b = Read_Point();
    122             CC[i].R = VectorLenth( Subt( CC[i].a, CC[i].center ) );
    123             CC[i].FaXian.a = CC[i].center;
    124             CC[i].FaXian.b = Subt( CC[i].center, NormalVector( CC[i].center, CC[i].a, CC[i].b ) );
    125         }
    126 
    127         bool ok = true;
    128         double MinAns = 1e200;
    129         for ( int i = 0; ok && i < N; ++i )
    130         for ( int j = i + 1; ok && j < N; ++j )
    131         {
    132             double tmp = LineToLine( CC[i].FaXian, CC[j].FaXian );
    133             if ( dcmp( tmp - ( CC[i].R + CC[j].R ) ) <= 0 )
    134             {
    135                 ok = false;
    136                 break;
    137             }
    138             MinAns = min( MinAns, tmp - ( CC[i].R + CC[j].R ) );
    139         }
    140 
    141         if ( ok ) printf( "%.2f
    ", MinAns );
    142         else puts("Lucky");
    143     }
    144     return 0;
    145 }
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  • 原文地址:https://www.cnblogs.com/GBRgbr/p/3217875.html
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