有向非循环图中两个顶点之间的最大加权路径

Question

Love some guidance on this problem:

G is a directed acyclic graph. You want to move from vertex c to vertex z. Some edges reduce your profit and some increase your profit. How do you get from c to z while maximizing your profit. What is the time complexity?

Thanks!

Answer 1

The problem has an optimal substructure. To find the longest path from vertex c to vertex z , we first need to find the longest path from c to all the predecessors of z . Each problem of these is another smaller subproblem (longest path from c to a specific predecessor).

Lets denote the predecessors of z as u1,u2,...,uk and dist[z] to be the longest path from c to z then dist[z]=max(dist[ui]+w(ui,z)) ..

Here is an illustration with 3 predecessors omitting the edge set weights:

So to find the longest path to z we first need to find the longest path to its predecessors and take the maximum over (their values plus their edges weights to z ).

This requires whenever we visit a vertex u , all of u 's predecessors must have been analyzed and computed.

So the question is: for any vertex u , how to make sure that once we set dist[u] , dist[u] will never be changed later on? Put it in another way: how to make sure that we have considered all paths from c to u before considering any edge originating at u ?

Since the graph is acyclic, we can guarantee this condition by finding a topological sort over the graph. topological sort is like a chain of vertices where all edges point left to right. So if we are at vertex vi then we have considered all paths leading to vi and have the final value of dist[vi] .

The time complexity: topological sort takes O(V+E) . In the worst case where z is a leaf and all other vertices point to it, we will visit all the graph edges which gives O(V+E) .

Answer 2

Let f(u) be the maximum profit you can get going from c to u in your DAG. Then you want to compute f(z) . This can be easily computed in linear time using dynamic programming/topological sorting.

Initialize f(u) = -infinity for every u other than c , and f(c) = 0 . Then, proceed computing the values of f in some topological order of your DAG. Thus, as the order is topological, for every incoming edge of the node being computed, the other endpoints are calculated, so just pick the maximum possible value for this node, ie f(u) = max( f(v) + cost(v, u) ) for each incoming edge (v, u) .

Answer 3

Its better to use Topological Sorting instead of Bellman Ford since its DAG.

Source:- http://www.utdallas.edu/~sizheng/CS4349.d/l-notes.d/L17.pdf

EDIT:-

G is a DAG with negative edges.

Some edges reduce your profit and some increase your profit

• Edges - increase profit - positive value
• Edges - decrease profit - negative value

After TS, for each vertex U in TS order - relax each outgoing edge.

dist[] = {-INF, -INF, ….}
dist[c] = 0 // source

for every vertex u in topological order
  if (u == z) break; // dest vertex
  for every adjacent vertex v of u
    if (dist[v] < (dist[u] + weight(u, v))) // < for longest path = max profit
      dist[v] = dist[u] + weight(u, v)

ans = dist[z];