Question

我创建了一个程序，在这些节点之间添加节点和边缘。现在我想找到哪个节点有最大连接边数。这是我的代码：

    const int N = 15;

struct graf {
    int key;
    graf *next;
};




void init(graf *gr[]) {
    for (int i = 0; i<N; i++) {
        gr[i] = NULL;
    }
}

int search_node(graf *gr[], int c) {
    int flag = 0;
    for (int i = 0; i<N; i++) {
        if (gr[i]) {
            if (gr[i]->key == c)flag = 1;
        }
    }
    return flag;
}

int search_arc(graf *gr[], int c1, int c2) {
    int flag = 0;
    if (search_node(gr, c1) && search_node(gr, c2)) {
        int i = 0;
        while (gr[i]->key != c1)
            i++;
        graf *p = gr[i];
        while (p->key != c2 && p->next != NULL)
            p = p->next;
        if (p->key == c2)
            flag = 1;
    }
    return flag;
}

void add_node(graf *gr[], int c) {
    if (search_node(gr, c))
        printf("Edge existing !\n");
    else {
        int j = 0;
        while (gr[j] && (j<N))j++;
        if (gr[j] == NULL) {
            gr[j] = new graf;
            gr[j]->key = c;
            gr[j]->next = NULL;
        }
        else
            printf("Overflow !\n");
    }
}

void add_arc(graf *gr[], int c1, int c2) {
    int i = 0;
    graf *p;
    if (search_arc(gr, c1, c2))
        printf("there is edge from node  %s to node %s", c1, c2);
    else {
        if (!(search_node(gr, c1)))
            add_node(gr, c1);
        if (!(search_node(gr, c2)))
            add_node(gr, c2);
        while (gr[i]->key != c1)
            i++;
        p = new graf;
        p->key = c2;
        p->next = gr[i]->next;
        gr[i]->next = p;
    }
}

int get_graph_rank(graf*& gr,int i=0)
{
    int max = NULL;

    if (gr != NULL)
    {
      //i think to use recursion but i don't know how to start ...
    }

    return max;
}
int main() {
    int c, k;
    int menu = NULL, br = NULL;
    graf *gr1[N];
    init(gr1);
    do
    {
        printf("1. add node\n");
        printf("2. add edge\n");
        printf("3.get the rang\n");
        printf("0. exit\n");
        do {
            scanf_s("%d",&menu);
        } while (menu<0 || menu>4);
        switch (menu) {
        case 1:
            printf("enter edge: ");
            scanf_s("%d", &c);
            add_node(gr1, c);
            break;
        case 2:
            printf("begining edge: ");
            scanf_s("%d", &c);
            printf("ending edge: ");
            scanf_s("%d", &k);
            add_arc(gr1, c, k);
            break;
        case 3:
            get_graph_rank(*gr1);
            break;
        }
    } while (menu != 0);

    delete gr1;
}

我希望你能帮助我编写函数的逻辑，该函数将返回节点的值，该节点的最大边连接到它，并且如果有多个连接边数相同的节点，则返回NULL.Thanks in advice .....

Answer 1

看起来代码使用单链接列表节点来表示图节点。这意味着您的节点不能超过1（传出）边缘。不确定这样的图表表示是否有用。

另一种方法是表示图形是使用数组作为图形节点值并将边缘表示为adjacency matrix N-by-N，其中每个元素都指定是否存在从节点A到节点B的边缘。

获取连接到图中节点的最大边数

1 个答案: