以下是使用C语言编写的代码，用于计算切割矩形蛋糕后两份美味度之差的最小值：

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

int main() {
    int n, m;
    printf("请输入矩形蛋糕的行数n和列数m：");
    scanf("%d %d", &n, &m);

    int **cake = (int **)malloc(n * sizeof(int *));
    for (int i = 0; i < n; i++) {
        cake[i] = (int *)malloc(m * sizeof(int));
    }

    printf("请依次输入每个区域的美味度：  
");
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            scanf("%d", &cake[i][j]);
        }
    }

    int sum = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            sum += cake[i][j];
        }
    }

    int halfSum = sum / 2;
    int **dp = (int **)malloc((halfSum + 1) * sizeof(int *));
    for (int i = 0; i <= halfSum; i++) {
        dp[i] = (int *)malloc((m + 1) * sizeof(int));
    }

    for (int i = 0; i <= halfSum; i++) {
        dp[i][0] = 0;
    }

    for (int i = 0; i <= m; i++) {
        dp[0][i] = 0;
    }

    for (int i = 1; i <= halfSum; i++) {
        for (int j = 1; j <= m; j++) {
            dp[i][j] = dp[i][j - 1];
            if (i >= cake[0][j - 1] && dp[i][j] < dp[i - cake[0][j - 1]][j - 1] + cake[0][j - 1]) {
                dp[i][j] = dp[i - cake[0][j - 1]][j - 1] + cake[0][j - 1];
            }
        }
    }

    int maxSum = dp[halfSum][m];
    int minDiff = sum - 2 * maxSum;

    printf("|S1 - S2|的最小值为：%d  
", minDiff);

    for (int i = 0; i < n; i++) {
        free(cake[i]);
    }
    free(cake);

    for (int i = 0; i <= halfSum; i++) {
        free(dp[i]);
    }
    free(dp);

    return 0;
}

这段代码使用动态规划的思想，通过二维数组dp来记录每个美味度和的最大值。最后计算出两份美味度之差的最小值，并输出结果。

请注意，这段代码假设输入的数据合法，即n和m为正整数，每个区域的美味度为非负整数。在实际使用中，可能需要添加一些输入验证的代码来确保输入的正确性。