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An example for lazy segment trees.. to read full topic visit http://se7so.blogspot.com
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/** | |
* In this code we have a very large array called arr, and very large set of operations | |
* Operation #1: Increment the elements within range [i, j] with value val | |
* Operation #2: Get max element within range [i, j] | |
* Build tree: build_tree(1, 0, N-1) | |
* Update tree: update_tree(1, 0, N-1, i, j, value) | |
* Query tree: query_tree(1, 0, N-1, i, j) | |
*/ | |
#include<iostream> | |
#include<algorithm> | |
using namespace std; | |
#include<string.h> | |
#include<math.h> | |
#define N 20 | |
#define MAX (1+(1<<6)) // Why? :D | |
#define inf 0x7fffffff | |
int arr[N]; | |
int tree[MAX]; | |
int lazy[MAX]; | |
/** | |
* Build and init tree | |
*/ | |
void build_tree(int node, int a, int b) { | |
if(a > b) return; // Out of range | |
if(a == b) { // Leaf node | |
tree[node] = arr[a]; // Init value | |
return; | |
} | |
build_tree(node*2, a, (a+b)/2); // Init left child | |
build_tree(node*2+1, 1+(a+b)/2, b); // Init right child | |
tree[node] = max(tree[node*2], tree[node*2+1]); // Init root value | |
} | |
/** | |
* Increment elements within range [i, j] with value value | |
*/ | |
void update_tree(int node, int a, int b, int i, int j, int value) { | |
if(lazy[node] != 0) { // This node needs to be updated | |
tree[node] += lazy[node]; // Update it | |
if(a != b) { | |
lazy[node*2] += lazy[node]; // Mark child as lazy | |
lazy[node*2+1] += lazy[node]; // Mark child as lazy | |
} | |
lazy[node] = 0; // Reset it | |
} | |
if(a > b || a > j || b < i) // Current segment is not within range [i, j] | |
return; | |
if(a >= i && b <= j) { // Segment is fully within range | |
tree[node] += value; | |
if(a != b) { // Not leaf node | |
lazy[node*2] += value; | |
lazy[node*2+1] += value; | |
} | |
return; | |
} | |
update_tree(node*2, a, (a+b)/2, i, j, value); // Updating left child | |
update_tree(1+node*2, 1+(a+b)/2, b, i, j, value); // Updating right child | |
tree[node] = max(tree[node*2], tree[node*2+1]); // Updating root with max value | |
} | |
/** | |
* Query tree to get max element value within range [i, j] | |
*/ | |
int query_tree(int node, int a, int b, int i, int j) { | |
if(a > b || a > j || b < i) return -inf; // Out of range | |
if(lazy[node] != 0) { // This node needs to be updated | |
tree[node] += lazy[node]; // Update it | |
if(a != b) { | |
lazy[node*2] += lazy[node]; // Mark child as lazy | |
lazy[node*2+1] += lazy[node]; // Mark child as lazy | |
} | |
lazy[node] = 0; // Reset it | |
} | |
if(a >= i && b <= j) // Current segment is totally within range [i, j] | |
return tree[node]; | |
int q1 = query_tree(node*2, a, (a+b)/2, i, j); // Query left child | |
int q2 = query_tree(1+node*2, 1+(a+b)/2, b, i, j); // Query right child | |
int res = max(q1, q2); // Return final result | |
return res; | |
} | |
int main() { | |
for(int i = 0; i < N; i++) arr[i] = 1; | |
build_tree(1, 0, N-1); | |
memset(lazy, 0, sizeof lazy); | |
update_tree(1, 0, N-1, 0, 6, 5); // Increment range [0, 6] by 5 | |
update_tree(1, 0, N-1, 7, 10, 12); // Incremenet range [7, 10] by 12 | |
update_tree(1, 0, N-1, 10, N-1, 100); // Increment range [10, N-1] by 100 | |
cout << query_tree(1, 0, N-1, 0, N-1) << endl; // Get max element in range [0, N-1] | |
} |
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Thanks for your effort. I really appreciate your work. this code has taught me implementation of lazy propagation very easily.