如何打印从一个节点到另一个节点的所有可能路径的成本？

我想打印从源节点到目标节点的所有路径，以及这些路径的成本。

到目前为止，我有以下代码：

// Find all paths from source to destination.
void search(Graph* graph, int src, int dst)
{
    // Mark the current node and store it in path.
    graph->visited[src] = true;
    graph->path[graph->index] = src;
    graph->index++;

    // If current vertex is same as destination.
    if (src == dst)
    {
        for (int i = 0; i < graph->index; i++)
        {
            if (i != graph->index - 1)
            {
                printf("%d -> ", graph->path[i] + 1);
            }
            else
            {
                printf("%d = %.3f\n\t", graph->path[i] + 1, graph->distance);
            }
        }
    }
    else
    {
        // For all unvisited vertices adjacent to current vertex.
        for (Node* adj = graph->array[src]; adj; adj = adj->next)
        {
            if (!graph->visited[adj->vertex])
            {
                graph->distance += adj->weight;
                search(graph,adj->vertex, dst);
            }
        }
    }

    // Remove current vertex from path and mark it as unvisited.
    graph->index--;
    graph->distance -= graph->array[graph->index]->weight;
    graph->visited[src] = false;
}

它可以部分工作，正确地打印路径和成本，直到需要返回到以前的节点。此时，程序以错误代码结束。

下面是一个如何发生的示例：

Paths from 1 to 5:
        1 -> 3 -> 5 = 39.576
        1 -> 3 -> 2 -> 5 = 46.172
        1 -> 3 -> 2 -> 4 -> 5 = 52.768

Process finished with exit code -1073741819 (0xC0000005)

预期的结果是算法返回节点2，并从那里搜索其他路径。然后按如下方式打印结果：

Paths from 1 to 5:
        1 -> 3 -> 5 = 39.576
        1 -> 3 -> 2 -> 5 = 46.172
        1 -> 3 -> 2 -> 4 -> 5 = 52.768
        1 -> 2 -> 5 = 39.576
        1 -> 2 -> 4 -> 5 = 46.172

Process finished with exit code 0

有人能帮我解决这个问题吗？