Question

因此，我写了一种算法，可以找到无向，无权图中唯一不同的最短路径的数量。我相信这是正确的，而且我很难弄清运行时间是多少。非常感谢所有帮助！

shortestPaths(undirected graph G, startNode v, endNode end)
    initialize  all levels of nodes to -1
    count = 1;
    initialize queue Q = {v}
    level(v) = 0

    while Q is not empty
        curr = pop(Q)

        if( curr == end)
            w = pop(Q)
            while(level(w)==level(curr))
                if(w == end)
                    count++
                w=pop(Q)
            return count

        for all neighbors (nxt) of node curr do 
            if( level(nxt) == -1 )
                level(nxt) = level(curr) + 1
                Push(Q, nxt)
            else if( level(nxt) == level(curr) + 1 )
                Push(Q, nxt)
            else if( level(nxt) > level(curr) + 1)
                Level(nxt) = Level(curr) + 1
    return count

Answer 1

算法的运行时间是指数的。您可以通过不将顶点多次推入堆中，而应将计数器与每个顶点相关联，并在到达该顶点的每个新路径中递增计数器来避免这种情况。

尝试这样的事情：

initialize  all counts of nodes to 0                     // added
counts(v) = 1                                            // added
...
while Q is not empty
    curr = pop(Q)

    if( curr == end)
        return counts(curr)

    for all neighbors (nxt) of node curr do 
        if( level(nxt) == -1 )
            level(nxt) = level(curr) + 1
            counts(nxt) = counts(curr)                   // added
            Push(Q, nxt)
        else if( level(nxt) == level(curr) + 1 )
            counts(nxt) += counts(curr)                  // added & removed
return 0

这与BFS-O(|E|+|V|)具有相同的复杂性。

最短路径算法的时间复杂度

1 个答案: