矩阵里元素交换的一个算法，如何利用C语言的编程的技术实现的办法

Question

Problem Description
There is a matrix M that has n rows and m columns (1≤n≤1000,1≤m≤1000).Then we perform q(1≤q≤100,000) operations:

1 x y: Swap row x and row y (1≤x,y≤n);

2 x y: Swap column x and column y (1≤x,y≤m);

3 x y: Add y to all elements in row x (1≤x≤n,1≤y≤10,000);

4 x y: Add y to all elements in column x (1≤x≤m,1≤y≤10,000);

Input
There are multiple test cases. The first line of input contains an integer T(1≤T≤20) indicating the number of test cases. For each test case:

The first line contains three integers n, m and q.
The following n lines describe the matrix M.(1≤Mi,j≤10,000) for all (1≤i≤n,1≤j≤m).
The following q lines contains three integers a(1≤a≤4), x and y.

Output
For each test case, output the matrix M after all q operations.

Sample Input
2
3 4 2
1 2 3 4
2 3 4 5
3 4 5 6
1 1 2
3 1 10
2 2 2
1 10
10 1
1 1 2
2 1 2

Sample Output
12 13 14 15
1 2 3 4
3 4 5 6
1 10
10 1

Answer 1

#include<stdio.h>
int main()
{
    int T, n, m, k, i, j, tmp;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d%d", &n, &m, &k);
        int M[n][m];
        int a, x, y;
        for(i=0; i<n; i++)
        {
            for(j=0; j<m; j++)
            {
                scanf("%d", &M[i][j]);
            }
        }
        while(k--)
        {
            scanf("%d%d%d", &a, &x, &y);
            if(a == 1)
            {
                for(i=0; i<m; i++)
                {
                    tmp = M[y-1][i];
                    M[y-1][i] = M[x-1][i];
                    M[x-1][i] = tmp;
                }
            }
            else if(a == 2)
            {
                for(i=0; i<n; i++)
                {
                    tmp = M[i][y-1];
                    M[i][y-1] = M[i][x-1];
                    M[i][x-1] = tmp;
                }
            }
            else if(a == 3)
            {
                for(i=0; i<m; i++)
                    M[x-1][i] += y;
            }
            else{
                for(i=0; i<n; i++)
                    M[i][x-1] += y;
            }
        }
        for(i=0; i<n; i++)
        {
            for(j=0; j<m; j++)
            {
                printf("%d ", M[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}