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  • LA 4973异面线段

    题目大意:给两条线段求他们间的最小距离的平方(以分数形式输出)。

    贴个模版吧!太抽象了。

    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    
    struct Point3 {
      int x, y, z;
      Point3(int x=0, int y=0, int z=0):x(x),y(y),z(z) { }
    };
    
    typedef Point3 Vector3;
    
    Vector3 operator + (const Vector3& A, const Vector3& B) { return Vector3(A.x+B.x, A.y+B.y, A.z+B.z); }
    Vector3 operator - (const Point3& A, const Point3& B) { return Vector3(A.x-B.x, A.y-B.y, A.z-B.z); }
    Vector3 operator * (const Vector3& A, int p) { return Vector3(A.x*p, A.y*p, A.z*p); }
    
    bool operator == (const Point3& a, const Point3& b) {
      return a.x==b.x && a.y==b.y && a.z==b.z;
    }
    
    Point3 read_point3() {
      Point3 p;
      scanf("%d%d%d", &p.x, &p.y, &p.z);
      return p;
    }
    
    int Dot(const Vector3& A, const Vector3& B) { return A.x*B.x + A.y*B.y + A.z*B.z; }
    int Length2(const Vector3& A) { return Dot(A, A); }
    Vector3 Cross(const Vector3& A, const Vector3& B) { return Vector3(A.y*B.z - A.z*B.y, A.z*B.x - A.x*B.z, A.x*B.y - A.y*B.x); }
    
    typedef long long LL;
    
    LL gcd(LL a, LL b) { return b ? gcd(b, a%b) : a; }
    LL lcm(LL a, LL b) { return a / gcd(a,b) * b; }
    
    struct Rat {
      LL a, b;
      Rat(LL a=0):a(a),b(1) { }
      Rat(LL x, LL y):a(x),b(y) {
        if(b < 0) a = -a, b = -b;
        LL d = gcd(a, b); if(d < 0) d = -d;
        a /= d; b /= d;
      }
    };
    
    Rat operator + (const Rat& A, const Rat& B) {
      LL x = lcm(A.b, B.b);
      return Rat(A.a*(x/A.b)+B.a*(x/B.b), x);
    }
    
    Rat operator - (const Rat& A, const Rat& B) { return A + Rat(-B.a, B.b); }
    Rat operator * (const Rat& A, const Rat& B) { return Rat(A.a*B.a, A.b*B.b); }
    
    void updatemin(Rat& A, const Rat& B) {
      if(A.a*B.b > B.a*A.b) A.a = B.a, A.b = B.b;
    }
    
    // 点P到线段AB的距离的平方
    Rat Rat_Distance2ToSegment(const Point3& P, const Point3& A, const Point3& B) {
      if(A == B) return Length2(P-A);
      Vector3 v1 = B - A, v2 = P - A, v3 = P - B;
      if(Dot(v1, v2) < 0) return Length2(v2);
      else if(Dot(v1, v3) > 0) return Length2(v3);
      else return Rat(Length2(Cross(v1, v2)), Length2(v1));
    }
    
    // 求异面直线p1+su和p2+tv的公垂线对应的s。如果平行/重合,返回false
    bool Rat_LineDistance3D(const Point3& p1, const Vector3& u, const Point3& p2, const Vector3& v, Rat& s) {
      LL b = (LL)Dot(u,u)*Dot(v,v) - (LL)Dot(u,v)*Dot(u,v);
      if(b == 0) return false;
      LL a = (LL)Dot(u,v)*Dot(v,p1-p2) - (LL)Dot(v,v)*Dot(u,p1-p2);
      s = Rat(a, b);
      return true;
    }
    
    void Rat_GetPointOnLine(const Point3& A, const Point3& B, const Rat& t, Rat& x, Rat& y, Rat& z) {
      x = Rat(A.x) + Rat(B.x-A.x) * t;
      y = Rat(A.y) + Rat(B.y-A.y) * t;
      z = Rat(A.z) + Rat(B.z-A.z) * t;
    }
    
    Rat Rat_Distance2(const Rat& x1, const Rat& y1, const Rat& z1, const Rat& x2, const Rat& y2, const Rat& z2) {
      return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2);
    }
    
    int main() {
      int T;
      scanf("%d", &T);
      LL maxx = 0;
      while(T--) {
        Point3 A = read_point3();
        Point3 B = read_point3();
        Point3 C = read_point3();
        Point3 D = read_point3();
        Rat s, t;
        bool ok = false;
        Rat ans = Rat(1000000000);
        if(Rat_LineDistance3D(A, B-A, C, D-C, s))
          if(s.a > 0 && s.a < s.b && Rat_LineDistance3D(C, D-C, A, B-A, t))
            if(t.a > 0 && t.a < t.b) {
              ok = true; // 异面直线/相交直线
              Rat x1, y1, z1, x2, y2, z2;
              Rat_GetPointOnLine(A, B, s, x1, y1, z1);
              Rat_GetPointOnLine(C, D, t, x2, y2, z2);
              ans = Rat_Distance2(x1, y1, z1, x2, y2, z2);
        }
        if(!ok) { // 平行直线/重合直线
          updatemin(ans, Rat_Distance2ToSegment(A, C, D));
          updatemin(ans, Rat_Distance2ToSegment(B, C, D));
          updatemin(ans, Rat_Distance2ToSegment(C, A, B));
          updatemin(ans, Rat_Distance2ToSegment(D, A, B));
        }
        printf("%lld %lld
    ", ans.a, ans.b);
      }
      return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/xiong-/p/3440395.html
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