**I think lots of resource in online about this topics. But i thought i can explain it with a better way.**

## When Centroid Decomposition comes?

- Suppose a problem says that how many paths have length exactly k in a tree.
- Or, how many paths have xor k.
- Sum of all xor path in a tree.
- update a node Black to white or white to black. now query the shortest path from a node to a white node. etc.

**So , it is clear , when a paths problem comes then we can use Centroid Decomposition.**

**For single node query or from a specific node , Dsu on Tree may do it.**

## Algorithm:

### 1. Find a centroid of current tree T.

`What is Centroid?`

Simply Centroid is a node, if we delete it. It makes some subtrees where every subtree size must be less than **sz/2** { sz is the size of current tree T.}

`How can we find it?`

=> Take a node random node of current tree T. Now if its every subtree size less than sz/2. Then it is a centroid. => If not, go to highest size of subtree.

**Note : In a tree only two Centroid possible From Jordan Theorem . If there is two centroid. you can take any. Cause those two centroid look like same**

=> Same thing must be done.

In this graph. If we start from node 1. Then Size[2]=7 which is larger than 11/2. so 1 is not centroid. Go to the 2. Now every subtree size less than 11/2. So for this graph centroid is 2.

**Here is the code for finding centroid of current Tree.**

/// 'u' will give us the centroid
int u = _u;
while (true) {
int nu = -1;
for (int v : e[u]) {
if (!tocheck[v] || v == p[u]) continue;
if (1 + size[v] > sz / 2) nu = v;
}
if (sz — 1 — size[u] > sz / 2 && p[u] != -1
&&tocheck[p[u]]) nu = p[u];
if (nu != -1) u = nu; else break;
}
/// tocheck array check whether this node already done as a centroid of any tree or not
/// For size array you need just a dfs call
void dfs(int u) {
for (int v : e[u]) {
if (v == p[u]) continue;
p[v] = u;
dfs(v);
size[u] += 1 + size[v];
}
}

### 2. Problem Solving part.

Before Deleting centroid. we must find the answer for current Tree T.

**You must solve this part with O(sz) time**

suppose we need to find how many paths length equal k. Then we need to compute for current tree T.

**how many paths go through the centroid of T which length equal k?**

Take your time and think about it. If we call a dfs from centroid node and find the length of all node from centoid , then it will be easiar for us.

```
void dfs2(int u,int p,int val,bool flag)
{
if(flag) cnt[val]++;
else cnt[val]--;
for(auto v : e[u])
{
if(tocheck[v] && v!=p)
{
dfs2(v,u,val+1,flag);
}
}
}
void solve(int u,int p,int val)
{
if(val>k) return ;
sol+=cnt[k-val];
for(auto v : e[u])
{
if(tocheck[v] && v!=p)
{
solve(v,u,val+1);
}
}
}
void func(int u,int par)
{
sol=0;
dfs2(u,par,0,1);
sol+=cnt[k];
for(auto X : e[u])
{
if(tocheck[X])
{
dfs2(X,u,1,0);
solve(X,u,1);
dfs2(X,u,1,1);
}
}
ans+=sol/2;
}
```

### 3. Delete the Centroid node C and find the new centroid of all subtrees.

### 4. Repeat same Thing from 1 to 3 for every subtree.

## Property:

- The main property here is every node comes logn times under a centroid. So total complexity
**NlogN**. - Any path property like distance/xor/sum/etc. we can compute (logn+logn)times. Cause Highest depth for centroid Tree is
**logn**. - LCA of any two node in centroid tree
**logn**.

**Full Code is here for finding all path length equal to k**

```
#define ll long long
#define vi vector<int >
#define vil vector<ll >
#define pb push_back
#define fi first
#define sc second
#define pii pair<int , int >
const int N = 500050;
const int INF = 1e9+100;
ll sol,k,ans;
struct CentroidDecomposition {
/// cd for Centroid Tree
/// e for Main tree
vector<vi> cd, &e;
/// tocheck for checking a node is already in centroid tree or not?
vector<bool> tocheck;
/// p for tracking parent of a node
/// cnt for counting length
vi size, p,cnt;
/// centroid Tree root
int root;
CentroidDecomposition(vector<vi> &tree) : e(tree) {
int sz = e.size();
tocheck.assign(sz, true);
col.assign(sz, false);
cd.assign(sz, vi());
p.assign(sz, -1);
cnt.assign(N, 0);
size.assign(sz, 0);
dfs(0);
root = decompose(0, sz,-1);
}
void dfs(int u) {
for (int v : e[u]) {
if (v == p[u]) continue;
p[v] = u;
dfs(v);
size[u] += 1 + size[v];
}
}
/// we can solve it for any amount of k
void dfs2(int u,int p,int val,bool flag)
{
if(flag) cnt[val]++;
else cnt[val]--;
for(auto v : e[u])
{
if(tocheck[v] && v!=p)
{
dfs2(v,u,val+1,flag);
}
}
}
void solve(int u,int p,int val)
{
if(val>k) return ;
sol+=cnt[k-val];
for(auto v : e[u])
{
if(tocheck[v] && v!=p)
{
solve(v,u,val+1);
}
}
}
/// finiding centroid and get answer for this centroid
int decompose(int _u, int sz,int par) {
int u = _u;
while (true) {
int nu = -1;
for (int v : e[u]) {
if (!tocheck[v] || v == p[u]) continue;
if (1 + size[v] > sz / 2) nu = v;
}
if (sz - 1 - size[u] > sz / 2 && p[u] != -1
&&tocheck[p[u]]) nu = p[u];
if (nu != -1) u = nu; else break;
}
for (int v = p[u]; v != -1 && tocheck[v]; v = p[v])
size[v] -= 1 + size[u];
sol=0;
dfs2(u,par,0,1);
sol+=cnt[k];
for(auto X : e[u])
{
if(tocheck[X])
{
dfs2(X,u,1,0);
solve(X,u,1);
dfs2(X,u,1,1);
}
}
ans+=sol/2;
dfs2(u,par,0,0);
tocheck[u] = false;
for (int v : e[u]) {
if (!tocheck[v]) continue;
int V2 = 1 + size[v];
if (v == p[u]) V2 = sz - 1 - size[u];
cd[u].push_back(decompose(v, V2,u));
}
return u;
}
};
int main(){
int n;
cin >> n>>k;
vector<vi> tree(n, vi());
for (int i = 0; i < n - 1; ++i) {
int a, b;
cin >> a >> b;
a--; b--;
tree[a].push_back(b);
tree[b].push_back(a);
}
CentroidDecomposition cd(tree);
cout<<ans<<endl;
return 0;
}
```

** More problem **

**You can also check this for better understanding**

Centroid Decomposition of a Tree by Tanuj Khattar

`Sorry for bad grammatical issues. And this is my first tutorial blog, so don't take seriously formy mistakes`

** Thank You **

Thanks a lot it helps me.

There can be atmost 2 centroids in a tree. One example is a path of length 4.

Thank you. But we need to choose one. And two node must have same subtree size.

editedAre you fan of youtube channel Mo Vlogs? YoYo-YoYo.

No. just for fun. Nothing else

more problems for practice :)

321C - Ciel the Commander

150E - Freezing with Style

715C - Digit Tree

293E - Close Vertices

776F - Sherlock's bet to Moriarty

833D - Red-Black Cobweb

914E - Palindromes in a Tree

809E - Surprise me!

342E - Xenia and Tree

Count It

Thank you.

EditedIn the problem you solved, each time you find a centroid, you traverse(dfs) its complete-subtree, to find the number of nodes at distance 'i' from the centroid , so is the total time complexity O(N*log(N)) ?

Yes. Cause every node visited at most logn time..

So for n node.. it takes nlogn :)

One more question. Say for each centroid-'x',I traverse its complete subtree,then I traverse all of its children's subtree and keep doing this until I reach the leaves. What will be the time complexity now !?

its nlogn,

ok lets take an example,,

suppose the tree is,

In this tree centroid is 2.. ->now we traverse whole tree for a specific answer,

->then delete 2 from tree.

-> now we got three subtree ,

now, if we take subtree which contain 1,3,6,7 node.. and we go through this process , we got 3 as a centroid.

now again we got three subtree 1 , 6 , 7. and we delete 3 from the tree.

so its clear every time we divide a tree = how many adjacent node a centroid have.. they must be smaller than sz/2 . so its easy to check the complexity.

if we thought about binary tree, ( cause binary tree creates highest compexity here and you can understand easily)

from 1 centroid we delete one node get two subtrees (2 centroid) ,, take complexity O(n)

from 2 centroid we delete two nodes get 4 subtrees ( 4 centroid) ,, take compexity O(n-1)

from 4 centroid we delete 4 nodes, get 8 subtrees (8 centroid) ,, take complexity O(n-3) // 3 nodes already deleted before this process start

from 8 -----> 16 subtrees----> 16 centroid----> complexity O(n-7) // 7 nodes deleted

so if we sum all we get = O(n) + O(n-1) + O(n-3) + O(n-7) + O(n-15)+....... == nlogn

if this is not a binary tree, this complexity will be less than nlogn..

I think you can understand it.

thank you

Doing centroid decomposition may result in completely another tree, and the paths count or their length in the centroid tree for the particular node can differ the paths count/lengths in the original tree. I just don't understand this part. Could you please share a bit more details about this?

If you draw a centroid tree for any tree, it may not look anything like the tree at all. Instead it is just a decomposition of the tree that has some different properties (like same LCAs, NlogN paths etc.). You can retain the general information about the nodes from the given tree and carry them to the decomposed tree, and with the given advantage of logN height, some queries can be answered fast.

So that means we should get and store this info (lengths, etc...) while building the centroid tree and not after it. Is that correct?

Yeah. While or before building the centroid tree.

Thank you, now I got the point!