lightoj1240：Point Segment Distance (3D)（三分）

最新推荐文章于 2023-02-27 19:13:49 发布

junior19

最新推荐文章于 2023-02-27 19:13:49 发布

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分类专栏：二分/三分

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本文介绍了一个计算三维空间中点到线段最短欧几里得距离的问题，并提供了一种通过三分法来求解该问题的有效算法实现。输入包括多个测试案例，每个案例包含线段两端点坐标及目标点坐标，输出则是各案例中点到线段的最短距离。

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1240 - Point Segment Distance (3D)

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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

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 <iostream>
# include <cstdio>
# include <cmath>
# include <vector>
# include <cstring>
# include <algorithm>
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;
}