c++ - 如何仅对 10 x 10 矩阵的右四分之一进行排序？

Question

10x10 矩阵分为 4 部分，右四分之一按行中的 any_method 排序:

我是初学者。我完全不明白选择正确的季度进行进一步排序的算法是如何真正起作用的。这是我目前所拥有的。

#include <iostream>
#include <ctime>
using namespace std;

int main()
{
srand(time(0));

const int size = 10;
int matrix[size][size];
int sum = 0;

system("mode con cols=100 lines=70");

cout << "Source matrix 10x10" << endl;
cout << endl;

for (int i = 0; i < size; i++)
{
    for (int j = 0; j < size; j++)
    {
        matrix[i][j] = rand() % 50;   
    }
}

for (int i = 0; i < size; i++)
{
    for (int j = 0; j < size; j++)
    {
        cout << matrix[i][j] << "\t";
    }
    cout << endl << endl;
}

cout << endl;
cout << "Transformed matrix 10x10" << endl;

int temp;
int k = 0;
int cond = 0;

for (int s = 0; s < size; s++)
{
    for (int i = 0; i < size; i++)
    {
        for (int j = size - 1; j > i; j--)
        {    
            if (matrix[s][j] < matrix[s][j - 1])
            {
                temp = matrix[s][j];
                matrix[s][j] = matrix[s][j - 1];
                matrix[s][j - 1] = temp;
            }
        }
    }
}

for (int i = 0; i < size; i++)
{
    for (int j = 0; j < size; j++)
    {
        cout << matrix[i][j] << "\t";
    }
    cout << endl << endl;
}

cout << endl << sum << endl;

return 0;
system("pause");

}

Answer 1

I have got some problems with how actually i am going to pick the region of the right quarter. So choosing the right quarter higher in priority then sorting

我用过std::sort演示对 10 x 10 矩阵的右四分之一进行排序。

代码注释中解释了我的逻辑:

void sortRightQuarter(int m[10][10])
{
    /*
    Indexes of the elements in the Right Quarter
                                    (0, 9) // already sorted
                            (1, 8), (1, 9) // i = 1, k = 8
                    (2, 7), (2, 8), (2, 9) // i = 2, k = 7
            (3, 6), (3, 7), (3, 8), (3, 9) // i = 3, k = 6
    (4, 5), (4, 6), (4, 7), (4, 8), (4, 9) // i = 4, k = 5
    (5, 5), (5, 6), (5, 7), (5, 8), (5, 9) // i = 5, k = 5
            (6, 6), (6, 7), (6, 8), (6, 9) // i = 6, k = 6
                    (7, 7), (7, 8), (7, 9) // i = 7, k = 7
                            (8, 8), (8, 9) // i = 8, k = 8
                                    (9, 9) // already sorted
    */

    // Sort the rows [1, 8]
    int k = 8;
    for (int i = 1; i <= 8; ++i)
    {
        // pick your favorite sorting function
        // to sort elements in the range [m[i][k], last element of the row]
        sort(&m[i][k], &m[i][10]);

        // change the value of k
        if (i < 4)
            --k;
        else
            ++k;
    }
}

我用你的代码来测试我的功能。效果很好。