Range Queries Part 3: Line Sweep, Subtree Queries, and Extra Topic in Segment Trees Lucca Siaudzionis and Jack Spalding-Jamieson 2020/02/11 University of British Columbia
Announcements • A3 is released. It’s due Sunday, March 1st. • You will have all the necessary material background by the end of Thursday, but you will be able to almost all of the questions after today. • Piazza on the due date: we will no longer be responding to Piazza problems on the day of the due-date. • Please start on the assignments earlier. (You have three weeks for the next one!) 1
Project/Written Report • A description and rubric for the project are posted on the course website! • Topics should be approved before Tuesday, March 3rd . There is a list of cool topics on the course website. • You should take a look at some of them today . 2
Rectangles: Problem Description Input : Up to n ≤ 10 6 axis-aligned rectangles with coordinates in the space − 10 5 ≤ x , y ≤ 10 5 , some of which may overlap. Output : The total area covered by any rectangle. 3
Line Sweep (1) Imagine a giant vertical line sweeping from left to right over the rectangles. 4
Line Sweep (2) Imagine a giant vertical line sweeping from left to right over the rectangles. 5
Line Sweep (3) Imagine a giant vertical line sweeping from left to right over the rectangles. 6
Line Sweep (4) Imagine a giant vertical line sweeping from left to right over the rectangles. 7
Line Sweep (5) Imagine a giant vertical line sweeping from left to right over the rectangles. 8
Line Sweep (6) Imagine a giant vertical line sweeping from left to right over the rectangles. 9
Line Sweep (7) Imagine a giant vertical line sweeping from left to right over the rectangles. 10
Line Sweep Step As we iterate through the x -coordinates, we will have a current set of intersecting rectangles, which are of the form of vertical segments. The total area we see at each step is the total length of our combined segments. How do we actually keep track of which vertical segments we’re intersecting with? 11
Line Sweep: Segment Tree Solution We can store our current vertical segments with a (lazy) segment tree! Whenever we encounter a rectangle with top coordinate y and height h , we add 1 to the range [ y , y + h ]. Whenever we encounter the end of such a rectangle, we remove 1 from the same range. We then use a new operation in our (lazy) segment tree: The total count of non-zero values. • This is very easy to do, but very hard to analyse (we will not be analysing it in class). • Create a RMQ segment tree that also stores the total length at each node, where length is computed to be the width of the node’s interval if the minimum value is > 0, and is computed recursively otherwise (sum of answer for sub-segments). • Additionally, do not push the lazy value when querying nodes contained fully within the query interval. 12
Line Sweep with a Segment Tree (1) 13
Line Sweep with a Segment Tree (2) 14
Line Sweep with a Segment Tree (3) 15
Line Sweep with a Segment Tree (4) 2 16
Line Sweep with a Segment Tree (5) 2 17
Line Sweep with a Segment Tree (6) 2 2 1 18
Line Sweep with a Segment Tree (7) 2 2 1 19
Line Sweep with a Segment Tree (8) 2 2 1 1 20
Line Sweep with a Segment Tree (9) 2 2 1 1 21
Line Sweep with a Segment Tree (10) 2 2 2 1 1 22
Line Sweep with a Segment Tree (11) 2 2 2 1 1 23
Line Sweep with a Segment Tree (12) 2 2 2 4 1 1 24
Line Sweep with a Segment Tree (13) 2 2 2 4 1 1 25
Line Sweep with a Segment Tree (14) 2 2 2 2 4 1 1 26
Line Sweep with a Segment Tree (15) 2 2 2 2 4 1 1 27
Line Sweep with a Segment Tree (16) 2 2 2 2 4 1 1 28
Discussion Problem (based on ICPC Pac-NW Regional 2018) Input : Given 1 ≤ n ≤ 10 6 rectangle coordinates with − 10 5 ≤ x , y ≤ 10 5 . Output : The total amount of area covered by an odd number of rectangles. 1 2 3 2 1 1 1 29
Discussion Problem (based on ICPC Pac-NW Regional 2018) - Insight You can almost run the exact same algorithm, but now your segment tree has to do a different operation: Output the sum of space that is covered by an odd number of intervals. Notice how the update becomes a range xor. 30
Subtree Queries Input : A tree T with 1 ≤ n ≤ 10 6 nodes, and a value stored at each node, and 1 ≤ q ≤ 10 5 queries of the following forms: Type A: Add x to a node t and everything in the subtree rooted at t . Type B: What is the sum of all nodes in the subtree rooted at t ? 31
Euler Tour: Review Recall Euler tours over trees. 32
Euler Tour: Review 33
Euler Tour: Review 34
Euler Tour: Review 35
Euler Tour: Review 36
Euler Tour: Review 37
Euler Tour: Review 38
Euler Tour: Review 39
Euler Tour: A Single Line This is our final Euler tour. It’s only a single line! 40
Euler Tour: Segment 41
Euler Tour: Segment 42
Euler Tour: Segment 43
Euler Tour: Segment 44
Euler Tour: Segment 45
Euler Tour: Segment 46
Euler Tour: Segment 47
Euler Tour: Segment 48
Euler Tour: Segment 49
Euler Tour: Segment 50
Euler Tour: Segment Tree Create a segment tree over the Euler tour segment! Now we can support range queries and range updates on this segment. We want to support subtree queries and updates. Do these translate to segments? Yes! The segment for the subtree rooted at t is exactly the segment [ l , r ], where l is the first appearance of t in the euler tour, and r is the last appearance. 51
Euler Tour: Subtree Segment 52
Euler Tour: Subtree Segment 53
Euler Tour: Subtree Segment 54
Euler Tour: Subtree Segment 55
Euler Tour: Subtree Segment 56
Euler Tour: Subtree Segment 57
Euler Tour: Subtree Segment 58
Euler Tour: Subtree Segment 59
Euler Tour: Subtree Segment 60
Euler Tour: Subtree Segment 61
Euler Tour: Subtree Segment 62
Euler Tour: Subtree Segment Code // after running, segment tree will be on the range [0,n-1] 1 // subtree queries for subtree root node i will be on the range [l[i],r[i]] 2 void euler(vector<vector<int>> &children, vector<int> &l, vector<int> &r, 3 int i, int &count) { 4 l[i] = count++; 5 for (int c : children[i]) 6 euler(children,l,r,c,count); 7 r[i] = count; 8 } 9 63
Discussion Problem: LCA (again) Input : A tree with 1 ≤ n ≤ 10 5 nodes, 1 ≤ q ≤ 10 5 queries asking for the lowest common ancestor (LCA) of two nodes u , v in the tree. Output : For each query, output the LCA of the two nodes. Can we solve this problem with a segment tree? 64
Discussion Problem: LCA (again) - Insight If we complete the Euler tour as normal (not how it was done in the code from before), then the LCA of u and v is the smallest-depth node appearing in between u and v in the Euler tour (the smallest depth along a segment!). 65
Extra Topic for Segment Trees: Problem Statement Input : An array of size 1 ≤ n ≤ 10 5 , followed by q queries of the following kinds: • Point updates to the array. • Range queries of the form ”How many entries are there in the range [ l , r ] with values in the range [ a , b ]?”. Output : The answer to each query. 66
Extra Topic for Segment Trees: Vectors (1) If we have a sorted list/vector of elements, it’s easy to find how many entries are in the corresponding range: Use binary search to find each endpoint of the range! [ a, b ] = [4 , 10] 4 6 7 7 9 11 15 17 This takes O (log n ). In order to support updates, we need to use an order statistics tree (or a treap) instead of a vector. 67
Extra Topic for Segment Trees: Vectors (2) If we have k sorted lists/vectors, we can repeat this process to find the total count in each range for a total of O ( k log n ) time. Inside each node of a segment tree, we can store the sorted vector of elements in the range of the node: 4 6 7 7 9 11 15 17 4 6 7 11 7 9 15 17 4 7 6 11 7 15 9 17 7 4 6 11 7 15 17 9 This table also gives us the amount of space this segment tree takes up: O ( n log n ). 68
Extra Topic for Segment Trees: Vectors (3) Given this table, we need only find the disjoint set of segments that correspond to our solution. This is done with standard querying for segment trees. There are O (log n ) = k such segments, so the overall complexity of our solution is O ( n log 2 n ) 69
Extra Topic for Segment Trees: Demo (1) 4 6 7 7 9 11 15 17 4 6 7 11 7 9 15 17 4 7 6 11 7 15 9 17 7 4 6 11 7 15 17 9 70
Extra Topic for Segment Trees: Demo (2) 4 6 7 11 7 9 15 17 4 7 6 11 7 15 9 17 7 4 6 11 7 15 17 9 71
Extra Topic for Segment Trees: Demo (3) 7 9 15 17 4 7 6 11 7 15 9 17 7 4 6 11 7 15 17 9 72
Extra Topic for Segment Trees: Demo (4) 7 9 15 17 6 11 7 15 9 17 7 4 6 11 7 15 17 9 73
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