拓扑排序以找到到 t 的路径数 [英] Topological sort to find the number of paths to t

问题描述

我必须开发与拓扑排序相关的 O(|V|+|E|) 算法，该算法在有向无环图 (DAG) 中确定从图的每个顶点到 t 的路径数(t 是出度为 0 的节点).我对 DFS 做了如下修改:

I have to develop an O(|V|+|E|) algorithm related to topological sort which, in a directed acyclic graph (DAG), determines the number of paths from each vertex of the graph to t (t is a node with out-degree 0). I have developed a modification of DFS as follow:

DFS(G,t):
    for each vertex u ∈ V do
        color(u) = WHITE
        paths_to_t(u) = 0
    for each vertex u ∈ V do
        if color(u) == WHITE then
            DFS-Visit(u,t)

DFS-Visit(u,t):
    color(u) = GREY
    for each v ∈ neighbors(u) do
        if v == t then
            paths_to_t(u) = paths_to_t(u) + 1
        else then
            if color(v) == WHITE then
                DFS-Visit(v)
            paths_to_t(u) = paths_to_t(u) + paths_to_t(v)
    color(u) = BLACK

但我不确定这个算法是否与拓扑排序有关，或者我是否应该从另一个角度重构我的工作.

But I am not sure if this algorithm is related to topological sort or if should I restructure my work with another point of view.

推荐答案

可以使用动态规划和拓扑排序来完成，如下所示:

It can be done using Dynamic Programming and topological sort as follows:

Topological sort the vertices, let the ordered vertices be v1,v2,...,vn
create new array of size t, let it be arr
init: arr[t] = 1
for i from t-1 to 1 (descending, inclusive):
    arr[i] = 0  
    for each edge (v_i,v_j) such that i < j <= t:
         arr[i] += arr[j]

完成后，对于[1,t]中的每个i，arr[i]表示从vi 到 vt

When you are done, for each i in [1,t], arr[i] indicates the number of paths from vi to vt

现在，证明上述主张很容易(与您的算法相比，我不知道它是否正确以及如何证明它)，它是通过归纳完成的:

Now, proving the above claim is easy (comparing to your algorithm, which I have no idea if its correct and how to prove it), it is done by induction:

Base: arr[t] == 1，并且确实有一条从 t 到 t 的路径，即空路径.
假设:对于 m <范围内的每个 k 的说法都是正确的.k <= t
证明:我们需要证明 m 的声明是正确的.
让我们看看 vm 的每个出边:(v_m,v_i).
因此，使用这条边(v_m,v_i)的从v_m开始到vt的路径数.正是 arr[i](归纳假设).将 v_m 中出边的所有可能性相加，得到从 v_m 到 v_t 的路径总数——这正是算法所做的.
因此，arr[m] = #paths from v_m to v_t

Base: arr[t] == 1, and indeed there is a single path from t to t, the empty one.
Hypothesis: The claim is true for each k in range m < k <= t
Proof: We need to show the claim is correct for m.
Let's look at each out edge from vm: (v_m,v_i).
Thus, the number of paths to vt starting from v_m that use this edge (v_m,v_i). is exactly arr[i] (induction hypothesis). Summing all possibilities of out edges from v_m, gives us the total number of paths from v_m to v_t - and this is exactly what the algorithm do.
Thus, arr[m] = #paths from v_m to v_t

QED

时间复杂度:
第一步(拓扑排序)需要O(V+E).
循环迭代所有边一次，所有顶点一次，所以也是O(V+E).
这给了我们O(V+E)

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