Floyd-warshall for longest path problem

Question

I have few doubts related to Longest distance problem.  From Wikipedia -->

The NP-hardness of the unweighted longest path problem can be shown using a reduction from the Hamiltonian path problem: a graph G has a Hamiltonian path if and only if its longest path has length n − 1, where n is the number of vertices in G. Because the Hamiltonian path problem is NP-complete, this reduction shows that the decision version of the longest path problem is also NP-complete. In this decision problem, the input is a graph G and a number k; the desired output is "yes" if G contains a path of k or more edges, and no otherwise.^[1]

If the longest path problem could be solved in polynomial time, it could be used to solve this decision problem, by finding a longest path and then comparing its length to the number k. Therefore, the longest path problem is NP-hard. It is not NP-complete, because it is not a decision problem.^[2]

In weighted complete graphs with non-negative edge weights, the weighted longest path problem is the same as the Travelling salesman path problem, because the longest path always includes all vertices.^[3] 

However, it has a linear time solution for directed acyclic graphs, which has important applications in finding the critical path in scheduling problems.

Means in some special cases Longest path problem comes under P-class. 

So my doubts are -->

• What is the algorithm  For finding Longest path in directed acyclic graphs ?
• Why modified Floyd-Warshall algorithm  could not be used for finding Longest path in general (means why does it fails) ?
  • Some modification like instead of taking min can not we take Max or 
  • Change positive edge weight to negative then use normal algorithm as asked in (https://stackoverflow.com/questions/42500120/floyd-warshall-for-longest-distance-for-undirected-graph)

Comments

For finding longest path in DAG, just negate all the edge weights and then find the shortest path. You can find this in more detail in CLRS book.