When submitting a solution in C++, please select either C++14 (GCC 6-32) or C++17 (GCC 7-32) as your compiler. ×

The_dreamer_'s blog

By The_dreamer_, history, 10 months ago, In English

Please help me with this problem: Given an undirected weighted graph have n vertices and m edges (n <= 1000, m <= 10000). How to check if there exists a simple cycle have total weight >= 0. Thanks for your help.

  • Vote: I like it
  • +21
  • Vote: I do not like it

»
10 months ago, # |
  Vote: I like it +10 Vote: I do not like it

You can multiply each edge weight by $$$-1$$$. Then use Bellman Ford's algorithm to find a negative cycle which iirc works in $$$O(n^2 + n \cdot m)$$$.

  • »
    »
    10 months ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    But as far as I know, it can only be used for directed graph. When use it in undirected graph, this algorithm can go on a loop in a negative weighted edge (cycle of length 2) but it not simple cycle.

    • »
      »
      »
      10 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Not sure. There's another approach you can try. It has an $$$O(m^2)$$$ complexity so not sure if it will pass. For eaxh edge from node $$$a$$$ to node $$$b$$$ with weight $$$w$$$, "remove" that edge, then find the shortest distance between $$$a$$$ and $$$b$$$ (assuming there is indeed a path between these after removing the edge). Let the distance found be $$$p$$$. You have a cycle with weight $$$p+w$$$.

      • »
        »
        »
        »
        10 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        But how can we find the shortest distance between a and b ? It still can go on a loop in a negative weighted edge.

»
10 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone help me please. I still can't solve this problem.

  • »
    »
    10 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Sorry I forgot to reply back. You can use modified Bellman Ford's: keep track of the node used to relax the distance for every node, this way you can avoid 2 node cycles.

»
10 months ago, # |
  Vote: I like it 0 Vote: I do not like it

problem link??

»
10 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I think you can use Bellman Ford's algorithm but with vertices {vertex, last edge}, and then you update distance only if new edge is not equal to the last edge used. I think it will still be O(m^2) because every edge creates only 2 new vertices.