HDU 4617 Weapon（三维几何）

Problem Description

　　Doctor D. are researching for a horrific weapon. The muzzle of the weapon is a circle. When it fires, rays form a cylinder that runs through the circle verticality in both side. If one cylinder of rays touch another, there will be an horrific explosion. Originally, all circles can rotate easily. But for some unknown reasons they can not rotate any more. If these weapon can also make an explosion, then Doctor D. is lucky that he can also test the power of the weapon. If not, he would try to make an explosion by other means. One way is to find a medium to connect two cylinder. But he need to know the minimum length of medium he will prepare. When the medium connect the surface of the two cylinder, it may make an explosion.

Input

　　The first line contains an integer T, indicating the number of testcases. For each testcase, the first line contains one integer N(1 < N < 30), the number of weapons. Each of the next 3N lines  contains three float numbers. Every 3 lines represent one weapon. The first line represents the coordinates of center of the circle, and the second line and the third line represent two points in the circle which surrounds the center. It is supposed that these three points are not in one straight line. All float numbers are between -1000000 to 1000000.

Output

　　For each testcase, if there are two cylinder can touch each other, then output 'Lucky', otherwise output then minimum distance of any two cylinders, rounded to two decimals, where distance of two cylinders is the minimum distance of any two point in the surface of two cylinders.

题目大意：给多个圆柱，若有任意两个圆柱相交，则输出Lucky，否则输出两个圆柱间的最短距离。

思路：已经算是模板题了，不多解释。

代码（0MS）：

  1 #include <cstdio>
  2 #include <algorithm>
  3 #include <cstring>
  4 #include <iostream>
  5 #include <cmath>
  6 using namespace std;
  7 typedef long long LL;
  8 
  9 const double EPS = 1e-8;
 10 const double INF = 1e50;
 11 const double PI = acos(-1.0);
 12 
 13 inline int sgn(double x) {
 14     return (x > EPS) - (x < EPS);
 15 }
 16 
 17 struct Point3D {
 18     double x, y, z;
 19     Point3D() {}
 20     Point3D(double x, double y, double z): x(x), y(y), z(z) {}
 21     void read() {
 22         scanf("%lf%lf%lf", &x, &y, &z);
 23     }
 24     double operator * (const Point3D &rhs) const {
 25         return x * rhs.x + y * rhs.y + z * rhs.z;
 26     }
 27     Point3D operator + (const Point3D &rhs) const {
 28         return Point3D(x + rhs.x, y + rhs.y, z + rhs.z);
 29     }
 30     Point3D operator - (const Point3D &rhs) const {
 31         return Point3D(x - rhs.x, y - rhs.y, z - rhs.z);
 32     }
 33     double length() const {
 34         return sqrt(x * x + y * y + z * z);
 35     }
 36 };
 37 
 38 struct Line3D {
 39     Point3D st, ed;
 40     Line3D() {}
 41     Line3D(Point3D st, Point3D ed): st(st), ed(ed) {}
 42 };
 43 
 44 struct Plane3D {
 45     Point3D a, b, c;
 46     Plane3D() {}
 47     Plane3D(Point3D a, Point3D b, Point3D c): a(a), b(b), c(c) {}
 48     void read() {
 49         a.read(), b.read(), c.read();
 50     }
 51 };
 52 
 53 double dist(const Point3D &a, const Point3D &b) {
 54     return (a - b).length();
 55 }
 56 //叉积
 57 Point3D cross(const Point3D &u, const Point3D &v) {
 58     Point3D ret;
 59     ret.x = u.y * v.z - u.z * v.y;
 60     ret.y = u.z * v.x - u.x * v.z;
 61     ret.z = u.x * v.y - u.y * v.x;
 62     return ret;
 63 }
 64 //点到直线距离
 65 double point_to_line(const Point3D &p, const Line3D &l) {
 66     return cross(p - l.st, l.ed - l.st).length() / dist(l.ed, l.st);
 67 }
 68 //求两直线间的距离
 69 double line_to_line(const Line3D u, const Line3D v) {
 70     Point3D n = cross(u.ed - u.st, v.ed - v.st);
 71     return fabs((u.st - v.st) * n) / n.length();
 72 }
 73 //取平面法向量
 74 Point3D vector_of_plane(const Plane3D &s) {
 75     return cross(s.a - s.b, s.b - s.c);
 76 }
 77 //判断两直线是否平行
 78 bool isParallel(const Line3D &u, const Line3D &v) {
 79     return sgn(cross(u.ed - u.st, v.ed - v.st).length()) <= 0;
 80 }
 81 
 82 const int MAXN = 35;
 83 
 84 Plane3D s[MAXN];
 85 Line3D l[MAXN];
 86 double r[MAXN];
 87 int T, n;
 88 
 89 int main() {
 90     scanf("%d", &T);
 91     while(T--) {
 92         scanf("%d", &n);
 93         for(int i = 0; i < n; ++i) s[i].read();
 94         for(int i = 0; i < n; ++i) {
 95             Point3D v = vector_of_plane(s[i]);
 96             l[i] = Line3D(s[i].a, s[i].a + v);
 97             r[i] = dist(s[i].a, s[i].b);
 98         }
 99         double ans = INF;
100         for(int i = 0; i < n; ++i) {
101             for(int j = i + 1; j < n; ++j) {
102                 double d;
103                 if(isParallel(l[i], l[j])) d = point_to_line(l[i].st, l[j]);
104                 else d = line_to_line(l[i], l[j]);
105                 ans = min(ans, d - r[i] - r[j]);
106             }
107         }
108         if(sgn(ans) <= 0) puts("Lucky");
109         else printf("%.2f
", ans);
110     }
111 }

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