请用c语言帮我写一个代码：从键盘输入矩阵的行、列和非零元素个数，以及若干组数据（分别表示系数矩阵非零元素的行、列和值），即三元组表对应的信息；为了便于观察结果，请将新三元组表中的每条记录均初值化为0，再开始完成矩阵的快速转置

#include <stdio.h>
#include <stdlib.h>

typedef struct {
    int row;
    int col;
    int value;
} Triple;

void printMatrix(Triple *matrix, int m, int n) {
    int p = 1;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[p].row == i && matrix[p].col == j) {
                printf("%d ", matrix[p].value);
                p++;
            } else {
                printf("0 ");
            }
        }
        printf("\n");
    }
}

void transpose(Triple *matrix, Triple *result, int m, int n, int count) {
    int *rowSize = (int*)calloc(n + 1, sizeof(int));
    int *rowStart = (int*)calloc(n + 1, sizeof(int));

    for (int i = 1; i <= count; i++) {
        rowSize[matrix[i].col]++;
    }
    rowStart[1] = 1;
    for (int i = 2; i <= n; i++) {
        rowStart[i] = rowStart[i - 1] + rowSize[i - 1];
    }

    for (int i = 1; i <= count; i++) {
        int j = rowStart[matrix[i].col]++;
        result[j].row = matrix[i].col;
        result[j].col = matrix[i].row;
        result[j].value = matrix[i].value;
    }

    free(rowSize);
    free(rowStart);
}

int main() {
    int m, n, count;
    printf("Enter the size of matrix (m n count): ");
    scanf("%d%d%d", &m, &n, &count);

    Triple *matrix = (Triple*)calloc(count + 1, sizeof(Triple));
    for (int i = 1; i <= count; i++) {
        printf("Enter the %d-th element (row col value): ", i);
        scanf("%d%d%d", &matrix[i].row, &matrix[i].col, &matrix[i].value);
    }

    printf("Original matrix:\n");
    printMatrix(matrix, m, n);

    Triple *result = (Triple*)calloc(count + 1, sizeof(Triple));
    transpose(matrix, result, m, n, count);

    printf("Transposed matrix:\n");
    printMatrix(result, n, m);

    free(matrix);
    free(result);

    return 0;
}