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UVa Ahoy, Pirates! The case starts with an integer M followed by M pairs of lines. The first line in each pair has an integer T , and the second line has a binary string fragment. For each of the M pairs, you need to concatenate T copies of the string fragment. These instructions tell you to concatenate 10 five times, followed by two times, which produces the binary string After the instructions for creating the binary string, the test case provides an integer Q that indicates the number of operations and queries that follow.
For each of Q lines, there is a capital letter and a range start position and end position in the string. A segment tree is a data structure that is designed to efficiently answer questions about ranges in an array. For an array of integers, a segment tree is often used to answer these questions about the elements in a range:.
The segment tree answers range questions efficiently by storing information about the underlying data in a binary tree. More precisely, it is a complete binary tree, meaning that each level of the tree is filled from left to right so that every level except possibly the last one is completely filled. A complete binary tree can be conveniently represented as an array. This is the same approach used for representing a binary heap.
Each node in a segment tree represents information about a segment in the underlying array. A segment is just a contiguous set of array elements.
The root node node 1 represents the entire array. So nodes 2 and 3, the children of the root node, represent the two halves of the original array. The leaf nodes of the segment tree represent segments of length 1 — i. A node can contain information that helps answer one or more of the questions listed above: the minimum value in the range, the sum of the elements of the range, or some other item of interest to the problem. For this problem, we only care about the sum. If the source data never changes, then the segment tree can be built once and never updated.
For this problem, updates are mixed in with queries. Therefore, we need a way to keep the tree up to date while the underlying binary string changes.
The problem is that keeping the tree completely up to date at all times eliminates all of the performance advantages of the tree, which means our solution will be too slow. The solution is a technique called lazy propagation. At other times, we can just store reminders that updates may need to happen eventually. With the general segment tree design in mind, we can now create a segment tree implementation that is specific to this problem.
The first step is to build the initial segment tree, which represents information about the binary string before any changes have been made. Like most tree algorithms, this one relies on recursion. If the input binary string has length len , then you can build the segment tree by calling build and passing in 1, 0, len-1 : the root node and its range.
Remember that you can calculate the range of each child from the range of the parent. To store node information, the simplest approach is to use an array of Node objects. Each Node object can then track the endpoints of the range, the sum, and any other useful information. Alternatively, you can use separate integer arrays indexed by node position. In my implementation, I used a set of arrays, since I needed some extra performance.
To use a segment tree in problems like this one, where the underlying data can change, we need a way to keep the tree up to date. To do that, we need the concept of lazy propagation. Lazy propagation works by storing a pending value for each node. This value serves as a reminder that updates may need to happen in the future. Without lazy propagation, updating a node could require one or both of its children to be updated as well to keep the tree up to date.
To avoid this costly chain reaction of updates propagating through the tree, we just update the node and two pending values for the left and right children. Instead of propagating these values immediately, we wait until we need the updated information. If we never receive a query that involves those nodes, then we never need to update them. These effects are encapsulated in the following method:. For 1 , the sum will equal the number of elements in the range. To do that propagation, we can use a helper function that knows how to change the data values for a single node.
For set and clear , we can just propagate the action directly. For example, if we clear all bits in a range, then the bits in the first half are cleared, and the bits in the second half are cleared. Therefore, the pending value for both child nodes gets set to clear. Propagating the flip action requires a bit of extra thinking. Since the outcome of a flip operation is different depending on what value it is operating on, we need to consider each child separately.
If a node has a pending value of clear , then at some point we need to clear all of the bits in its range. But if we find that its parent has a flip value that needs to be propagated, we need to reverse that pending operation. So clear becomes set , set becomes clear , and flip becomes do nothing flipping a bit twice returns it to its original value. They could be any sequence of binary digits from the original string. So we set its pending value to flip , which essentially postpones the operation until a later time.
Normally the purpose of a range update is to update all of the values in a range to a single value. UVa adds the complication that we may need to flip the bits rather than just setting or clearing them.
To update a target range to a target value, we use these three steps, always starting with the root node:. We will always call the range sum query by passing in the root node and the target range:. In my experience, most starred uHunt problems use a three-second time limit. There are also some that use one- or two-second limits. This problem has a five-second limit. Nevertheless, I had to do some performance tuning to get my Java implementation to complete in under the time limit.
Here are some ideas on getting your code to run faster:. Although segment trees may not be a standard topic in Computer Science curricula, there is plenty of information about them online. Here are some sources for further reading:. If you made it this far, you should have a good start on creating your own implementation. If you need some help after you have given that a try, you can find my implementation on GitHub.
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Solving UVa 11402 with Segment Trees
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