在 c 中获取距离 2 图邻接列表中的邻居

Question

我在 c 中使用邻接列表创建了一个图问题是，我只想获取并打印给定节点距离为 2 的邻居。我试过 BFS 但它得到了一个节点可以去的每个节点。我知道如果我使用了邻接矩阵，我可以将矩阵与自身相乘。

我使用的代码结构如下

// BFS algorithm in C

#include <stdio.h>
#include <stdlib.h>
#define SIZE 40

struct queue {
  int items[SIZE];
  int front;
  int rear;
};

struct queue* createQueue();
void enqueue(struct queue* q, int);
int dequeue(struct queue* q);
void display(struct queue* q);
int isEmpty(struct queue* q);
void printQueue(struct queue* q);

struct node {
  int vertex;
  struct node* next;
};

struct node* createNode(int);

struct Graph {
  int numVertices;
  struct node** adjLists;
  int* visited;
};

// BFS algorithm
void bfs(struct Graph* graph, int startVertex) {
  struct queue* q = createQueue();

  graph->visited[startVertex] = 1;
  enqueue(q, startVertex);

  while (!isEmpty(q)) {
    printQueue(q);
    int currentVertex = dequeue(q);
    printf("Visited %d\n", currentVertex);

    struct node* temp = graph->adjLists[currentVertex];

    while (temp) {
      int adjVertex = temp->vertex;

      if (graph->visited[adjVertex] == 0) {
        graph->visited[adjVertex] = 1;
        enqueue(q, adjVertex);
      }
      temp = temp->next;
    }
  }
}

// Creating a node
struct node* createNode(int v) {
  struct node* newNode = malloc(sizeof(struct node));
  newNode->vertex = v;
  newNode->next = NULL;
  return newNode;
}

// Creating a graph
struct Graph* createGraph(int vertices) {
  struct Graph* graph = malloc(sizeof(struct Graph));
  graph->numVertices = vertices;

  graph->adjLists = malloc(vertices * sizeof(struct node*));
  graph->visited = malloc(vertices * sizeof(int));

  int i;
  for (i = 0; i < vertices; i++) {
    graph->adjLists[i] = NULL;
    graph->visited[i] = 0;
  }

  return graph;
}

// Add edge
void addEdge(struct Graph* graph, int src, int dest) {
  // Add edge from src to dest
  struct node* newNode = createNode(dest);
  newNode->next = graph->adjLists[src];
  graph->adjLists[src] = newNode;

  // Add edge from dest to src
  newNode = createNode(src);
  newNode->next = graph->adjLists[dest];
  graph->adjLists[dest] = newNode;
}

// Create a queue
struct queue* createQueue() {
  struct queue* q = malloc(sizeof(struct queue));
  q->front = -1;
  q->rear = -1;
  return q;
}

// Check if the queue is empty
int isEmpty(struct queue* q) {
  if (q->rear == -1)
    return 1;
  else
    return 0;
}

// Adding elements into queue
void enqueue(struct queue* q, int value) {
  if (q->rear == SIZE - 1)
    printf("\nQueue is Full!!");
  else {
    if (q->front == -1)
      q->front = 0;
    q->rear++;
    q->items[q->rear] = value;
  }
}

// Removing elements from queue
int dequeue(struct queue* q) {
  int item;
  if (isEmpty(q)) {
    printf("Queue is empty");
    item = -1;
  } else {
    item = q->items[q->front];
    q->front++;
    if (q->front > q->rear) {
      printf("Resetting queue ");
      q->front = q->rear = -1;
    }
  }
  return item;
}

// Print the queue
void printQueue(struct queue* q) {
  int i = q->front;

  if (isEmpty(q)) {
    printf("Queue is empty");
  } else {
    printf("\nQueue contains \n");
    for (i = q->front; i < q->rear + 1; i++) {
      printf("%d ", q->items[i]);
    }
  }
}

int main() {
  struct Graph* graph = createGraph(6);
  addEdge(graph, 0, 1);
  addEdge(graph, 0, 2);
  addEdge(graph, 1, 2);
  addEdge(graph, 1, 4);
  addEdge(graph, 1, 3);
  addEdge(graph, 2, 4);
  addEdge(graph, 3, 4);

  bfs(graph, 0);

  return 0;
}

Answer 1

设N为您感兴趣的节点，求距离。
将每个可直接从 N 访问的元素放入 list/queue.
将每个元素标记为不可访问 (false)。
对于第 2 步中添加的每个元素，检查它们的邻接列表并将可访问的节点标记为可访问（真）。
检查哪些节点可以访问。

伪代码：

 queue<nodes> Q = EMPTY
 for each node N reachable from START NODE:
      enqueue N in Q
 array<bool> IS_REACHABLE = FALSE (for all nodes)
 for each node N in Q:
      go through the adjacency list of N and mark TRUE in array iS_REACHABLE for every element reachable from N
      remove N from Q

数组 IS_REACHABLE 中为真的元素将是您的答案。

Answer 2

您需要一个 distance 数组来跟踪它。

将起始节点的距离设置为0。每次入队一个新节点时，将该节点的距离更新为其父节点的距离+1。打印距离[顶点] == 2的节点。

void bfs( ... ) {
  ...
  distance[startVertex] = 0;
  while (!isEmpty(q)) { 
    int currentVertex = dequeue(q);
    ....
    while (temp) {
      ....
      if (graph->visited[adjVertex] == 0) {
        graph->visited[adjVertex] = 1;
        enqueue(q, adjVertex);
        distance[adjVertex] = distance[currentVertex] + 1;
      }
    }
  }
}

在 c 中获取距离 2 图邻接列表中的邻居

get neighbors in distance 2 graph adjacency list in c

c

adjacency-list