In the last lecture of **Algorithm Gym** (Data Structures), I introduced you Segment trees.

In this lecture, I want to tell you more about its usages and we will solve some serious problems together. Segment tree types :

## Classic Segment Tree

Classic, is the way I call it. This type of segment tree, is the most simple and common type. In this kind of segment trees, for each node, we should keep some simple elements, like integers or boolians or etc.

This kind of problems don't have update queries on intervals.

Example 1 (Online):

Problem 380C - Sereja and Brackets :

For each node (for example *x*), we keep three integers : 1.`t[x]`

= Answer for it's interval. 2. `o[x]`

= The number of $($s after deleting the brackets who belong to the correct bracket sequence in this interval whit length `t[x]`

. 3. `c[x]`

= The number of $)$s after deleting the brackets who belong to the correct bracket sequence in this interval whit length `t[x]`

.

Lemma : For merging to nodes 2*x* and 2*x* + 1 (children of node 2*x* + 1) all we need to do is this :

```
tmp = min(o[2 * x], c[2 * x + 1])
t[x] = t[2 * x] + t[2 * x + 1] + tmp
o[x] = o[2 * x] + o[2 * x + 1] - tmp
c[x] = c[2 * x] + c[2 * x + 1] - tmp
```

So, as you know, first of all we need a *build* function which would be this : (as above) (C++ and [*l*, *r*) is inclusive-outclusive )

```
void build(int id = 1,int l = 0,int r = n){
if(r - l < 2){
if(s[l] == '(')
o[id] = 1;
else
c[id] = 1;
return ;
}
int mid = (l+r)/2;
build(2 * id,l,mid);
build(2 * id + 1,mid,r);
int tmp = min(o[2 * id],c[2 * id + 1]);
t[id] = t[2 * id] + t[2 * id + 1] + tmp;
o[id] = o[2 * id] + o[2 * id + 1] - tmp;
c[id] = c[2 * id] + c[2 * id + 1] - tmp;
}
```

For queries, return value of the function should be 3 values : *t*, *o*, *c* which is the values I said above for the intersection of the node's interval and the query's interval (we consider query's interval is [*x*, *y*) ), so in C++ code, return value is a `pair<int,pair<int,int> >`

(`pair<t, pair<o,c> >`

) :

```
typedef pair<int,int>pii;
typedef pair<int,pii> node;
node segment(int x,int y,int id = 1,int l = 0,int r = n){
if(l >= y || x >= r) return node(0,pii(0,0));
if(x <= l && r <= y)
return node(t[id],pii(o[id],c[id]));
int mid = (l+r)/2;
node a = segment(x,y,2 * id,l,mid), b = segment(x,y,2 * id + 1,mid,r);
int T, temp, O, C;
temp = min(a.y.x , b.y.y);
T = a.x + b.x + temp;
O = a.y.x + b.y.x - temp;
C = a.y.y + b.y.y - temp;
return node(T,pii(O,C));
}
```

Example 2 (Offline): Problem KQUERY

Imagine we have an array *b*_{1}, *b*_{2}, ..., *b*_{n} which, and *b*_{i} = 1 if an only if *a*_{i} > *k*, then we can easily answer the query (*i*, *j*, *k*) in *O*(*log*(*n*)) using a simple segment tree (answer is *b*_{i} + *b*_{i + 1} + ... + *b*_{j} ).

We can do this ! We can answer the queries offline.

First of all, read all the queries and save them somewhere, then sort them in increasing order of *k* and also the array *a* in increasing order (compute the permutation *p*_{1}, *p*_{2}, ..., *p*_{n} where *a*_{p1} ≤ *a*_{p2} ≤ ... ≤ *a*_{pn})

At first we'll set all array *b* to 1 and we will set all of them to 0 one by one.

Consider after sorting the queries in increasing order of their *k*, we have a permutation *w*_{1}, *w*_{2}, ..., *w*_{q} (of 1, 2, ..., *q*) where *k*_{w1} ≤ *k*_{w2} ≤ *k*_{w2} ≤ ... ≤ *k*_{wq} (we keep the answer to the *i* - *th* query in *ans*_{i} .

Pseudo code : (all 0-based)

```
po = 0
for j = 0 to q-1
while po < n and a[p[po]] <= k[w[j]]
b[p[po]] = 0, po = po + 1
```

So, build function would be like this (*s*[*x*] is the sum of *b* in the interval of node *x*) :

```
void build(int id = 1,int l = 0,int r = n){
if(r - l < 2){
s[id] = 1;
return ;
}
int mid = (l+r)/2;
build(2 * id, l, mid);
build(2 * id + 1, mid, r);
s[id] = s[2 * id] + s[2 * id + 1];
}
```

et An update function for when we want to st `b[p[po]] = 0`

to update the segment tree:

```
void update(int p,int id = 1,int l = 0,int r = n){
if(r - l < 2){
s[id] = 0;
return ;
}
int mid = (l+r)/2;
if(p < mid)
update(p, 2 * id, l, mid);
else
update(p, 2 * id + 1, mid, r);
s[id] = s[2 * id] + s[2 * id + 1];
}
```

Finally, a function for sum of an interval

```
int sum(int x,int y,int id = 1,int l = 0,int r = n){// [x, y)
if(x >= r or l >= y) return 0;// [x, y) intersection [l,r) = empty
if(x <= l && r <= y) // [l,r) is a subset of [x,y)
return s[id];
int mid = (l + r)/2;
return sum(x, y, id * 2, l, mid) +
sum(x, y, id*2+1, mid, r) ;
}
```

So, in main function instead of that pseudo code, we will use this :

```
build();
int po = 0;
for(int y = 0;y < q;++ y){
int x = w[y];
while(po < n && a[p[po]] <= k[x])
update(p[po ++]);
ans[x] = sum(i[x], j[x] + 1); // the interval [i[x], j[x] + 1)
}
```

## Lazy Propagation

I told you enough about lazy propagation in the last lecture. In this lecture, I want to solve ans example .

Example : Problem POSTERS.

We don't need all elements in the interval [1, 10^{7}]. The only thing we need is the set *s*_{1}, *s*_{2}, ..., *s*_{k} where for each *i*, *s*_{i} is at least *l* or *r* in one of the queries.

We can use interval 1, 2, ..., *k* instead of that (each query is running in this interval, in code, we use 0-based, I mean [0, *k*) ). For the *i* - *th* query, we will paint all the interval [*l*, *r*] whit color *i* (1-based).

For each interval, if all it's interval is from the same color, I will keep that color for it and update the nodes using lazy propagation.

So,we will have a value *lazy* for each node and there is no any build function (if *lazy*[*i*] ≠ 0 then all the interval of node *i* is from the same color (color *lazy*[*i*]) and we haven't yet shifted the updates to its children. Every member of *lazy* is 0 at first).

A function for shifting the updates to a node, to its children using lazy propagation :

```
void shift(int id){
if(lazy[id])
lazy[2 * is] = lazy[2 * id + 1] = lazy[id];
lazy[id] = 0;
}
```

Update (paint) function (for queries) :

```
void upd(int x,int y,int color, int id = 0,int l = 0,int r = n){//painting the interval [x,y) whith color "color"
if(x >= r or l >= y) return ;
if(x <= l && r <= y){
lazy[id] = color;
return ;
}
int mid = (l+r)/2;
shift(id);
upd(x, y, color, 2 * id, l, mid);
upd(x, y, color, 2*id+1, mid, r);
}
```

So, for each query you should call *upd*(*x*, *y* + 1, *i*) (*i* is the query's 1-base index) where *s*_{x} = *l* and *s*_{y} = *r* .

At last, for counting the number of different colors (posters), we run the code below (it's obvious that it's correct) :

```
set <int> se;
void cnt(int id = 1,int l = 0,int r = n){
if(lazy[id]){
se.insert(lazy[id]);
return ; // there is no need to see the children, because all the interval is from the same color
}
if(r - l < 2) return ;
int mid = (l+r)/2;
cnt(2 * id, l, mid);
cnt(2*id+1, mid, r);
}
```

And answer will be `se.size()`

.

## Segment tree with vectors

In this type of segment tree, for each node we have a vector (we may also have some other variables beside this) .

Example : Online approach for problem KQUERYO (I added this problem as the online version of KQUERY):

It will be nice if for each node, with interval [*l*, *r*) such that *i* ≤ *l* ≤ *r* ≤ *j* + 1 and this interval is maximal (it's parent's interval is not in the interval [*i*, *j* + 1) ), we can count the answer.

For that propose, we can keep all elements of *a*_{l}, *a*_{l + 1}, ..., *a*_{r} in increasing order and use binary search for counting. So, memory will be *O*(*n*.*log*(*n*)) (each element is in *O*(*log*(*n*)) nodes ). We keep this sorted elements in verctor *v*[*i*] for *i* - *th* node. Also, we don't need to run sort on all node's vectors, for node *i*, we can merge *v*[2 * *i*] and *v*[2 * *id* + 1] (like merge sort) .

So, build function is like below :

```
void build(int id = 1,int l = 0,int r = n){
if(r - l < 2){
v[id].push_back(a[l]);
return ;
}
int mid = (l+r)/2;
build(2 * id, l, mid);
build(2*id+1, mid, r);
merge(v[2 * id].begin(), v[2 * id].end(), v[2 * id + 1].begin(), v[2 * id + 1].end(), back_inserter(v[id])); // read more about back_inserter in http://www.cplusplus.com/reference/iterator/back_inserter/
}
```

And function for solving queries :

```
int cnt(int x,int y,int k,int id = 1,int l = 0,int r = n){// solve the query (x,y-1,k)
if(x >= r or l >= y) return 0;
if(x <= l && r <= n)
return v[id].size() - (upper_bound(v[id].begin(), v[id].end(), k) - v[id].begin());
int mid = (l+r)/2;
return cnt(x, y, k, 2 * id, l, mid) +
cnt(x, y, k, 2*id+1, mid, r) ;
}
```

Another example : Component Tree

## Segment tree with sets

In this type of segment tree, for each node we have a `set`

or `multiset`

or `hash_map`

(here) or `unorderd_map`

or etc (we may also have some other variables beside this) .

Consider this problem :

We have *n* vectors, *a*_{1}, *a*_{2}, ..., *a*_{n} and all of them are initially empty. We should perform *m* queries on this vectors of two types :

*A**p**k*Add number*k*at the end of*a*_{p}*C**l**r**k*print the number where*count*(*a*_{i},*k*) is the number of occurrences of*k*in*a*_{i}.

For this problem, we use a segment tree where each node has a `multiset`

, node *i* with interval [*l*, *r*) has a `multiset`

*s*[*i*] that contains each number *k* exactly times (memory would be *O*(*q*.*log*(*n*)) ) .

For answer query *C* *x* *y* *k*, we will print the sum of all *s*_{x}.*count*(*k*) where if the interval of node *x* is [*l*, *r*), *x* ≤ *l* ≤ *r* ≤ *y* + 1 and its maximal (its parent doesn't fulfill this condition) .

We have no build function (because vectors are initially empty). But we need an add function :

```
void add(int p,int k,int id = 1,int l = 0,int r = n){// perform query A p k
s[id].insert(k);
if(r - l < 2) return ;
int mid = (l+r)/2;
if(p < mid)
add(p, k, id * 2, l, mid);
else
add(p, k, id*2+1, mid, r);
}
```

And the function for the second query is :

```
int ask(int x,int y,int k,int id = 1,int l = 0,int r = n){// Answer query C x y-1 k
if(x >= r or l >= y) return 0;
if(x <= l && r <= y)
return s[id].count(k);
int mid = (l+r)/2;
return ask(x, y, k, 2 * id, l, mid) +
ask(x, y, k, 2*id+1, mid, r) ;
}
```

## Segment tree with other data structures in each node

From now, all the other types of segments, are like the types above.

#### 2*D* Segment trees

In this type of segment tree, for each node we have another segment tree (we may also have some other variables beside this) .

#### Segment trees with tries

In this type of segment tree, for each node we have a trie (we may also have some other variables beside this) .

#### Segment trees with DSU

In this type of segment tree, for each node we have a disjoint set (we may also have some other variables beside this) .

Example : Problem 76A - Gift, you can read my source code (8613428) with this type of segment trees .

#### Segment trees with Fenwick

In this type of segment tree, for each node we have a Fenwick (we may also have some other variables beside this) . Example :

Consider this problem :

We have *n* vectors, *a*_{1}, *a*_{2}, ..., *a*_{n} and all of them are initially empty. We should perform *m* queries on this vectors of two types :

*A**p**k*Add number*k*at the end of*a*_{p}*C**l**r**k*print the number for each*j*≤*k*where*count*(*a*_{i},*k*) is the number of occurrences of*k*in*a*_{i}.

For this problem, we use a segment tree where each node has a `vector`

, node *i* with interval [*l*, *r*) has a `set`

*v*[*i*] that contains each number *k* if and only if (memory would be *O*(*q*.*log*(*n*)) ) (in increasing order).

First of all, we will read all queries, store them and for each query of type *A*, we will insert *k* in *v* for all nodes that contain *p* (and after all of them, we sort these vectors using merge sort and run unique function to delete repeated elements) .

Then, for each node *i*, we build a vector *fen*[*i*] with size |*s*[*i*]| (initially 0).

Insert function :

```
void insert(int p,int k,int id = 1,int l = 0,int r = n){// perform query A p k
if(r - l < 2){
v[id].push_back(k);
return ;
}
int mid = (l+r)/2;
if(p < mid)
insert(p, k, id * 2, l, mid);
else
insert(p, k, id*2+1, mid, r);
}
```

Sort function (after reading all queries) :

```
void SORT(int id = 1,int l = 0,int r = n){
if(r - l < 2)
return ;
int mid = (l+r)/2;
SORT(2 * id, l, mid);
SORT(2*id+1, mid, r);
merge(v[2 * id].begin(), v[2 * id].end(), v[2 * id + 1].begin(), v[2 * id + 1].end(), back_inserter(v[id])); // read more about back_inserter in http://www.cplusplus.com/reference/iterator/back_inserter/
v[id].resize(unique(v[id].begin(), v[id].end()) - v[id].begin());
fen[id] = vector<int> (v[id].size() + 1, 0);
}
```

Then for all queries of type *A*, for each node *x* containing *p* we will run :

```
for(int i = a + 1;i < fen[x].size(); i += i & -i) fen[x][i] ++;
```

Where *v*[*x*][*a*] = *k* . Code :

```
void upd(int p,int k, int id = 1,int l = 0,int r = n){
int a = lower_bound(v[id].begin(), v[id].end(), k) - v[id].begin();
for(int i = a + 1; i < fen[id].size(); i += i & -i )
fen[id][i] ++ ;
if(r - l < 2) return;
int mid = (l+r)/2;
if(p < mid)
upd(p, k, 2 * id, l, mid);
else
upd(p, k, 2*id+1, mid, r);
}
```

And now we can easily compute the answer for queries of type *C* :

```
int ask(int x,int y,int k,int id = 1,int l = 0,int r = n){// Answer query C x y-1 k
if(x >= r or l >= y) return 0;
if(x <= l && r <= y){
int a = lower_bound(v[id].begin(), v[id].end(), k) - v[id].begin();
int ans = 0;
for(int i = a + 1; i > 0; i -= i & -i)
ans += fen[id][i];
return ans;
}
int mid = (l+r)/2;
return ask(x, y, k, 2 * id, l, mid) +
ask(x, y, k, 2*id+1, mid, r) ;
}
```

## Segment tree on a rooted tree

As you know, segment tree is for problems with array. So, obviously we should convert the rooted tree into an array. You know DFS algorithm and starting time (the time when we go into a vertex, starting from 1). So, if *s*_{v} is starting time of *v*, element number *s*_{v} (in the segment tree) belongs to the vertex number *v* and if *f*_{v} = *max*(*s*_{u}) + 1 where *u* is in subtree of *v*, the interval [*s*_{v}, *f*_{v}) shows the interval of subtree of *v* (in the segment tree) .

Example : Problem 396C - On Changing Tree

Consider *h*_{v} height if vertex *v* (distance from root).

For each query of first of type, if *u* is in subtree of *v*, its value increasing by *x* + (*h*_{u} - *h*_{v}) × - *k* = *x* + *k*(*h*_{v} - *h*_{u}) = *x* + *k* × *h*_{v} - *k* × *h*_{u}. So for each *u*, if *s* is the set of all queries of first type which *u* is in the subtree of their *v*, answer to query 2 *u* is , so we should calculate two values and , we can answer the queries. So, we for each query, we can store values in all members of its subtree ( [*s*_{v}, *f*_{v}) ).

So for each node of segment tree, we will have two variables and (we don't need lazy propagation, because we only update maximal nodes).

Source code of update function :

```
void update(int x,int k,int v,int id = 1,int l = 0,int r = n){
if(s[v] >= r or l >= f[v]) return ;
if(s[v] <= l && r <= f[v]){
hkx[id] = (hkx[id] + x) % mod;
int a = (1LL * h[v] * k) % mod;
hkx[id] = (hkx[id] + a) % mod;
sk[id] = (sk[id] + k) % mod;
return ;
}
int mid = (l+r)/2;
update(x, k, v, 2 * id, l, mid);
update(x, k, v, 2*id+1, mid, r);
}
```

Function for 2nd type query :

```
int ask(int v,int id = 1,int l = 0,int r = n){
int a = (1LL * h[v] * sk[id]) % mod;
int ans = (hkx[id] + mod - a) % mod;
if(r - l < 2) return ans;
int mid = (l+r)/2;
if(s[v] < mid)
return (ans + ask(v, 2 * id, l, mid)) % mod;
return (ans + ask(v, 2*id+1, mid, r)) % mod;
}
```

## Persistent Segment Trees

In the last lecture, I talked about this type of segment trees, now I just want to solve an important example.

Example : Problem MKTHNUM

First approach : *O*((*n* + *m*).*log*^{2}(*n*))

I won't discuss this approach, it's using binary search an will get TLE.

Second approach : *O*((*n* + *m*).*log*(*n*))

This approach is really important and pretty and too useful :

Sort elements of *a* to compute permutation *p*_{1}, *p*_{2}, ..., *p*_{n} such that *a*_{p1} ≤ *a*_{p2} ≤ ... ≤ *a*_{pn} and *q*_{1}, *q*_{2}, ..., *q*_{n} where, for each *i*, *p*_{qi} = *i*.

We have an array *b*_{1}, *b*_{2}, ..., *b*_{n} (initially 0) and a persistent segment tree on it.

Then *n* step,for each *i*, starting from 1, we perform *b*_{qi} = 1 .

Lest *sum*(*l*, *r*, *k*) be *b*_{l} + *b*_{l + 1} + ... + *b*_{r} after *k* - *th* update (if *k* = 0, it equals to 0)

As I said in the last lecture, we have an array *root* and the root of the empty segment tree, *ir* . So for each query *Q*(*x*, *y*, *k*), we need to find the first *i* such that *sum*(1, *i*, *r*) - *sum*(1, *i*, *l* - 1) > *k* - 1 and answer will be *a*_{pi}. (I'll explain how in the source code) :

Build function (*s* is the sum of the node's interval):

```
void build(int id = ir,int l = 0,int r = n){
s[id] = 0;
if(r - l < 2)
return ;
int mid = (l+r)/2;
L[id] = NEXT_FREE_INDEX ++;
R[id] = NEXT_FREE_INDEX ++;
build(L[id], l, mid);
build(R[id], mid, r);
s[id] = s[L[id]] + s[R[id]];
}
```

Update function :

```
int upd(int p, int v,int id,int l = 0,int r = n){
int ID = NEXT_FREE_INDEX ++; // index of the node in new version of segment tree
s[ID] = s[id] + 1;
if(r - l < 2)
return ID;
int mid = (l+r)/2;
L[ID] = L[id], R[ID] = R[id]; // in case of not updating the interval of left child or right child
if(p < mid)
L[ID] = upd(p, v, L[ID], l, mid);
else
R[ID] = upd(p, v, R[ID], mid, r);
return ID;
}
```

Ask function (it returns *i*, so you should print *a*_{pi} :

```
int ask(int id, int ID, int k, int l = 0,int r = n){// id is the index of the node after l-1-th update (or ir) and ID will be its index after r-th update
if(r - l < 2) return l;
int mid = (l+r)/2;
if(s[L[ID]] - s[L[id]] >= k)// answer is in the left child's interval
return ask(L[id], L[ID], k, l, mid);
else
return ask(R[id], R[ID], k - (s[L[ID]] - s[L[id]] ), mid, r);// there are already s[L[ID]] - s[L[id]] 1s in the left child's interval
}
```

As you can see, this problem is too tricky.

**If there is any error or suggestion let me know.**

Another nice blog. Thanks. And if possible write a blog on Graph + DP.

YES,I would love it to learn DP from PrinceOfPersia's blog

Why there is no section only for algorithms and data structures on CF? I keep 2-3 tabs open all the time to avoid losing posts like this one.

Keep up the good work!

Nice idea. Meanwhile until the idea is implemented, you can click on the star at the end of the post so that it is added to your favorite blogs and you can always get back to it in future. You can also mark you favorite users(concept of friends) , problems, and solutions.

Yes, favorite! Nice feature, thanks (Also it was a memento to read them if I want to close the tab, now I will keep only one tab open just for that). I forgot about it. Now, how do I solve the problem of finding older posts about algorithms and data structures?

here are some

Asnwer -> Answer :-)

Thanks, fixed.

why downvote? He's an author))

Thanks a lot for the wonderful article :)

A very good summary on segment tree! Thanks.

prince of persia did your online for kquery get AC?

I didn't submit it. It was just an example of segment tree with vector .

And by the way, prince, not price.

typo fixed. :)

This was my first idea when I was solving the problem and it didn't get AC(I don't remember it was because of TLE or MLE).

i couldn't get AC with online. i dont know if any online submission would pass.

Use

`getchar_unlocked`

or`buffer`

for reading inputs and`printf`

for printing the output.tried fread(). also, instead of vector, used static arrays. still TLE

it looks like

O((n+m).log^{3}(n)) solution for problem MKTHNUM can fit in time limit! code : )What's your purpose ? Learning algorithm or getting AC ?

Why not both?

If you want both, don't let it go when you got AC, go and learn the best way, instead of kicking my ass with your crap algorithm (we call it TOF in Persian).

The best algorithm is the one that works (i.e. gets AC) and easier to think of and to write.

PrinceOfPersia i'm so pissed watching your comment getting down voted -_-

people can be so ungrateful :/

Thanks again to PrinceOfPersia for this great tutorial

Just a tip,when I was solving 380 Problem C with guidance from the code above, I got snagged at the querying input format.

To get the value of a specific cell 0, the function to call is segment(0,1); Hence since the input adheres to 1-based indexing, segment(l-1,r) will do it.

Nice tutorial!! It will take some time reading all of it though B).

this tutorial extracted the fear of segment tree out of me... thanx :)

Great tutorial!! PrinceOfPersia can u write a blog on BIT? That would be a lot of help!!

Great tutorial! Thanks for your effort.

Did you submit the solution you described here for SPOJ POSTERS? I ask because I am not convinced that a solution using segment trees and co-ordinate compression is possible. The only implementations I have found that try to do this fail on the following test case:

The expected output is

`3`

but segment tree/coordinate compression solutions give`2`

.Can you point me to an implementation using segment trees and co-ordinate compression that can deal with this case? My own attempt is here.

compression can be done for this problem.but you have to be implement it a bit differently. author missed to point it out.

while compressing the highest relative difference has to be 2 or greater, not 1 like general compression.

otherwise failure will occur like you've stated above.

So while compressing you'll have to include the numbers before and after of every 'special' numbers (l or r for every queries)

Then it will work just fine.

here is my implementation. :)

There is a bug in the author's code for POSTERS. Change id = 0 to id = 1 in the upd function. Here's my implementation. Code

Very grateful keep on doing your job!

great job! Can you give me more problems solved by offline SegmentTree/BIT ? Thanks in advance.

According to C++ reference, multiset::count is linear in number of matches. Hence in your subsection — Segment tree with sets — if say 1 is inserted into the first vector n times and the query is to count the number of occurences of 1 in all the vectors, then that query actually takes O(n) time. There can be n such queries resulting in O(n^2) overall running time. Is there a workaround to this?

It is possible to combine AVL tree vs Segment tree to have a new structure called Segment tree with insert and delete operators.

http://codeforces.com/blog/entry/12285

Just for fun! It is quite useless.

PrinceOfPersia this is a wonderful article on segment tree and thanks a lot for this . But seriously i am having hard time understanding the logic used in the solution of very first. :( Can you please EXPLAIN ?

I got the logic . Thanks

how to scale values in a given range say [L,R] with a constant say C,by segment tree or BIT.

Thanks in advance :D

Values are integers → brute force, scaling more than 64 times would cause overflow anyway

Values are floating-point →

could u provide a link to implementation of code and problems on same :D

I don't have those things; since I assume you're talking about the floating-point case, I'll go into a little more detail:

When you have an array

A, instead of addingA[i] to your BIT, add . When you want to scale byCin the range [l,r], instead add in the range [l,r]. When you want to find the value ofA[i], the value is given bye^{A[i]}.Such an helpful post. Thank you very much. If you have enough time please write about line sweeping with segment tree :)

PrinceOfPersia Should'nt it be- build(2 * id + 1,mid+1,r); I think probably you have missed "mid+1" part.

Actually, it is correct as written (it should not be mid+1). The reason is the way the code is written is for the interval [

l,r), and he splits the interval into [l,mid) and [mid,r).Great editorial

AmirMohammad Dehghan. Thanks a lot :)nice editorial

getting WA for posterS ![user:amd] Used seg trees to store all posters from position 1 to max of queries.

Then Lazy propogation is used and after all poster queries did a last passing of lazy values till bottom.

Then after all poster queries counted values at bottom nodes (stored in wall[])

Can anyone come up with test cases where my code getting WA?

sol: http://pastebin.com/KYcUvpku

Thanks for the lecture. I was wondering what is the good approach if the input values are too large to fit in an array (e.g.the order 10^18). The updates are in ranges e.g. add x -> a,b The queries are also in ranges e.g. find sum -> c,d

Compress the numbers :) .

Thanks a lot. That was really helpful.

Shouldn't it be "y"?? Under the section "Segment tree with vectors"??

awesome tutorial, thanks amd!

A great blog on segment tree. But could not figure out Persistent Segment Tree very well :( .it seems for me, i Need more details.

You can read it from here : https://blog.anudeep2011.com/persistent-segment-trees-explained-with-spoj-problems/

shouldn't in sereja and brackets 380 c it should be t[x] = t[2 * x] + t[2 * x + 1] + 2*tmp ?

It depends on the merging step. If you do t[x] = t[2 * x] + t[2 * x + 1] + 2*tmp, then you don't need to print 2*answer. Otherwise you have to print 2*answer because of the merging of both the opening and closing brackets.

How to get number of elements in a range [l,r] which are

greater than xandless then ybysegment tree?you can save numbers [l,r] sorted in each node this can be done O(n.lgn) with merge sort and use binary search to find how many numbers are greater than x and less then y in nodes . each query can be done in O(lg^2 (n) ).

Really nice tutorial.

In "Segment tree with sets" section of the blog, if we use segment tree with each node as multiset, can we handle range updates along with range queries?

For example, In this problem,

We have n vectors, a1, a2, ..., an and all of them are initially empty. We should perform m queries on this vectors of two types :

1) A l r k Add number k to the elements of each a[i], l<=i<=r.

2) C l r k print the sum of the number of occurrences of k in each a[i], l<=i<=r.

How to solve this kind of problem?

Great blog! In case someone had problems for the persistent segment tree (implementation or MKTHNUM problem) they can check this out: https://blog.anudeep2011.com/persistent-segment-trees-explained-with-spoj-problems/

In the problem Sereja and Brackets , the segment tree approach is giving TLE on test case 13

http://codeforces.com/contest/380/problem/C

typedef pair<int,int> pii; typedef pair<int,pii> node;

void buildTree(int * ansArr , int *OArr , int *CArr , string arr , int index , int s , int e){ if(s>e){ return; }

}

node query(int * ansArr , int *OArr , int * CArr , int index , int s , int e , int qs , int qe){ No Overlap if(qs>e || qe<s){ return node(0,pii(0,0)); } Complete Overlap if(s>=qs && e<=qe){ return node(ansArr[index] , pii(OArr[index] , CArr[index])); }

}

int main(){ std::ios::sync_with_stdio(false); string str; cin>>str; int * ansArr = new int[4*str.size()+1]; int * OArr = new int[4*str.size()+1]; int * CArr = new int[4*str.size()+1]; int q; cin>>q; buildTree(ansArr,OArr,CArr ,str ,1 ,0,str.size()-1);

}`

I am trying to solve the problem Component Tree mentioned above but can't find a solution. The brute force approach i.e simulating the problem will take O(NQ) time. The another approach I can think of is to create a vector where each node stores the properties of its ancestor and keeps the latest one in case of a duplicate. But that will take O(N^2).

I can't think of a segment tree kind of solution. I can't find it through web. Any help? Thanks.

For problem KQUERYO I followed the approach mentioned in here, storing sorted vector for each node of the segment tree and using binary search for query. But I'm getting TLE.

Here is my solution.

How can I optimise my solution to make it pass? Any help is appreciated.

Thanks!

Nevermind, I found the problem and corrected it to get AC. My query function was too slow because it was merging nodes for every query. I changed it to return the answer directly by using binary search instead.Here is the AC solution.

I couldn't understand why the reasoning behind this statement sum(1, i, r) - sum(1, i, l - 1) > k - 1 and answer will be api. How do we arrive at the fact that we have to check tree after the rth and l-1th updates?

Never mind, I was confused by the multiple arrays that had been considered. Got it now.

How to find the number of contiguous subsequences in a given range ?

Example :

4 (N)

1 2 3 4 (array[i])

2 4 (Query)

Need to find the number of contiguous subsequences from 2 to 4 whose value is a perfect square.

Explanation :

{ 2 & 3 & 4 } = 0 (Perfect Square)

{ 3 & 4 } = 0 (Perfect Square )

{4 } = 4 (Perfect Square )

Range:

N <=10^5 Q<=5*10^5 (Query)

wait for codechef's long challenge to end, You will find your answer in editorial :)