By Steven Halim
This ebook is a must-have for each aggressive programmer to grasp in the course of their heart section in their programming profession in the event that they desire to take a breakthrough from being simply one other usual coder to being between one of many best most interesting programmer within the world.
Typical readers of this booklet may be:
1. college scholars who're competing within the annual ACM foreign Collegiate Programming Contest (ICPC)  local Contests (including the area Finals),
2. Secondary or highschool scholars who're competing within the annual foreign Olympiad
in Informatics (IOI)  (including the nationwide level),
3. Coaches who're searching for accomplished education fabric for his or her scholars ,
4. somebody who loves fixing difficulties via laptop courses. there are lots of programming contests in case you are not any longer eligible for ICPC like TopCoder Open, Google
CodeJam, net challenge fixing Contest (IPSC), and so on.
Read or Download Competitive Programming 2 PDF
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Additional resources for Competitive Programming 2
In this section, we discuss the ideas and the implementation examples of these data structures. 1 Graph Graph is a pervasive data structure which appears in many Computer Science problems. Graph in its basic form is simply a collection of vertices and edges (that store connectivity information between those vertices). Later in Chapter 3, 4, and 8, we will explore many important graph problems and algorithms. To get ourselves ready, we discuss three basic ways (there are a few others) to store the information of a graph G with V vertices and E edges in this subsection.
With the segment tree ready, answering an RMQ can now be done in O(log n). The answer for RMQ(i, i) is trivial, which is i itself. But for general cases, RMQ(i, j), further check is needed. Let p1 = RMQ(i, (i + j) / 2) and p2 = RMQ((i + j) / 2 + 1, j). Then RMQ(i, j) is p1 if A[p1] ≤ A[p2], p2 otherwise. For example, we want to answer RMQ(1, 3). 4 (solid lines) is as follows: Start from the root which represents segment [0, 6]. We know that the answer for RMQ(1, 3) cannot be the stored minimum value of segment [0, 6] = 5 as it is the minimum value over a larger23 segment [0, 6] than the RMQ(1, 3).
Thus RMQ(4, 6) = 5. We do not have to traverse the unnecessary parts of the tree! In the worst case, we have two root-to-leaf paths which is just O(2 × log n) = O(log n). For example in RMQ(3, 4) = 4, we have one root-to-leaf path from [0, 6] to [3, 3] and another root-to-leaf path from [0, 6] to [4, 4]. If the array A is static, then using Segment Tree to solve RMQ is overkill as there exists a Dynamic Programming (DP) solution that requires O(n log n) one-time pre-processing and O(1) per RMQ.
Competitive Programming 2 by Steven Halim