Question

我有一个文本文件，提供用于在城市之间创建通信网络的输入数据。

N=12

C=[72,107,65,99,20,11,31,7,42,28,4, 31,10,57,37,71,18,17,73,39,14, 91,77,19,39,33,59,31,44,98, 71,66,36,23,10,100,93,41, 21,80,22,43,78,22,93, 77,81,49,62,89,19, 21,19,51,40,18, 102,92,28,51, 31,47,51, 73,38, 50]

R=[0.81,0.88,0.83,0.90,0.98,0.93,0.86,0.91,0.92,0.86,0.80, 0.91,0.87,0.78,0.83,0.90,0.93,0.87,0.79,0.99,0.94, 0.88,0.98,0.97,0.96,0.92,0.81,0.99,0.92,0.91, 0.80,0.88,0.87,0.93,0.90,0.99,0.98,0.91, 0.89,0.80,0.92,0.93,0.83,0.87,0.99, 0.97,0.91,0.89,0.99,0.94,0.90, 0.87,0.89,0.95,0.85,0.96, 0.99,0.79,0.88,0.81, 0.84,0.97,0.77, 0.83,0.93, 0.99]

a_b=1

Req_Reliability =  0.5

Req_Cost = 267

其中： N 是网络中的城市数量和 C ＆amp; R 对称N×N矩阵分别给出了在两个城市之间实现连接的成本和可靠性。

N 可以变化，只要它是大于2的整数。并且给定的矩阵C＆amp; R会相应变化。

从文本文件开始，由于矩阵是对称的，所以只给出了矩阵的上半部分，所有索引为i = j的元素都应该为0，这样文本文件中矩阵C的格式就是这样：

C=[C(1-2),C(1-3),...,C(1-12),whitespaceC(2-3),C(2-4),...,C(2-12),...,whitespaceC(11-12)]
with C(IDstartCity-IDstopCity)

文本文件中的空格表示对下一行的更改，相同的结构适用于矩阵R.

我想要做的是在我的C程序中将每个变量分配给它的匹配。虽然我对如何将12分配给N有了一个想法，但我对于如何将文本文件中的矩阵C和R的奇怪结构分配给C程序中的矩阵感到无能为力。

int main(int argc, const char * argv[]) {

    FILE *finput;
    char var_name[5];
    int value = 0;
    finput = fopen("Prj1_input.txt", "r");

    //Inputs Start
    int N;                              //Number of cities
        //Cost Array
    int **C = malloc(N * sizeof(int*));
    for (i = 0; i < N; i++) {
        C[i] = malloc(N * sizeof(int));
        memset(C[i], 0, N * sizeof(int));
    }

    //Reliability Array
    double **R = malloc(N * sizeof(double*));
    for (i = 0; i < N; i++) {
        R[i] = malloc(N * sizeof(double));
        memset(R[i], 0, N * sizeof(double));
    }
    int con_cost;                       //Connection cost between 2 cities
    double con_reliability;             //Connection reliability between 2 cities

    int a_b;                            //0 is type a, 1 is type b

    double Req_reliability, Req_cost;   //reliability for type a, cost for type b
    //Inputs Stop


    fscanf(finput, "%c=%d", var_name, &value);

    printf("%s %d\n", var_name, value);



    fclose(finput);

    return 0;
}

我想我可以使用两个for循环，当我在文本文件中找到一个空格时，我会增加行，但我完全不知道如何处理文本文件。

现在，如果我在//输入开始和//输入停止之间注释掉所有内容，那么输出就是 N 12

由于

Answer 1

以C为例，首先制作一个大小为N的数组：

int **C = malloc(N * sizeof(int*));
for (i = 0; i < N; i++) {
    C[i] = malloc(N * sizeof(int));
    memset(C[i], 0, N * sizeof(int));
}

然后循环并分配给矩阵：

char *s = "<the string inside the [...]>";
char *p = s, *e;
for (i = 0; i < N; i++) {
    for (j = i + 1; j < N; j++) {
        e = strchr(p, ',');
        *e = '\0';
        C[i][j] = atoi(p);
        p = e + 1;
    }
}

从文本文件中填充C中的2D数组

1 个答案: