hdu 4741 Save Labman No.004（2013杭州网络赛）

最新推荐文章于 2021-07-28 18:48:22 发布

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最新推荐文章于 2021-07-28 18:48:22 发布

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原文链接：http://www.cnblogs.com/pangblog/p/3325143.html

本文介绍了一种计算空间中两条直线之间最近点对的方法，并提供了完整的C++代码实现。该方法通过向量运算找到最短距离及其对应的点。

http://blog.sina.com.cn/s/blog_a401a1ea0101ij9z.html

空间两直线上最近点对。

这个博客上给出了很好的点法式公式了。。。其实没有那么多的tricky。。。不知到别人怎么错的。。。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<algorithm>
#include<iostream>
#include<cstring>
#include<fstream>
#include<sstream>
#include<vector>
#include<string>
#include<cstdio>
#include<bitset>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define debug puts("**debug**")
#define LL long long
#define PB push_back
#define MP make_pair
#define eps 1e-10
using namespace std;

struct Point
{
    double x, y, z;
    Point(double x=0, double y=0, double z=0) : x(x), y(y),z(z){}
};
typedef Point Vector;

Vector operator + (Vector a, Vector b) { return Vector(a.x+b.x, a.y+b.y, a.z+b.z); };
Vector operator - (Vector a, Vector b) { return Vector(a.x-b.x, a.y-b.y, a.z-b.z); };
Vector operator * (Vector a, double p) { return Vector(a.x*p, a.y*p, a.z*p); }
Vector operator / (Vector a, double p) { return Vector(a.x/p, a.y/p, a.z/p); }

double Dot(Vector a, Vector b) { return a.x*b.x + a.y*b.y + a.z*b.z; }
double Length(Vector a) { return sqrt(Dot(a, a)); }
Vector Cross(Point a, Point b)
{
    return Vector(a.y*b.z-a.z*b.y, a.z*b.x-a.x*b.z, a.x*b.y-a.y*b.x);
}

Point a1, b1, a2, b2;
int main()
{
    int n;
    scanf("%d", &n);
    while(n--)
    {
        scanf("%lf%lf%lf", &a1.x, &a1.y, &a1.z);
        scanf("%lf%lf%lf", &b1.x, &b1.y, &b1.z);
        scanf("%lf%lf%lf", &a2.x, &a2.y, &a2.z);
        scanf("%lf%lf%lf", &b2.x, &b2.y, &b2.z);
        Vector v1 = (a1-b1), v2 = (a2-b2);
        Vector N = Cross(v1, v2);
        Vector ab = (a1-a2);
        double ans = Dot(N, ab) / Length(N);
        Point p1 = a1, p2 = a2;
        Vector d1 = b1-a1, d2 = b2-a2;
        Point ans1, ans2;
        double t1, t2;
        t1 = Dot((Cross(p2-p1, d2)), Cross(d1, d2));
        t2 = Dot((Cross(p2-p1, d1)), Cross(d1, d2));
        double dd = Length((Cross(d1, d2)));
        t1 /= dd*dd;
        t2 /= dd*dd;
        ans1 = (a1 + (b1-a1)*t1);
        ans2 = (a2 + (b2-a2)*t2);
        printf("%.6f\n", fabs(ans));
        printf("%.6f %.6f %.6f ", ans1.x, ans1.y, ans1.z);
        printf("%.6f %.6f %.6f\n", ans2.x, ans2.y, ans2.z);
    }
    return 0;
}

转载于:https://www.cnblogs.com/pangblog/p/3325143.html