Блог пользователя PR_0202

How to use segment tree in another tree
Автор PR_0202, история, 4 года назад, По-английски

I am confused about how to get sum from a node y to its ancestor x using a segment tree. there is query of 10^5 order consists of update and find sum of the nodes between them.!

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4 года назад, # |
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Auto comment: topic has been updated by PR_0202 (previous revision, new revision, compare).

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4 года назад, # |
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Please link the problem.

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    4 года назад, # ^ |
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    This problem is a subproblem of another problem

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      4 года назад, # ^ |
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      You can link that. It’s good to link problems so people know they’re not helping with a problem from an ongoing contest.

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    4 года назад, # ^ |
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    what I did is doing some thing like prefix sum from the root node and for the sum I printed \n tree[y-1].second-tree[x-1].second

    and for update I did this \n

    void update(int y,int z){ tree[y].second+=z; for(int x: tree[y].first){ update(x,z); } } \n I am confused how to do this if O(long) time complexity please help me

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Maintain an array which for every $$$i$$$ will store the sum of all the ancestors(including itself) in $$$array[i]$$$. Now to find the answer for $$$x$$$ and it's anscestor $$$y$$$ it would be equal to $$$array[x]$$$ — $$$array[parentOfY]$$$(similiar as we do in prefix sum from l to r).

Now coming to the update part. When a node value is updated, then what happens? Let $$$x$$$ is updated. Then only the nodes which are in the subtree of $$$x$$$(including itself) are affected, because $$$x$$$ can only be an ancestor of nodes in it's subtree or itself. So when when $$$x$$$ is update do a range update in it's subtree.

Suppose $$$x$$$ was holding value 10($$$a[x]$$$ not $$$array[x]$$$) earlier now it's an update and asks to change it to 5. since all the nodes in it's subtree were holding sum of all of it's ancestor which also includes $$$x$$$ so $$$5 - 10$$$ needs to be added to every nodes in subtree of $$$x$$$ incuding itself. For subtree update you can use preorder traversal.

Fenwick tree would be easier to use here.

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