Question

我有一个作业来实现 Bellman Ford 的算法并在一些图表上对其进行测试。我实现了该算法，在 3 个图中的 2 个上对其进行了测试，并且它有效。但是在第三张图中，我在调用函数时没有输出。

Graph* temaTrecuta = createGraph(10, 12);
addOrientedEdge(temaTrecuta, 0, 0, 1, 5);
addOrientedEdge(temaTrecuta, 1, 1, 2, 3);
addOrientedEdge(temaTrecuta, 2, 2, 3, 5);
addOrientedEdge(temaTrecuta, 4, 3, 9, 5);
addOrientedEdge(temaTrecuta, 5, 1, 9, 1);
addOrientedEdge(temaTrecuta, 6, 3, 4, 1);
addOrientedEdge(temaTrecuta, 7, 4, 8, 5);
addOrientedEdge(temaTrecuta, 8, 8, 7, 1);
addOrientedEdge(temaTrecuta, 9, 7, 5, 1);
addOrientedEdge(temaTrecuta, 10, 7, 6, 3);
addOrientedEdge(temaTrecuta, 11, 6, 0, 1);

这部分创建图形及其边。 createGraph 函数将顶点数和边数作为参数。

void addOrientedEdge(Graph* graph, int index, int source, int destination, int cost) {
    graph->edge[index].src = source;
    graph->edge[index].dest = destination;
    graph->edge[index].cost = cost;

    graph->matrix[source][destination] = cost;
}

这是添加新边的函数。

以下是我对 Bellman Ford 算法的实现。

void bellmanFord(Graph* gr, int src) {
    int* dist = (int*)malloc(sizeof(int) * gr->V);
    int* path = (int*)malloc(sizeof(int) * gr->V);

    if (!path || !dist) {
        printf("Nu am putut aloca.\n");
        exit(1);
    }

    for (int i = 0; i < gr->V; ++i) {
        dist[i] = INT_MAX;
        path[i] = 0;
    }

    path[src] = -1;

    dist[src] = 0;

    for (int i = 1; i <= gr->V - 1; ++i) {
        for (int j = 0; j < gr->E; ++j) {
            int m = gr->edge[j].src;
            int n = gr->edge[j].dest;
            int cost = gr->edge[j].cost;

            if (dist[m] != INT_MAX && dist[m] + cost < dist[n]) {
                dist[n] = dist[m] + cost;
                path[n] = m; 
            }
        }
    }

    for (int i = 0; i < gr->E; ++i) {
        int m = gr->edge[i].src;
        int n = gr->edge[i].dest;
        int cost = gr->edge[i].cost;

        if (dist[m] != INT_MAX && dist[m] + cost < dist[n]) {
            printf("Exista un ciclu negativ.");
            return;
        }
    }

    printBellmanSol(dist, gr->V, path);

    free(dist);
    free(path);
}

Answer 1

由于没有引用边缘索引，只要它是唯一且连续的，就应该考虑自动递增。除了边索引E，还有一个新的边容量。这是传递给 createGraph 的数字，最初设置计数器 E = 0。你可以用少一个参数来写你的 addOrientedEdge ；取下一个边索引。

static void addOrientedEdge(struct Graph* graph, int source, int destination, int cost) {
    const int index = graph->E;
    assert(graph && graph->E < graph->E_capacity);
    graph->edge[index].src = source;
    graph->edge[index].dest = destination;
    graph->edge[index].cost = cost;
    graph->E++;
    graph->matrix[source][destination] = cost;
}

这样您就不必担心边数了。

Answer 2

缺少导致问题的边缘。感谢@user3386109 看到它。

贝尔曼福特图算法

2 个答案: