Floyd Warshall 算法和网格图

Floyd Warshall Algorithm and Grid Graphs

使用维基百科上的伪代码在邻接表表示上实现 floyd warshall 算法,创建了以下代码。该图是一个网格,因此如果它是一个 3 x 3 网格,则顶点 0 有两条边,顶点 1 有 3 条,而顶点 2 有两条,依此类推。

self-> V = 图中的顶点数!!

void floyd(Graph *self, int** dist, int** next)
{
    int i, j, k;
    EdgeNodePtr current;

    current =  malloc(sizeof(current));

    for (i = 0; i < self->V; i++)
    {
        for (j = 0; j < self->V; j++) {
            dist[i][j] = INT_MAX; // Sets all minimun distances to infintiy
            next[i][j] = -1; // Sets all next vertex to a non existant vertex
        }
    }

    for (i = 0; i < self->V; i++)
    {
        for (j = 0; j < self->V; j++)
        {
            current = self->edges[i].head;

            while (current != NULL) // Cycles through all the edges in edgelist
            {
                if (current->edge.to_vertex == j) // If an edge to the correct vertex is found then adds distance
                {
                    dist[i][j] = current->edge.weight;
                    next[i][j] = j; // Updates next node
                }
                current =  current->next;
            }

        }
    }

    PRINT

    // Standard implemnation of floyds algorithm
    for (k = 0; k < self->V; k++)
    {
        for (i = 0; i < self->V; i++)
        {
            for (j = 0; j < self->V; j++)
            {
                if (dist[i][j] > dist[i][k] + dist[k][j])
                {
                    dist[i][j] = dist[i][k] + dist[k][j];
                    next[i][j] = next[i][k];
                }
            }
        }
    }
 PRINT
}

发生的事情是边缘都被正确地插入到距离数组中,通过简单的打印检查。当算法运行时会遇到问题,它将所有距离变成 INT_MINS 或类似的数字。虽然实际上没有计算距离。

我相信网格的结束距离图应该在数组中填充所有可能的距离,除了从顶点到自身的距离为无穷大。

打印列表图的输出图片,其中显示 PRINT

你需要小心int溢出。 Wikipedia pseudocode(在这个答案的底部)使用 "infinity",其中 "infinity + (anything finite) = infinity"。但是,当你用INT_MAX表示"infinity"时,由于溢出就不是这样了。尝试将您的 if 语句条件更改为:

if (dist[i][k] != INT_MAX &&
         dist[k][j] != INT_MAX &&
         dist[i][j] > dist[i][k] + dist[k][j]) {

这将避免 int 溢出向 INT_MAX 添加正距离。

维基百科伪代码:

1 let dist be a |V| × |V| array of minimum distances initialized to ∞ (infinity)
2 for each edge (u,v)
3    dist[u][v] ← w(u,v)  // the weight of the edge (u,v)
4 for each vertex v
5    dist[v][v] ← 0
6 for k from 1 to |V|
7    for i from 1 to |V|
8       for j from 1 to |V|
9          if dist[i][j] > dist[i][k] + dist[k][j] 
10             dist[i][j] ← dist[i][k] + dist[k][j]
11         end if