如何为无环有向图的顶点分配“层”？

Question

I have an acyclic directed graph. I would like to assign levels to each vertex in a manner that guarantees that if the edge (v1,v2) is in the graph, then level(v1) > level(v2). I would also like it if level(v1) = level(v3) whenever (v1,v2) and (v3,v2) are in the graph. Also, the possible levels are discrete (might as well take them to be the natural numbers). The ideal case would be that level(v1) = level(v2) + 1 whenever (v1,v2) is in the graph and there is no other path from v1 to v2, but sometimes that isn't possible with the other constraints - eg, consider a graph on five vertices with the edges (a,b) (b,d) (d,e) (a,c) (c,e).
Does anyone know a decent algorithm to solve this? My graphs are fairly small (|V| <= 25 or so), so I don't need something blazing fast - simplicity is more important.

My thinking so far is to just find a least element, assign it level 0, find all parents, assign them level 1, and resolve contradictions by adding +0.5 to the appropriate vertices, but this seems pretty awful.

Also, I get the feeling that it might be helpful to remove all "implicit" edges (ie, remove (v1,v3) if the graph contains both (v1,v2) and (v2,v3).

Answer 1

I think letting the level of v be the length of the longest directed path from v might work well for you. In Python:

# the level of v is the length of the longest directed path from v
def assignlevel(graph, v, level):
    if v not in level:
        if v not in graph or not graph[v]:
            level[v] = 0
        else:
            level[v] = max(assignlevel(graph, w, level) + 1 for w in graph[v])
    return level[v]

g = {'a': ['b', 'c'], 'b': ['d'], 'd': ['e'], 'c': ['e']}
l = {}
for v in g:
    assignlevel(g, v, l)
print l

Output:

{'a': 3, 'c': 1, 'b': 2, 'e': 0, 'd': 1}

Answer 2

您可以使用拓扑排序为具有所需属性的每个顶点分配唯一的编号。类似地，您可以按拓扑顺序遍历节点并分配max（parents）+ 1