Bfs遍历使用C [英] Bfs traversal using C

问题描述

我正在研究图表，我正在使用C.当我代表一个带有邻接列表的图形时，我需要一个BFS遍历的队列。但是，我在代码中遇到了一些问题 - 我不确定我是否掌握了排队的bfs遍历的概念。



我粘贴了评论下面的代码，我希望它是可读的。有人可以检查一下，或者至少提供一些关于如何正确推出这个问题的信息吗？



程序崩溃并且还显示分段错误



我的尝试：

I'm studying graphs at the moment, and I'm using C. When I represent a graph with an adjacency list, I need a queue for a BFS traversal. However, I'm having some issues with the code - I'm not sure if I grasped the concept of a bfs traversal with queues well.

I pasted the commented code below, I hope it's readable. Can someone check it out, or at least provide some info on how to pull this off the right way?

program is crashing and also showing segmentation fault

What I have tried:

    #include<stdio.h>
    #include<stdlib.h>
    struct Queue{
        int rear;
        int front;
        int capacity;
        int* array;
    };
    struct adjlistnode{
        int dest;
        struct adjlistnode* next;
    };
    struct adjlist{
        struct adjlistnode* head;
    };
    struct graph{
        int V;
        struct adjlist* array;
    };
    int visited[100];

    struct Queue* createqueue(int capacity){
        struct Queue* queue=(struct Queue*)malloc(sizeof(struct Queue));
        queue->rear = -1;
        queue->front = -1;
    queue->capacity=capacity;
    queue->array=(int*)malloc(sizeof(int)*capacity);
    return queue;
    }
    int isempty(struct Queue* queue){
        return(queue->front==-1 && queue->rear==-1);
    }
    void enqueue(struct Queue* queue,int data){
        if(isempty(queue)){
            queue->rear=0;
            queue->front=0;
            queue->array[queue->rear]=data;
            printf("%d",queue->array[queue->rear]);
            return;
        }
    queue->rear=(queue->rear+1)%queue->capacity;
    queue->array[queue->rear]=data;
    } 
    int dequeue(struct Queue* queue){
        if(isempty(queue))
            return -1;
        int temp=queue->front;
        queue->front=(queue->front+1)%queue->capacity;
        return (queue->array[temp]); 
    }
    int isfront(struct Queue* queue){
        return(queue->array[queue->front]);
    }
/// GRAPH FUNCTIONS
    struct adjlistnode* getnewnode(int dest){
        struct adjlistnode* newnode =(struct adjlistnode*)malloc(sizeof(struct adjlistnode));
        newnode->dest=dest;
        newnode->next=NULL;
        return newnode;
    }
    struct graph* creategraph(int V){
        struct graph* G = (struct graph*)malloc(sizeof(struct graph));
        G->V=V;
        G->array=(struct adjlist*)malloc(V*sizeof(struct adjlist));
        for(int i=0;i<V;i++){
            G->array[i].head=NULL;
        }
     return G;
    }
    void addedge(struct graph* G,int src ,int dest){
        struct adjlistnode* newnode=getnewnode(dest);

        newnode->next=G->array[src].head;
        G->array[src].head=newnode; 

        newnode=getnewnode(src);
 
        newnode->next=G->array[dest].head;
        G->array[dest].head=newnode;
    }
    void printgraph(struct graph* G){
        for(int i=0;i<G->V;i++){
            struct adjlistnode* temp=G->array[i].head;
            while(temp!=NULL){
                printf("%d",temp->dest);
                temp=temp->next;
            }
            printf("\n");
        }
    }
    void bfs(struct graph* G,struct Queue* queue,int startvertex){


        enqueue(queue,startvertex);


        while(!isempty(queue)){
           int u=dequeue(queue);
           visited[u]=1;
           printf(" \n %d ",u);
           struct adjlistnode* temp=G->array[u].head;
           while(temp){
             if(visited[temp->dest]==0){
                 visited[temp->dest]=1;
                 enqueue(queue,temp->dest);
              }
           temp=temp->next;
           }

        }

    }
    void bfstraversal(struct graph* G,struct Queue *queue){
        int i;
        for(i=0;i<G->V;i++)
            visited[i]=0;
        for(i=0;i<G->V;i++){
            if(!visited[i]){
                bfs(G,queue,i);
            }
        }
    }
     int main(){
        struct Queue* queue=createqueue(100);
        enqueue(queue,1);
        enqueue(queue,2);
        printf("\n%d",dequeue(queue));
        printf("\n%d",dequeue(queue));
        struct graph* G=creategraph(5);
        addedge(G,1,2);
        addedge(G,1,1);
        printgraph(G);
        bfstraversal(G,queue);
      //  printf("\n%d",dequeue(queue));
    }

推荐答案

编译并不意味着你的代码是对的！ ：笑：

将开发过程想象成编写电子邮件：成功编译意味着您使用正确的语言编写电子邮件 - 例如英语而不是德语 - 而不是电子邮件包含您的邮件想发送。



所以现在你进入第二阶段的发展（实际上它是第四或第五阶段，但你将在之后的阶段进入）：测试和调试。



首先查看它的作用，以及它与你想要的有何不同。这很重要，因为它可以为您提供有关其原因的信息。例如，如果程序旨在让用户输入一个数字并将其翻倍并打印答案，那么如果输入/输出是这样的：

Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.

So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.

Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:

Input   Expected output    Actual output
  1            2                 1
  2            4                 4
  3            6                 9
  4            8                16

然后很明显问题出在将它加倍的位 - 它不会将自身加到自身上，或者将它乘以2，它会将它自身相乘并返回输入的平方。

所以，你可以查看代码和很明显，它在某处：

Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:

int Double(int value)
   {
   return value * value;
   }

一旦你知道可能出现的问题，就开始使用调试器找出原因。在方法的第一行放置一个断点，然后运行你的应用程序。当它到达断点时，调试器将停止，并将控制权移交给您。您现在可以逐行运行代码（称为单步执行）并根据需要查看（甚至更改）变量内容（哎呀，您甚至可以更改代码并在需要时再试一次）。 />
在执行代码之前，请考虑代码中的每一行应该做什么，并将其与使用Step over按钮依次执行每一行时实际执行的操作进行比较。它符合您的期望吗？如果是这样，请转到下一行。

如果没有，为什么不呢？它有什么不同？

希望这可以帮助你找到代码的哪个部分有问题，以及问题是什么。

这是一项技能，它是一个值得开发的，因为它可以帮助你在现实世界和发展中。和所有技能一样，它只能通过使用来改善！

Once you have an idea what might be going wrong, start using the debugger to find out why. Put a breakpoint on the first line of the method, and run your app. When it reaches the breakpoint, the debugger will stop, and hand control over to you. You can now run your code line-by-line (called "single stepping") and look at (or even change) variable contents as necessary (heck, you can even change the code and try again if you need to).
Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?
Hopefully, that should help you locate which part of that code has a problem, and what the problem is.
This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!

这篇关于Bfs遍历使用C的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！