arrch's blog

By arrch, history, 10 years ago, In English

Hello cf! I am solving this problem http://www.spoj.com/problems/HORRIBLE/ . And I don't know how to implement BIT with range modification. Please help me. Thanks.

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You should read about Segment Tree. This blog relly helped me to understand it — segment tree!

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there are great tutorial on russian: tutorial, you can translate it with any translator.

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There is a variant of BIT Trees that does range_update & point_query we will use this one

To update a range [l, r] with value V (incrementing every element in range by V), we add V to bit[l] and subtract it from bit[r + 1], since bit tree returns for we always get the right results for point queries

Now in order to make range_queries and range_updates, lets have a look at our array a:

[0, 0, 0, 0, 0, 0]

Calling update(l = 3, r = 5, v = 5) should make it like this:

[0, 0, 5, 5, 5, 0]

And prefix_sum array: [0, 0, 5, 10, 15, 15]

Let's divide the array into 3 parts:

a[1, ..., l - 1] query(x) should return 0

a[l, ..., r] query(x) should return V × (x - (l - 1)) = Vx - V(l - 1)

a[r + 1, ..., N] query(x) should return V × (r - (l - 1)) = Vr - V(l - 1)

So basically, query(x) is now like a linear function of index x: f(x) = Ax + B

The terms that depend on x can be stored in a bit tree and other terms in another bit tree

Pseudo_code:


update_point(bit, x, v): while(x <= N): bit[x] += v x += x&-x update_range(l, r, v): update_point(bitAX, l, v) update_point(bitAX, r+1, -v) update_point(bitB, l, -v * (l-1)) update_point(bitB, r+1, v * r) query_point(bit, x) sum = 0 while(x > 0): sum += bit[x] x += x&-x query(x): return x * query_point(bitAX, x) + query_point(bitB, x) query(l, r): return query(r) + query(l-1)
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Read this Blog : Various Usage Of BIT