c - C 中的 Dijkstra 算法

Question

在这里寻求帮助。我必须为 Djikstra 算法编写一个程序。不必从用户那里获取输入或任何东西，只需硬编码。这是我第一次用 C 编写任何代码。并不是那么好用。我得到了我的代码，它在我的脑海中逻辑上是如何工作的。我的问题是，当我运行它时，我什么也得不到。什么都没有打印出来。有人可以帮助我和我一起解决这个问题，那就太好了。我会在尝试查找问题时及时通知您最新情况，但 C 语言比我聪明一点的人可能更容易解决问题。

#include <stdio.h>
void main (){
        int ab = 3;//path from a to b
        int ac = 7;//path from a to c
        int ad = 9;//path from a to d
        int bc = 2;//path from b to c
        int bd = 4;//path from b to d
        int cd = 1;//path from c to d
        int a = 10;//number values for position
        int b = 20;
        int c = 30;
        int d = 40;

        int position = 10;//starting position a
        int currenttravel = 0;
        //starting at a
        //if (position == 10){
        int checker = 40;//check for when at d
        do
        {
            //check for if at a
            if (position == 10){
                //if path a to b is shortest
                if (ab < ac && ab < ad){
                    position = b;//go to b
                    printf("%d", &position);
                    currenttravel+=ab;
                }
                //or if path a to c is shortest
                else if (ac < ad){
                    position = c;//go to c
                    printf("%d", &position);
                    currenttravel+=ac;
                }
                else{
                    position = d;
                    printf("%d", &position);
                    currenttravel+=ad;
                }

            }
            if (position == 20)//at b
            {
                if (bc < bd){
                    position = c;
                    printf("%d", &position);
                    currenttravel+=bc;
                }
                else{
                    position = d;
                    printf("%d", &position);
                    currenttravel+=bd;
                }
            }
            if (position == 30){
                position = d;
                printf("%d", &position);
                currenttravel+=cd;
            }
        }
        while(position != checker);
    //  }//end if start position is a
        printf("%d", currenttravel);
        return;  //leave function

     }

我已尽我所能发表评论，所以希望可以遵循我的逻辑。我可能把它复杂化了，但这应该是一种可行的方法。

有效的固定代码!

#include <stdio.h>
int main (){
        int ab = 3;//path from a to b
        int ac = 7;//path from a to c
        int ad = 9;//path from a to d
        int bc = 2;//path from b to c
        int bd = 4;//path from b to d
        int cd = 1;//path from c to d
        int a = 10;//number values for position
        int b = 20;
        int c = 30;
        int d = 40;

        int position = 10;//starting position a
        int currenttravel = 0;
        //starting at a
        //if (position == 10){
        int checker = 40;//check for when at d
        do
        {
        printf("starting at a \n");
            //check for if at a
            if (position == a){
                //if path a to b is shortest
                if (ab < ac && ab < ad){
                    position = b;//go to b
                    printf("b \n");
                    currenttravel+=ab;
                }
                //or if path a to c is shortest
                else if (ac < ad){
                    position = c;//go to c
                    printf("c \n");
                    currenttravel+=ac;
                }
                else{
                    position = d;
                    printf("d \n");
                    currenttravel+=ad;
                }

            }
            if (position == b)//at b
            {
                if (bc < bd){
                    position = c;
                    printf("c \n");
                    currenttravel+=bc;
                }
                else{
                    position = d;
                    printf("d \n");
                    currenttravel+=bd;
                }
            }
            if (position == c){
                position = d;
                printf("d \n");
                currenttravel+=cd;
            }
        }
        while(position != checker);
    //  }//end if start position is a
        printf("%d", currenttravel);
    //  return;  //leave function

     }

谢谢大家的帮助。现在我只需要将它转换为 Prim 的算法(这将非常简单，因为我只是不把所有东西都加起来)。也可以尝试不同的起始位置，但现在这可能就足够了。

Answer 1

这是我用 gcc 编译时得到的输出:

main.cpp:2:12: error: ‘::main’ must return ‘int’
 void main (){
           ^

main.cpp: In function ‘int main()’:
main.cpp:64:2: error: return-statement with no value, in function returning ‘int’ [-fpermissive]
  return;
  ^

main.cpp: In function ‘int main()’:
main.cpp:26:43: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int*’ [-Wformat=]
                     printf("%d", &position);
                                           ^
main.cpp:32:43: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int*’ [-Wformat=]
                     printf("%d", &position);
                                           ^
main.cpp:37:43: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int*’ [-Wformat=]
                     printf("%d", &position);
                                           ^
main.cpp:46:43: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int*’ [-Wformat=]
                     printf("%d", &position);
                                           ^
main.cpp:51:43: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int*’ [-Wformat=]
                     printf("%d", &position);
                                           ^
main.cpp:57:39: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int*’ [-Wformat=]
                 printf("%d", &position);

所以首先，它应该是 int main 和 printf("%d", position) 到处都是，你也应该删除 return;。该函数应返回一个 int。

然后代码执行并打印一些东西:

2030406

你可能想要在中间换行，使用 printf("%d\n", position)。然后:

20
30
40
6

虽然我还没有检查输出的正确性。