Graph serialization

问题

I'm looking for a simple algorithm to 'serialize' a directed graph. In particular I've got a set of files with interdependencies on their execution order, and I want to find the correct order at compile time. I know it must be a fairly common thing to do - compilers do it all the time - but my google-fu has been weak today. What's the 'go-to' algorithm for this?

回答1:

Topological Sort (From Wikipedia):

In graph theory, a topological sort or topological ordering of a directed acyclic graph (DAG) is a linear ordering of its nodes in which each node comes before all nodes to which it has outbound edges. Every DAG has one or more topological sorts.

Pseudo code:

L ← Empty list where we put the sorted elements
Q ← Set of all nodes with no incoming edges
while Q is non-empty do
    remove a node n from Q
    insert n into L
    for each node m with an edge e from n to m do
        remove edge e from the graph
        if m has no other incoming edges then
            insert m into Q
if graph has edges then
    output error message (graph has a cycle)
else 
    output message (proposed topologically sorted order: L)

回答2:

I would expect tools that need this simply walk the tree in a depth-first manner and when they hit a leaf, just process it (e.g. compile) and remove it from the graph (or mark it as processed, and treat nodes with all leaves processed as leaves).

As long as it's a DAG, this simple stack-based walk should be trivial.

回答3:

If the graph contains cycles, how can there exist allowed execution orders for your files? It seems to me that if the graph contains cycles, then you have no solution, and this is reported correctly by the above algorithm.

回答4:

I've come up with a fairly naive recursive algorithm (pseudocode):

Map<Object, List<Object>> source; // map of each object to its dependency list
List<Object> dest; // destination list

function resolve(a):
    if (dest.contains(a)) return;
    foreach (b in source[a]):
        resolve(b);
    dest.add(a);

foreach (a in source):
    resolve(a);

The biggest problem with this is that it has no ability to detect cyclic dependencies - it can go into infinite recursion (ie stack overflow ;-p). The only way around that that I can see would be to flip the recursive algorithm into an interative one with a manual stack, and manually check the stack for repeated elements.

Anyone have something better?

来源：https://stackoverflow.com/questions/4168/graph-serialization