Блог пользователя Tensor

Автор Tensor, 10 лет назад, По-английски
  • have been trying this problem but in vain .

here is my code so far. I have considered the graph as DAG.

any help plese. thanks in advance

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10 лет назад, # |
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The first thing to notice is that you will use at most N-1 edges, otherwise you would create a cycle and that wouldn't be a shortest path, because you could eliminate the cycle from the path and that would result in a shorter one. So now we have at most 150 nodes and 150 edges for query.

Second, if we are at a node X coming from a path with Y edges, we can go to any adjacent node of X using a path with Y+1 edges.

How can we use this in our advantage?

Let's sort the edges by increasing weight and try to relax all of them in this order; each time we relax an edge (u,v) we are getting the shortest path to v using only an increasing path, because the edges we've relaxed so far have a small weight.

But there is still the issue of the amount of edges used, so we can consider each node being a pair <node,num_edges>, then we try to relax all <u,x>, 0 <= x < n-1, reaching all <v,x+1>.

We can notice that if one query starts at node A, any other query that also starts at node A will have already been computed, so we can save the result of each query for further use.

When the relaxing is done, we can see all <B,x>, 0 <= x <= c, and get the small distance.

I'm not sure if I've made myself clear, so feel free to ask anything.

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    10 лет назад, # ^ |
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    thank you for explanation, if you don't mind what do you mean by sorting all edges?

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      10 лет назад, # ^ |
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      I mean that instead of creating a graph just store the edges in an array/vector, then sort them by increasing weight.