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Lazy
By _Bunny, history, 4 hours ago, In English

Problem: Given an array a1, a2, ..., an (1 <= n <= 1e6) and q queries, for each query has two types.

1 l r x: a[i] = min(a[i], x) (l <= i <= r).

2 i : print value in position i.

I dont know use lazy update to solve operator 1, help me pls (sorry i'm poor E).

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4 hours ago, # |
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Auto comment: topic has been updated by _Bunny (previous revision, new revision, compare).

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4 hours ago, # |
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instead of setting tree[id].min value to x just set it to min(tree[id].min, x)

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4 hours ago, # |
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tr[x].val = min (tr[x] . val , v);

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4 hours ago, # |
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Above problem can be solved using a simple segment tree, without doing lazy propagation since the operation 1 is commutative.

To learn "Lazy propagation" you can refer to "Segment Tree, part-2" of Edu section.

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3 hours ago, # |
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From now, lazy is also a new category just like greedy

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34 minutes ago, # |
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You don't need lazy, Use the approach of the cses Range Update Queries.

Here: Code

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