C语言中的Djikstra算法

Question

C语言中的Djikstra算法

3

需要帮助。我必须编写Djikstra算法的程序，不必从用户处获取任何输入，只需在程序中硬编码即可。这是我第一次用C语言编写任何东西，对它并不是很擅长。我已经在头脑中想好了代码的逻辑运行方式，但问题是当我运行它时，什么也没有输出。如果有人能帮我找到问题，那就太好了。我会继续更新进展情况，但比我更聪明的C语言专家可能更容易找到问题。

#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

     }

我已尽力进行注释，希望您能够理解我的逻辑。我可能过于复杂化了这个问题，但这应该是一种可能的解决方法。

修复后可正常工作的代码！

#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

     }

感谢大家的帮助。现在我只需要将它转换为普里姆算法（这将非常简单，因为我只是不将所有内容相加）。也许会尝试不同的起始位置，但现在这已经足够了。

- Flint Coal

如果在printf中加入\n，它们将会显示出来。而且不要传递指针，例如 printf("%d\n", position);。 - bruceg

1

那部分已经修好了，谢谢。是的，我现在正在学习指针，当我读到那里时，我就像“嗯，我真是太蠢了。”哈哈 - Flint Coal

1个回答

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- user4407569 · Accepted Answer

使用gcc编译的输出结果如下：

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);

首先，每个地方都应该是int main和printf("%d", position)，还需要删除return;。函数应该返回一个int。

然后代码执行并打印出一些东西：

你可能希望在它们之间添加换行符，使用 printf("%d\n", position)。然后：

虽然我还没有检查输出的正确性。