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Auto comment: topic has been updated by Diego.Gabriel (previous revision, new revision, compare). |
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oh, no worries. Do you think the solution I explained above works? |
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+1
But in sample test all edges are bridges... :(... I was thinking something like run 2 dfs to compute for every node the number of outgoing paths to S and T, one on the original graph (S->T) and the other on the inverted graph (T->S), then multiply both results for each node... Then the answer would be all nodes having the same values than S and T. But i'm not really sure if that really works... |
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The answer is 2 xD |
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