update : added another method to code this technique (easy to code but n log ^ 2).

update2 : added another method to code this technique (easy to code and n log).

update3 : bugs in style 2 have fixed. And 2 new problems added.

update4 (15 August 2016) : A new problem (sgu507) added to list.

update5 (20 September) : A new problem added to list.

Hi!

Most of people know about dsu but what is the "dsu on tree" ?

I will explain it and post ends with several problems in CF that can be solved by this technique.

What is the dsu on tree?

With dsu on tree we can answer queries of this type:

How many vertices in subtree of vertice v has some property in O(n lg n) time (for all of the queries).

For example:

Given a tree, every vertice has color. Query is how many vertices in subtree of vertice v are colored with color c?

Lets see how we can solve this problem and similar problems.

First, we have to calculate size of subtree of every vertice. It can be done with simple dfs:

int sz[maxn];
void getsz(int v, int p){
    sz[v] = 1;  // every vertice has itself in its subtree
    for(auto u : g[v])
        if(u != p){
            getsz(u, v);
            sz[v] += sz[u]; // add size of child u to its parent(v)
        }
}

Now we have size of subtree of vertice v in sz[v].

The naive method for solving that problem is this code(that works in O(N ^ 2) time)

int cnt[maxn];
void add(int v, int p, int x){
    cnt[ col[v] ] += x;
    for(auto u: g[v])
        if(u != p)
            add(u, v, x)
}
void dfs(int v, int p){
    add(v, p, 1);
    //now cnt[c] is the number of vertices in subtree of vertice v that has color c. You can answer the queries easily.
    add(v, p, -1);
    for(auto u : g[v])
        if(u != p)
            dfs(u, v);
}

Now, how to improve it? there is two style of coding for this technique.

1. easy to code but O(n log ^ 2).

map<int, int> *cnt[maxn];
void dfs(int v, int p){
    int mx = -1, bigChild = -1;
    for(auto u : g[v])
       if(u != p){
           dfs(u, v);
           if(sz[u] > mx)
               mx = sz[u], bigChild = u;
       }
    if(bigChild != -1)
        cnt[v] = cnt[bigChild];
    (*cnt[v])[ col[v] ] ++;
    for(auto u : g[v])
       if(u != p && u != bigChild){
           for(auto x : *cnt[u])
               (*cnt[v])[x.first] += x.second;
       }
    //now (*cnt)[c] is the number of vertices in subtree of vertice v that has color c. You can answer the queries easily.

}

2. easy to code and O(n lg n).

vector<int> *vec[maxn];
int cnt[maxn];
void dfs(int v, int p, bool keep){
    int mx = -1, bigChild = -1;
    for(auto u : g[v])
       if(u != p && sz[u] > mx)
           mx = sz[u], bigChild = u;
    for(auto u : g[v])
       if(u != p && u != bigChild)
           dfs(u, v, 0);
    if(bigChild != -1)
        dfs(bigChild, v, 1), vec[v] = vec[bigChild];
    vec[v]->push_back(v);
    cnt[ col[v] ]++;
    for(auto u : g[v])
       if(u != p && u != bigChild)
           for(auto x : *vec[u]){
               cnt[ col[x] ]++;
               vec[v] -> push_back(x);
           }
    //now (*cnt)[c] is the number of vertices in subtree of vertice v that has color c. You can answer the queries easily.
    // note that in this step *vec[v] contains all of the subtree of vertice v.
    if(keep == 0)
        for(auto u : *vec[v])
            cnt[ col[u] ]--;
}

3. heavy-light decomposition style : O(n lg n).

int cnt[maxn];
bool big[maxn];
void add(int v, int p, int x){
    cnt[ col[v] ] += x;
    for(auto u: g[v])
        if(u != p && !big[u])
            add(u, v, x)
}
void dfs(int v, int p, bool keep){
    int mx = -1, bigChild = -1;
    for(auto u : g[v])
       if(u != p && sz[u] > mx)
          mx = sz[u], bigChild = u;
    for(auto u : g[v])
        if(u != p && u != bigChild)
            dfs(u, v, 0);  // run a dfs on small childs and clear them from cnt
    if(bigChild != -1)
        dfs(bigChild, v, 1), big[bigChild] = 1;  // bigChild marked as big and not cleared from cnt
    add(v, p, 1);
    //now cnt[c] is the number of vertices in subtree of vertice v that has color c. You can answer the queries easily.
    if(bigChild != -1)
        big[bigChild] = 0;
    if(keep == 0)
        add(v, p, -1);
}

But why it is O(n log n)? You know that why dsu has O(q log n) time (for q queries); the code uses same method. Merge smaller to greater.

If you have heard heavy-light decomposition you will see that function add will go light edges only, because of this, code works in O(n log n) time.

Any problems of this type can be solved with same dfs function and just differs in add function.

Hmmm, this is what you want, problems that can be solved with this technique:

(List is sorted by difficulty and my code for each problem is given, my codes has heavy-light style)

600E - Lomsat gelral : heavy-light decomposition style : 14607801, easy style : 14554536. (I think this is the easiest problem of this technique in CF and it's good to start coding with this problem)

570D - Деревянные запросы : 17961189 (Thanks to Sora233; this problem is also good for start coding.)

sgu507 (This problem is also good for start.)

246E - Братья по крови возвращаются : 15409328

208E - Братья по крови : 16897324

291E - Древесно-строковая задача : See bhargav104's comment below : link

343D - Водяное дерево : 15063078 (Note that problem is not easy and my code doesn't use this technique (dsu on tree), but my friend, AmirAz 's solution to this problem uses this technique : 14904379).

375D - Дерево и запросы : 15449102 (Again note that problem is not easy :)) )

For persian users there is another problem in Shaazzz contest round #4 (season 2016-2017) problem 3 that is hardest problem with this technique I have ever seen!

If you have another problems with this tag, give me to complete the list :)).

And after all, special thanks from PrinceOfPersia who taught me this technique.