1240 - Point Segment Distance (3D)
PDF (English) | Statistics | Forum |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
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 |
Output for Sample Input |
2 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 |
Case 1: 1 Case 2: 0.8164965809 |
PROBLEM SETTER: JANE ALAM JAN
题意:求点到线段的最短距离。
思路:距离变化为凸函数,可以用三分法解决。
# include
# include
# include
# include
# include
# include
using namespace std;
double fun(double x, double y, double z, double a, double b, double c)
{
return sqrt((x-a)*(x-a)+(y-b)*(y-b)+(z-c)*(z-c));
}
int main()
{
int t, cas=1;
double x1, y1, z1, x2, y2, z2, x, y, z;
scanf("%d",&t);
while(t--)
{
int k=100;
scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf",&x1, &y1, &z1, &x2, &y2,&z2,&x, &y, &z);
double lx=x1, ly=y1,lz=z1, rx=x2,ry=y2,rz=z2, midx, midy, midz;
while(k--)
{
midx=(lx+rx)/2; midy=(ly+ry)/2; midz=(lz+rz)/2;
double s1 = fun(midx, midy, midz, x, y, z);
double mmidx=(midx+rx)/2, mmidy=(midy+ry)/2, mmidz=(midz+rz)/2;
double s2 = fun(mmidx, mmidy, mmidz, x, y, z);
if(s2 > s1)
{
rx = mmidx;
ry = mmidy;
rz = mmidz;
}
else
{
lx = midx;
ly = midy;
lz = midz;
}
}
printf("Case %d: ",cas++);
printf("%.10f\n", fun(midx, midy, midz, x, y, z));
}
return 0;
}