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Now you are asked to add x to all in the range i=1 and j=4. A solution can be if you update indexes 1, 2, 3, 4 etc separately as before. Then complexity per update is logn and nlogn time for at most n updates. This is not a very good solution, we can update logn here.
Some definitions:
Leaf nodes are the lowest nodes in any tree structure. All subnodes except leaves are internal nodes.
What are the leaf nodes in the segment tree? There is an original value of an index.
What are the internal (except leaf) nodes in the segment tree? There is the merged result of all the leaf nodes below.
The range of a node in the segment tree is the number of nodes whose indexes contain the merged result. For example, the range of node number 3 in the image is 5 to 7.
For example, in the above image, node number 10 has the value of index number 3 and node number 2 has the sum of nodes number 1,2,3,4. What is the need to update so many leaf nodes? From that, can we not come to node number 2 and put this information there “x will be added to all the indices below the current node”? That means when I see that the range of a node is completely inside the query, I can save the propagation value there without going under that node. The propagation value of a node is the sum of all indices within the range of that node.
In addition to the sum, each node has another variable called propagate. The function of this variable is to keep track of how much to add to the leaf nodes below it. The value of propagate is initially zero. Then go to the “relevant nodes” of the range that you want to update and add x with propagate. (As you already know in the previous section, “relevant nodes” are the nodes whose range is inside the entire query)
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Another example, if nodes 2 to 6 need to be updated, I would go to the yellow nodes and tell them to add x to the following nodes:
As with the previous update function, after updating a node, all the nodes in that path should be updated. Let’s look at the code:
struct info {
i64 prop, sum;
} tree[mx * 3]; //In addition to sum, I will keep the additional amount added below in prop
void update(int node, int b, int e, int i, int j, i64 x)
{
if (i > e || j < b)
return;
if (b >= i && e <= j) //Range of nodes inside range update
{
tree[node].sum += ((e – b + 1) * x); //There are e-b+1 nodes below, so e-b+1 times x will be added to this range
tree[node].prop += x; // x will be added to the following nodes
return;
}
int Left = node * 2;
int Right = (node * 2) + 1;
int mid = (b + e) / 2;
update(left, b, mid, i, j, x);
update(right, mid + 1, e, i, j, x);
tree[node].sum = tree[Left].sum + tree[Right].sum + (e – b + 1) * tree[node].prop;
//Updating the path nodes will be updated
// In addition to the left and right side sum, the added value below will be added
}
The difference with the previous update function is that now we are updating a range and if the range of a node is inside the update range, instead of going down, we say that x will be added to the following indices.
Now suppose we call the update function several times. The propagated values of the nodes are updated as follows:
Assume that the nodes whose values have not been entered are null. So the above picture means that x will be added to index 1-4, z will be added to 5-7 and so on.
Now the question is how to query?
Suppose our query range is 1-3. Normally we would add values from our relevant nodes 4 and 10. But we know x is added to everyone below node 2, so x is also added to everyone below node 4! So we add 2*x to the value in range 4, because there are 2 indices in range 4 of the node and x is added to all of them. Similarly, since nodes 2 and 5 have 10 under them, the range of node 10 has index 1 and will be added to (x+y).
That is, it is clear that the sum of the above propagated values should be carried along when visiting a node when querying. Then add a parameter called carry to the query function that will carry the value:
int query(int node, int b, int e, int i, int j, int carry = 0)
{
if (i > e || j < b)
return 0;
if (b >= i and e <= j)
return tree[node].sum + carry * (e – b + 1); // Sum will be added to the range as much as the addition is asked to add
int Left = node << 1;
int Right = (node << 1) + 1;
int mid = (b + e) >> 1;
int p1 = query(Left, b, mid, i, j, carry + tree[node].prop); //carry variable carrying the propagate value
int p2 = query(right, mid + 1, e, i, j, carry + tree[node].prop);
return p1 + p2;
}
The difference with the previous code is in the carry parameter and the return value. The additional value added is added to the initially calculated sum.
This is roughly the story of lazy propagation. Samari stands for:
1. If you want to update the range in O(logn), you have to use lazy propagation.
2. The work of lazy propagation is how the following indexes will be updated if a node falls within the update range without updating the leaf node.
3. When querying the propagation values saved in the above nodes should be fetched by the relevant node and return values accordingly.
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Many times the problem may say to replace all the numbers in a range with x (x is not additive), in that case the value to replace the subnode below should be kept as the propagation value. There will be a slight change in the carry value while doing the query, I leave the responsibility to you to figure out what kind of change will happen.
The main segment tree tutorial ends here. I will discuss 2D segment tree and some problems in next episode. If you have trouble understanding any part of the tutorial, you can contact us by email or better yet, let us know in the comment section.
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