judaybhaskerreddy's blog

help needed for cses prbm(calculating the time complexity for k shortest loopy paths)
By judaybhaskerreddy, history, 2 years ago, In English

i solved cses(graph) flight paths qn( https://cses.fi/problemset/task/1196 ), there is an explanation given in gfg(https://www.geeksforgeeks.org/1st-to-kth-shortest-path-lengths-from-node-1-to-n-in-given-graph/) i couldnt find out how they calculated the time complexity as O((n+m)klogk)) cuz i am getting a diff time complexity...can some one explain how did they come up with that time complexity.....thanks in advance..

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2 years ago, # |
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Your URL broken, you put ) at the end of URL

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2 years ago, # |
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In standard dijkstra's, we maintain the shortest route. In this problem, we'll only consider the k shortest path to reach each vertex. It could be save in 2-dimension array dist[n][k] where dist[i][j] is the j shortest path to reach vertex i. The initial state is dist[1][1] = 0. We can find the result by modifying dijkstra a little bit. My submission

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    2 years ago, # ^ |
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    i got the AC for the qn but the thing is how did the time complexity for the problem became O((n+m)klogk) as given in gfg i got a different result...what is the time complexity did you get

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