题解 | #矩阵交换#

#include <stdio.h>

int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    int arr1[11][11];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            scanf("%d", &arr1[i][j]);
        }
    }
    int k;
    scanf("%d", &k);
    int arr2[5][3];
    char arr3[5];
    for (int i = 0; i < k; i++) {
            scanf(" %c %d %d", &arr3[i],&arr2[i][1],&arr2[i][2]);   
    }
// -----------------上面位读取数据 -----------------------
    for (int i = 0; i < k; i++) { // 操作次数
        if (arr3[i] == 'r') { // 行变化
            for (int j = 0; j < m; j ++) { // 循环对按行数赋值
                int a = arr1[arr2[i][1] - 1][j]; //第arr2[i][1](1)行的第j个值
                int b = arr1[arr2[i][2] - 1][j]; //第arr2[i][1](2)行的第j个值
                arr1[arr2[i][2] - 1][j] = a; // 交换
                arr1[arr2[i][1] - 1][j] = b;
            }
        }
        else if (arr3[i] == 'c') { // 列变化
            for (int j = 0; j < n; j ++) { // 循环对按列数赋值
                int c = arr1[j][arr2[i][1] - 1]; //第arr2[i][1](1)列的第j个值
                int d = arr1[j][arr2[i][2] - 1]; //第arr2[i][1](2)列的第j个值
                arr1[j][arr2[i][2] - 1] = c; // 交换
                arr1[j][arr2[i][1] - 1] = d;
            }
        }
    }
    for(int i = 0;i < n;i++){
        for(int j = 0;j < m;j++){
            printf("%d ",arr1[i][j]);
        }
        printf("\n");
    }

    return 0;
}