将矩阵转换为下 Hessenberg 矩阵所需的最小步骤

给定一个NxN阶矩阵，找到将给定矩阵转换为下 Hessenberg 矩阵的最小步数。在每一步中，唯一允许的操作是将任何元素值减少或增加 1。
例子：

Input: mat[][] = { 
{1, 2, 8}, 
{1, 3, 4}, 
{2, 3, 4}} 
Output: 8 
Decrease the element a[0][2] 8 times. 
Now the matrix is lower Hessenberg.
Input: mat[][] = { 
{9, 2, 5, 5}, 
{12, 3, 4, 5}, 
{13, -3, 4, 2}, 
{-1, 10, 1, 4}} 
Output: 15 
 

方法：

• 要使矩阵成为下 Hessenberg 矩阵，其超对角线以上的所有元素都必须为零，即对于所有 j > i+1 ， A _ij = 0。 .
• 在下 Hessenberg 矩阵中转换给定矩阵所需的最小步骤数等于所有 j > i + 1 的所有 A _ij的绝对值之和。
• 单元的模量值被考虑在内，因为单元的增加和减少都算作一个步骤。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
#define N 4
using namespace std;
 
// Function to count steps in
// conversion of matrix into upper
// Hessenberg matrix
int stepsRequired(int arr[][N])
{
    int result = 0;
    for (int i = 0; i < N; i++) {
 
        for (int j = 0; j < N; j++) {
            if (j > i + 1)
                result += abs(arr[i][j]);
        }
    }
    return result;
}
 
int main()
{
    int arr[N][N] = { 1, 2, 3, 2,
                      3, 1, 0, 3,
                      3, 2, 1, 3,
                      -3, 4, 2, 1 };
 
    cout << stepsRequired(arr);
    return 1;
}

Java

// Java implementation of above approach
import java.io.*;
 
class GFG
{
     
static int N = 4;
 
// Function to count steps in
// conversion of matrix into upper
// Hessenberg matrix
static int stepsRequired(int arr[][])
{
    int result = 0;
    for (int i = 0; i < N; i++)
    {
 
        for (int j = 0; j < N; j++)
        {
            if (j > i + 1)
                result += Math.abs(arr[i][j]);
        }
    }
    return result;
}
 
// Driver code
public static void main (String[] args)
{
 
    int [][]arr = { {1, 2, 3, 2},
                    {3, 1, 0, 3},
                    {3, 2, 1, 3},
                    {-3, 4, 2, 1}
                    };
 
    System.out.println (stepsRequired(arr));
}
}
 
// The code is contributed by ajit.

Python3

# Python3 implementation of above approach
N = 4
 
# Function to count steps in
# conversion of matrix into upper
# Hessenberg matrix
def stepsRequired(arr):
    result = 0
    for i in range(N):
        for j in range(N):
            if (j > i + 1):
                result += abs(arr[i][j])
    return result
 
 
# Driver code
arr = [[1, 2, 3, 2],
        [3, 1, 0, 3],
        [3, 2, 1, 3],
        [-3, 4, 2, 1]]
 
print(stepsRequired(arr))
 
# This code is contributed by mohit kumar 29

C#

// C# implementation of above approach
using System;
 
class GFG
{
 
static int N = 4;
 
// Function to count steps in
// conversion of matrix into upper
// Hessenberg matrix
static int stepsRequired(int [,]arr)
{
    int result = 0;
    for (int i = 0; i < N; i++)
    {
 
        for (int j = 0; j < N; j++)
        {
            if (j > i + 1)
                result += Math.Abs(arr[i,j]);
        }
    }
    return result;
}
 
// Driver code
static public void Main ()
{
     
    int [,]arr = { {1, 2, 3, 2},
                    {3, 1, 0, 3},
                    {3, 2, 1, 3},
                    {-3, 4, 2, 1}
                    };
 
    Console.Write(stepsRequired(arr));
}
}
 
// The code is contributed by Tushil..

Javascript

// Java script implementation of above approach
let N = 4;
 
// Function to count steps in
// conversion of matrix into upper
// Hessenberg matrix
function stepsRequired(arr)
{
    let result = 0;
    for (let i = 0; i < N; i++)
    {
 
        for (let j = 0; j < N; j++)
        {
            if (j > i + 1)
                result += Math.abs(arr[i][j]);
        }
    }
    return result;
}
 
// Driver code
    let arr= [[1, 2, 3, 2],
                    [3, 1, 0, 3],
                    [3, 2, 1, 3],
                    [-3, 4, 2, 1]
                    ];
 
    document.write (stepsRequired(arr));
 
// This code is contributed by mohan pavan
 

输出：

8

时间复杂度： O(N*N)