找到一个点，使得曼哈顿距离之和最小

给定K维空间中的N个点，其中$2\leq N\leq 10^{5}$和$1\leq K\leq 5$ .任务是确定点，使得从该点到N个点的曼哈顿距离之和最小。
曼哈顿距离是沿直角轴测量的两点之间的距离。在p1位于(x1, y1)且p2位于(x2, y2)的平面中，它是|x1 – x2| + |y1 – y2| .

例子：

Input: N = 3, K = 3, Points = {1, 1, 1}, {2, 2, 2}, {3, 3, 3} 
Output: 2 2 2 
Input: N = 4, K = 4, Points = {1, 6, 9, 6}, {5, 2, 5, 7}, {2, 0, 1, 5}, {4, 6, 3, 9} 
Output: 2 2 3 6 

办法：尽量减少曼哈顿距离，我们所要做的仅仅是那种在所有K尺寸和输出所述K尺寸的中间元素的点。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function to print the required points which
// minimizes the sum of Manhattan distances
void minDistance(int n, int k, vector >& point)
{
 
    // Sorting points in all k dimension
    for (int i = 0; i < k; ++i)
        sort(point[i].begin(), point[i].end());
 
    // Output the required k points
    for (int i = 0; i < k; ++i)
        cout << point[i][(ceil((double)n / 2) - 1)] << " ";
}
 
// Driver code
int main()
{
 
    int n = 4, k = 4;
    vector > point = { { 1, 5, 2, 4 },
                                   { 6, 2, 0, 6 },
                                   { 9, 5, 1, 3 },
                                   { 6, 7, 5, 9 } };
 
    // function call to print required points
    minDistance(n, k, point);
 
    return 0;
}

Java

// Java implementation of above approach
import java.util.Arrays;
 
class GFG
{
 
// Function to print the required
// points which minimizes the sum
// of Manhattan distances
static void minDistance(int n, int k,
                        int point[][])
{
     
    // Sorting points in all k dimension
    for (int i = 0; i < k; i++)
        Arrays.sort(point[i]);
     
    // Output the required k points
    for (int i = 0; i < k; i++)
        System.out.print(point[i][(int)
               Math.ceil((double)(n / 2) - 1)] + " ");
}
 
// Driver code
public static void main(String[] args)
{
    int n = 4;
    int k = 4;
    int point[][] = { { 1, 5, 2, 4 },
                      { 6, 2, 0, 6 },
                       { 9, 5, 1, 3 },
                      { 6, 7, 5, 9 } };
     
    // function call to print required points
    minDistance(n, k, point);
}
}
 
// This code is contributed by Bilal

Python

# Python implementation of above approach
 
# Function to print the required points which
# minimizes the sum of Manhattan distances
def minDistance(n, k, point):
 
    # Sorting points in all dimension
    for i in range(k):
        point[i].sort()
 
    # Output the required k points
    for i in range(k):
        print(point[i][((n + 1) // 2) - 1], end =" ")
 
 
# Driver code
n = 4
k = 4
point = [[1, 5, 2, 4],
         [6, 2, 0, 6],
         [9, 5, 1, 3],
         [6, 7, 5, 9]]
 
# function call to print required points
minDistance(n, k, point)

C#

// C# implementation of above approach
using System;
 
class GFG
{
 
// Function to print the required
// points which minimizes the sum
// of Manhattan distances
static void minDistance(int n, int k,
                        int[][] point)
{
     
    // Sorting points in all k dimension
    for (int i = 0; i < k; i++)
        Array.Sort(point[i]);
     
    // Output the required k points
    for (int i = 0; i < k; i++)
        System.Console.Write(point[i][(int)
            Math.Ceiling((double)(n / 2) - 1)] + " ");
}
 
// Driver code
public static void Main()
{
    int n = 4;
    int k = 4;
    int[][] point = new int[][]{ new int[]{ 1, 5, 2, 4 },
                    new int[]{ 6, 2, 0, 6 },
                    new int[]{ 9, 5, 1, 3 },
                    new int[]{ 6, 7, 5, 9 } };
     
    // function call to print required points
    minDistance(n, k, point);
}
}
 
// This code is contributed by mits

Javascript

// Javascript implementation of above approach
     
// Function to print the required
// points which minimizes the sum
// of Manhattan distances
    function minDistance(n,k,points)
    {
        // Sorting points in all k dimension
        for (let i = 0; i < k; i++)
            (point[i]).sort(function(a,b){return a-b;});
       
        // Output the required k points
        for (let i = 0; i < k; i++)
            document.write(point[i][
               Math.ceil((n / 2) - 1)] + " ");
    }
     
    // Driver code
    let n = 4;
    let k = 4;
    let point = [[1, 5, 2, 4],
         [6, 2, 0, 6],
         [9, 5, 1, 3],
         [6, 7, 5, 9]];
     
    // function call to print required points
    minDistance(n, k, point);
         
// This code is contributed by rag2127

输出：

2 2 3 6

时间复杂度： O(k*nlog(n)

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