The_dreamer_'s blog

Check if there exists a simple cycle have total weight >= 0
By The_dreamer_, history, 20 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.

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

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    20 months ago, # ^ |
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    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.

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      20 months ago, # ^ |
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      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 $$$p+w$$$.

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        20 months ago, # ^ |
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        But how can we find the shortest distance between a and b ? It still can go on a loop in a negative weighted edge.

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20 months ago, # |
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Can someone help me please. I still can't solve this problem.

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    20 months ago, # ^ |
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    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.

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20 months ago, # |
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problem link??

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20 months ago, # |
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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.

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