lightoj 1240 计算机图论（计算三维点到线段的距离）

Description

Given a segment in 3D space, identified by A(x₁, y₁, z₁), B(x₂, y₂, z₂) and another point P(x, y, z) your task is to find the minimum possible Euclidean distance between the point P and the segment AB.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing nine integers x₁, y₁, z₁, x₂, y₂, z₂, x, y, z. The magnitude of any integer will not be greater than 100.

Output

For each case, print the case number and the distance. Errors less than 10^-6 will be ignored.

Sample Input

2

0 0 1 0 1 1 0 1 0

0 0 0 1 1 1 0 0 1

Sample Output

Case 1: 1

Case 2: 0.8164965809

   这个题目是一个几何问题，题目大意之灾三维空间中求一个点到一个线段的最短距离，我们可以采用向量点积的方法，首先设线段的端点为p1，p2，线段外的点是p，首先是计算向量p1-->p,  p1-->p2,的点积，如果两个数异号的话就说明p向线段做垂线垂足在线段外，所以最短距离为点到两个端点的最小值，如果两个同号的话就是求点到线段的垂线距离，我们可以将三个点连接起来，通过海伦公式求得三角形的面积，然后*2除底，就可以得到答案，但是还有一个重要的wa点就是要将数据类型全部转化为double进行运算，这样就可以减少int开方转化为double时候精度的损失（表示已经wa到哭）；

代码如下：

#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<cmath>using namespace std;struct Point3D {            //三维点或矢量    double x, y, z;};struct Line3D {             // 三维的直线或线段    Point3D p1, p2;};double cross( Point3D &a,Point3D &b, Point3D &c){    return (a.x-c.x)*(b.x-c.x) + (a.y-c.y)*(b.y-c.y) +(a.z-c.z)*(b.z-c.z);}double dis(Point3D &a,Point3D &b){    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z);}int main(){    //freopen("in.txt","r",stdin);    int T;    Point3D p,p1,p2;    scanf("%d",&T);    for(int i=1;i<=T;i++){  scanf("%lf %lf %lf %lf %lf %lf %lf %lf %lf",&p1.x,&p1.y,&p1.z,&p2.x,&p2.y,&p2.z,&p.x,&p.y,&p.z);  double d1 = dis(p,p1);  double d2 = dis(p,p2);  //cout << d1 << " " << d2 << endl;  double ans;  double c1 = cross(p,p1,p2);  double c2 = cross(p,p2,p1);  if ( c1 * c2 <= 0 )///判断垂足位置          ans = sqrt(double (min(d1,d2)));      else      {          double d3 = dis(p1,p2);          double s1 = sqrt(d1);          double s2 = sqrt(d2);          double s3 = sqrt(d3);          double c = (s1+s2+s3)/2;          double s = sqrt(c*(c-s1)*(c-s2)*(c-s3));///海伦公式          ans = 2*s/s3;      }      printf("Case %d: %.10lf\n",i,ans);}      return 0;}