Below is the syntax highlighted version of SegmentTree.java
from §9.9 Miscellaneous.
/****************************************************************************** * Compilation: javac SegmentTree.java * Execution: java SegmentTree * * A segment tree data structure. * ******************************************************************************/ import java.util.Arrays; /** * The {@code SegmentTree} class is an structure for efficient search of cummulative data. * It performs Range Minimum Query and Range Sum Query in O(log(n)) time. * It can be easily customizable to support Range Max Query, Range Multiplication Query etc. * <p> * Also it has been develop with {@code LazyPropagation} for range updates, which means * when you perform update operations over a range, the update process affects the least nodes as possible * so that the bigger the range you want to update the less time it consumes to update it. Eventually those changes will be propagated * to the children and the whole array will be up to date. * <p> * Example: * <p> * SegmentTreeHeap st = new SegmentTreeHeap(new Integer[]{1,3,4,2,1, -2, 4}); * st.update(0,3, 1) * In the above case only the node that represents the range [0,3] will be updated (and not their children) so in this case * the update task will be less than n*log(n) * * Memory usage: O(n) * * @author Ricardo Pacheco */ public class SegmentTree { private Node[] heap; private int[] array; private int size; /** * Time-Complexity: O(n*log(n)) * * @param array the Initialization array */ public SegmentTree(int[] array) { this.array = Arrays.copyOf(array, array.length); //The max size of this array is about 2 * 2 ^ log2(n) + 1 size = (int) (2 * Math.pow(2.0, Math.floor((Math.log((double) array.length) / Math.log(2.0)) + 1))); heap = new Node[size]; build(1, 0, array.length); } public int size() { return array.length; } //Initialize the Nodes of the Segment tree private void build(int v, int from, int size) { heap[v] = new Node(); heap[v].from = from; heap[v].to = from + size - 1; if (size == 1) { heap[v].sum = array[from]; heap[v].min = array[from]; } else { //Build childs build(2 * v, from, size / 2); build(2 * v + 1, from + size / 2, size - size / 2); heap[v].sum = heap[2 * v].sum + heap[2 * v + 1].sum; //min = min of the children heap[v].min = Math.min(heap[2 * v].min, heap[2 * v + 1].min); } } /** * Range Sum Query * * Time-Complexity: O(log(n)) * * @param from from index * @param to to index * @return sum */ public int rsq(int from, int to) { return rsq(1, from, to); } private int rsq(int v, int from, int to) { Node n = heap[v]; //If you did a range update that contained this node, you can infer the Sum without going down the tree if (n.pendingVal != null && contains(n.from, n.to, from, to)) { return (to - from + 1) * n.pendingVal; } if (contains(from, to, n.from, n.to)) { return heap[v].sum; } if (intersects(from, to, n.from, n.to)) { propagate(v); int leftSum = rsq(2 * v, from, to); int rightSum = rsq(2 * v + 1, from, to); return leftSum + rightSum; } return 0; } /** * Range Min Query * * Time-Complexity: O(log(n)) * * @param from from index * @param to to index * @return min */ public int rMinQ(int from, int to) { return rMinQ(1, from, to); } private int rMinQ(int v, int from, int to) { Node n = heap[v]; //If you did a range update that contained this node, you can infer the Min value without going down the tree if (n.pendingVal != null && contains(n.from, n.to, from, to)) { return n.pendingVal; } if (contains(from, to, n.from, n.to)) { return heap[v].min; } if (intersects(from, to, n.from, n.to)) { propagate(v); int leftMin = rMinQ(2 * v, from, to); int rightMin = rMinQ(2 * v + 1, from, to); return Math.min(leftMin, rightMin); } return Integer.MAX_VALUE; } /** * Range Update Operation. * With this operation you can update either one position or a range of positions with a given number. * The update operations will update the less it can to update the whole range (Lazy Propagation). * The values will be propagated lazily from top to bottom of the segment tree. * This behavior is really useful for updates on portions of the array * <p> * Time-Complexity: O(log(n)) * * @param from from index * @param to to index * @param value value */ public void update(int from, int to, int value) { update(1, from, to, value); } private void update(int v, int from, int to, int value) { //The Node of the heap tree represents a range of the array with bounds: [n.from, n.to] Node n = heap[v]; /** * If the updating-range contains the portion of the current Node We lazily update it. * This means We do NOT update each position of the vector, but update only some temporal * values into the Node; such values into the Node will be propagated down to its children only when they need to. */ if (contains(from, to, n.from, n.to)) { change(n, value); } if (n.size() == 1) return; if (intersects(from, to, n.from, n.to)) { /** * Before keeping going down to the tree We need to propagate the * the values that have been temporally/lazily saved into this Node to its children * So that when We visit them the values are properly updated */ propagate(v); update(2 * v, from, to, value); update(2 * v + 1, from, to, value); n.sum = heap[2 * v].sum + heap[2 * v + 1].sum; n.min = Math.min(heap[2 * v].min, heap[2 * v + 1].min); } } //Propagate temporal values to children private void propagate(int v) { Node n = heap[v]; if (n.pendingVal != null) { change(heap[2 * v], n.pendingVal); change(heap[2 * v + 1], n.pendingVal); n.pendingVal = null; //unset the pending propagation value } } //Save the temporal values that will be propagated lazily private void change(Node n, int value) { n.pendingVal = value; n.sum = n.size() * value; n.min = value; array[n.from] = value; } //Test if the range1 contains range2 private boolean contains(int from1, int to1, int from2, int to2) { return from2 >= from1 && to2 <= to1; } //check inclusive intersection, test if range1[from1, to1] intersects range2[from2, to2] private boolean intersects(int from1, int to1, int from2, int to2) { return from1 <= from2 && to1 >= from2 // (.[..)..] or (.[...]..) || from1 >= from2 && from1 <= to2; // [.(..]..) or [..(..).. } //The Node class represents a partition range of the array. static class Node { int sum; int min; //Here We store the value that will be propagated lazily Integer pendingVal = null; int from; int to; int size() { return to - from + 1; } } /** * Read the following commands: * init n v Initializes the array of size n with all v's * set a b c... Initializes the array with [a, b, c ...] * rsq a b Range Sum Query for the range [a, b] * rmq a b Range Min Query for the range [a, b] * up a b v Update the [a,b] portion of the array with value v. * exit * <p> * Example: * init * set 1 2 3 4 5 6 * rsq 1 3 * Sum from 1 to 3 = 6 * rmq 1 3 * Min from 1 to 3 = 1 * input up 1 3 * [3,2,3,4,5,6] * * @param args the command-line arguments */ public static void main(String[] args) { SegmentTree st = null; String cmd = "cmp"; while (true) { String[] line = StdIn.readLine().split(" "); if (line[0].equals("exit")) break; int arg1 = 0, arg2 = 0, arg3 = 0; if (line.length > 1) { arg1 = Integer.parseInt(line[1]); } if (line.length > 2) { arg2 = Integer.parseInt(line[2]); } if (line.length > 3) { arg3 = Integer.parseInt(line[3]); } if ((!line[0].equals("set") && !line[0].equals("init")) && st == null) { StdOut.println("Segment Tree not initialized"); continue; } int array[]; if (line[0].equals("set")) { array = new int[line.length - 1]; for (int i = 0; i < line.length - 1; i++) { array[i] = Integer.parseInt(line[i + 1]); } st = new SegmentTree(array); } else if (line[0].equals("init")) { array = new int[arg1]; Arrays.fill(array, arg2); st = new SegmentTree(array); for (int i = 0; i < st.size(); i++) { StdOut.print(st.rsq(i, i) + " "); } StdOut.println(); } else if (line[0].equals("up")) { st.update(arg1, arg2, arg3); for (int i = 0; i < st.size(); i++) { StdOut.print(st.rsq(i, i) + " "); } StdOut.println(); } else if (line[0].equals("rsq")) { StdOut.printf("Sum from %d to %d = %d%n", arg1, arg2, st.rsq(arg1, arg2)); } else if (line[0].equals("rmq")) { StdOut.printf("Min from %d to %d = %d%n", arg1, arg2, st.rMinQ(arg1, arg2)); } else { StdOut.println("Invalid command"); } } } }