假设我创建了一个名为 graph-file.txt 的文件,其中包含
的数据有向图如下:
7
{5, 2, 3}, {1,5}, {}, { }, {3}, { }, { }
文件的第一行显示顶点数(在这种情况下为7)。 第二行描述了每个顶点的邻居N +(v)。例如,从顶点编号1边缘指向顶点5,2和3,对于顶点编号7,没有边缘指向该图中的任何其他顶点。
我想问一下从该文件中读取BFS算法所需边缘信息(如上所述)的方法是什么?
我对BFS使用以下功能:
#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;
};
struct Graph* createGraph(int vertices);
void addEdge(struct Graph* graph, int src, int dest);
void printGraph(struct Graph* graph);
void bfs(struct Graph* graph, int startVertex);
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;
}
}
}
struct node* createNode(int v)
{
struct node* newNode = malloc(sizeof(struct node));
newNode->vertex = v;
newNode->next = NULL;
return newNode;
}
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;
}
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;
}
struct queue* createQueue() {
struct queue* q = malloc(sizeof(struct queue));
q->front = -1;
q->rear = -1;
return q;
}
int isEmpty(struct queue* q) {
if(q->rear == -1)
return 1;
else
return 0;
}
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;
}
}
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;
}
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]);
}
}
}
答案 0 :(得分:0)
假设输入有效,作为一个有趣的练习,我试图想出所需的最小解析器。对于x86_64,它使用少量寄存器编译为less than 40 instructions:
void edge(int from, int to);
void parse(void)
{
int v = 0;
for (;;) {
const int c = getchar();
if (c == EOF)
return;
// Next 'from' vertex?
if (c == '{') {
++v;
continue;
}
// Ignore vertex count
if (v == 0)
continue;
// Found 'to' vertex?
int w = c - '0';
if (w < 0 || w > 9)
continue;
// Parse 'to' vertex
for (;;) {
const int d = getchar() - '0';
if (d < 0 || d > 9)
break;
w = w * 10 + d;
}
edge(v, w);
}
}
当然,如果你需要验证输入,那么你需要更多的状态来处理逗号,关闭括号,不匹配的顶点数,超出范围的顶点数,无效字符等。