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Автор Arpa, история, 19 месяцев назад, По-английски,

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Hi!

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

In Iran, we call this technique "Guni" (the word means "sack" in English), instead of "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 the subtree of vertex v has some property in time (for all of the queries)?

For example:

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

Let's see how we can solve this problem and similar problems.

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

int sz[maxn];
void getsz(int v, int p){
    sz[v] = 1;  // every vertex 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 the size of the subtree of vertex 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 vertex 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 are several styles of coding for this technique.

1. easy to code but .

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];
    else
        cnt[v] = new map<int, int> ();
    (*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[v])[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.

}

2. easy to code and .

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];
    else
        vec[v] = new vector<int> ();
    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[v])[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.
    // note that in this step *vec[v] contains all of the subtree of vertex v.
    if(keep == 0)
        for(auto u : *vec[v])
            cnt[ col[u] ]--;
}

3. heavy-light decomposition style .

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 vertex v that has color c. You can answer the queries easily.
    if(bigChild != -1)
        big[bigChild] = 0;
    if(keep == 0)
        add(v, p, -1);
}

4. My invented style .

This implementation for "Dsu on tree" technique is new and invented by me. This implementation is easier to code than others. Let st[v] dfs starting time of vertex v, ft[v] be it's finishing time and ver[time] is the vertex which it's starting time is equal to time.

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

But why it is ? You know that why dsu has time (for q queries); the code uses the 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 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 - Tree Requests : 17961189 Thanks to Sorasorasora; this problem is also good for start coding.

Sgu507 (SGU is unavailable, read the problem statements here) This problem is also good for the start.

HackerEarth, The Grass Type This problem is also good for start (See bhishma's comment below).

246E - Blood Cousins Return : 15409328

208E - Blood Cousins : 16897324

IOI 2011, Race (See SaSaSaS's comment below).

291E - Tree-String Problem : See bhargav104's comment below.

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

375D - Tree and Queries : 15449102 Again note that problem is not easy :)).

716E - Digit Tree : 20776957 A hard problem. Also can be solved with centroid decomposition.

741D - Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths : 22796438 A hard problem. You must be very familiar with Dsu on tree to solve it.

For Persian users, there is another problem in Shaazzz contest round #4 (season 2016-2017) problem 3 that is a very hard problem with this technique.

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

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

 
 
 
 
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19 месяцев назад, # |
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A2OJ's DSU Section has quite a few tree DSU problems.

Thank you for this post, it explains the theory well and is very easy to read.

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19 месяцев назад, # |
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What does the variable "keep" denote ?

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    19 месяцев назад, # ^ |
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    The way I understand HLD here is basically if a child is the big child, we don't want to recompute answer for it to reduce computation. So we just store the answer for it in the cnt array already so that it's parent doesn't need to re-dfs this subtree. keep denotes whether or not this child is that big child. Please correct me if I'm wrong. :)

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    19 месяцев назад, # ^ |
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    Look at last two lines:

        if(keep == 0)
            add(v, p, -1);
    

    It means that if keep == false after dfs clear vs subtree information from cnt. And if keep == true, don't clear vs subtree information from cnt. In other word if keep == trueafter calling dfs, for each u from subtree of vertice v, col[u] is in cnt (cnt[ col[u] ]++).

    And NibNalin is right. keep is true if and only if v is biggest child of it's parent.

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      15 месяцев назад, # ^ |
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      Hi Arpa, thanks a ton for this awesome post. I have really learnt lot from it.

      I do have one question though. What is the advantage of having keep=false? If that part is kept as it is, without clearing, doesnt the computation become faster? Can you please help clearing this doubt?

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        15 месяцев назад, # ^ |
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        Hi, Thanks.

        Consider vertex v has two children, q and p. If you call dfs for both of them with keep = true, they will mixed up their information, and queries will be incorrectly answered.

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    19 месяцев назад, # ^ |
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    Observations to understand the complexity:

    1. The dfs function visits each node exactly once.
    2. The problem might seem with the add function. You might think that it is making the algorithm n^2. Note that in the add function, we only go down from a vertex to its children if the edge connecting the vertex to the child is a light edge.

    You can think of it in this way, each vertex v will be visited by a call to the add function for any ancestor of v that is connected to a light edge. Since there are at most log(n) light edges going up from any vertex to the root, each vertex will be visited at most log(n) times.

    So the algorithm is: Say you are at a vertex v, first you find the bigchild, then you run dfs on small childs, passing the value of keep as 0. Why? So they are cleared from cnt. Then you run a dfs on bigchild, and you do not clear it from cnt. Now, cnt stores the results for all vertices in the subtree of bigchild (since we cleared cnt for small childs and didn't do so for the bigchild), so we call the add function to "add" the information of children of current vertex that are connected to it via a light edge. Now we are ready to compute the answer

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      5 месяцев назад, # ^ |
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      As you said "add" function goes down using only light edges,Don't these two lines if(bigChild != -1) big[bigChild] = 0; of heavy light decomposition implementation would affect it as if you call "add" after all dfs are done and returned to the root then we only have one heavy edge marked that of root itself others are zero so as "add" goes below it traverses whole tree. Help me here.

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        5 месяцев назад, # ^ |
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        Got it.

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          5 месяцев назад, # ^ |
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          For those who had same doubt as I had: First lets understand why it is wrong to remove it,consider you are at particular node(let it be called A) in recursion,above line is not there,you have 3 children one of them is big child(3rd) while others are normal so you traversed inside 1st and came back to A then you traversed inside 2nd child if you do not have above line then while going inside this children you would have all answer for big children of 1st child which would mess your answer. Now let's understand why complexity is O(n(log(n)):

          Note 1: To calculate complexity you need to measure how many times add function visits the ever y node.

          Note 2: For first add called at a node: A node will visited by add only through its ancestors which are connected by light edges so n nodes log(n) light edges above it this gives us O(n(log(n)))

          Note 3: For second add called at a node: Now somebody may protest that that after above mentioned line(big[bigChild]=0) we are unmarking heavy edge and also calling add after that which may mess up complexity as it travels every node below it which is O(n) but keep==0 condition ensures that for each node there atmost log(n) nodes above in ancestor which have keep=0 function is called.which again gives O(n(log(n)).

          Giving us finally O(n(log(n))) complexity. Follow this link to understand heavy light decomposition's properties: https://blog.anudeep2011.com/heavy-light-decomposition/

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19 месяцев назад, # |
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In second easy with O(n*lg n)

Why if(keep==false) we delete only vertex from main vector

for(auto u : *vec[v])
            cnt[ col[u] ]--;

but we don't delete vertex from cnt which we changed here:

for(auto u : g[v])
       if(u != p && u != bigChild){
           for(auto u : *vec[u])
               cnt[ col[u] ]++;
       }
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    18 месяцев назад, # ^ |
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    There was a mistake in writing. I'm sorry.

    Thanks for reporting this problem.

    code should be:

           if(u != p && u != bigChild){
               for(auto x : *vec[u])
                   cnt[ col[x] ]++;
           }
    

    Instead of:

           if(u != p && u != bigChild){
               for(auto u : *vec[u])
                   cnt[ col[u] ]++;
           }
    

    I have edited that.

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18 месяцев назад, # |
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can someone tell me how 208E is solved with this technique? thanks a lot.

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    18 месяцев назад, # ^ |
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    You need to compute for each pair v p the p-th cousin of v. That is equivalent to finding the number of p-th descendants of the p-th ancestor of v — 1.

    So for each query, replace v p with p_th_ancestor_of_v p. Now you need to store in cnt the number of nodes at a certain depth. In other words, cnt[x] should be equal to number of nodes at depth x in the current subtree.

    Code for Reference: http://codeforces.com/contest/208/submission/17513471

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      18 месяцев назад, # ^ |
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      can't understand why for every vertex v we ans[depth[v]] increase by 1 when we call add function,why must we do it?or why it must be ans[deth[v]] when depth[v] means distance from root to v?

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        18 месяцев назад, # ^ |
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        ans[h] is equal to number of vertices with height h, (with distance h from root).

        Let par, p'th ancestor of v, the answer to query is:

        Consider only subtree of vertice par, print ans[ height[v] ] — 1.

        So with method above we can process all of the queries.

        See my code for better understanding.

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18 месяцев назад, # |
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If i haven't read this article, i wouldn't get ac on this problem. It is another problem which can be solved easily by dsu.

here is my code in HLD-style.

Thanks!

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18 месяцев назад, # |
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I can't understand why the second code is correct...

Consider this example:

We wanna calculate the cnt for Vertex 8. These are the steps:

Going to Vertex 1

Going to Vertex 2 by keep=0

Going to Vertex 3 by keep=0,Vec[3]={3}

Going to Vertex 4 by keep=1,Vec[4]={4},Cnt[color[4]]=1

Going back to Vertex 2,Vec[2]={2,4},Cnt[color[4]]=0,Cnt[color[3]]=1

And then when we go to Vertices 5,6,7,8 still Cnt[color[3]]=1.

Also sorry if I did the steps wrong...

UPD Thank you for editing the blog. My problem fixed.

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17 месяцев назад, # |
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Great post. If you explained the idea before showing the code, it would be better to understand. Also commenting the variables meaning in the code would be of great help.

It would be good to mention that most solutions will answer the queries offline, which may be a problem sometime (maybe someone didn't notice this lol).

Also, it would be nice to post hints about the solutions.

Proving explicitly why it is nlogn would be good too (ie. as each node's subtree set gets merged into one set of size equal or greater, and the base set has size 1 and the last set has size n, then we take logn steps to go from 1 to n. Summarizing, each node gets merged logn times, so the total complexity is O(nlogn)).

Here's my solution to 343D, maybe it will be of help to someone: 18958875. A lot easier to code than the one in the tutorial.

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    16 месяцев назад, # ^ |
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    Thanks for your suggestions first !

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

    And about your code (18958875), anyone has a different opinion !

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      16 месяцев назад, # ^ |
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      Thanks for the reply!

      Yeah, you did prove. People who remember DSU's proof will most likely understand. I stated a more extensive proof would be better thinking about people who don't exactly know the proof. Don't take me wrong, but they may get a little confused reading this proof.

      I mentioned my code exactly because everyone has a different opinion,. Maybe it'll help a later reader, that's all.

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        14 месяцев назад, # ^ |
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        Sorry this might be a stupid question to bring up, but why is the complexity of the heavy-light decomposition style one in O(nlogn)?

        In the case where each node has at most two children: Denote the root node of the tree as u, which is of size s. The child of u connected to the lighter edge is of size at most . So the total number of nodes on which we run the "add" function would be at most . So I don't understand where the log(n) factor comes from.

        The online tutorial for HLD says a new chain is built when we arrive at a child node via a lighter edge, where each chain is stored as a segment tree, and so I can see there is indeed a O(logn) factor involved.

        Regardless can you perhaps elaborate a little bit more on the time complexity of the dsu structure? Thank you!

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          14 месяцев назад, # ^ |
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          Hi !

          The online tutorial for HLD says a new chain is built when we arrive at a child node via a lighter edge, where each chain is stored as a segment tree, and so I can see there is indeed a O(logn) factor involved.

          As you know, if you use segment tree in heavy-light decomposition, each query time complexity will be O(log2(n)). Because in each query you will go O(log(n)) chains and in each chain it will spend O(log(n)) time.

          Now, I will proof that "heavy-light decomposition style implementation" of "dsu on tree" is O(n.log(n)):

          Consider a complete binary tree with n vertices. In dfs function you will run another dfs function in child (T(n / 2) * 2) and you will call add function and it will spend O(n / 2) time. So, T(n) = n / 2 + 2 * T(n / 2) = O(n.log(n))

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      11 месяцев назад, # ^ |
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      You know that why dsu has O(q log n) time (for q queries); the code uses same method. Merge smaller to greater.

      Pardon me , but I don't follow. Which dsu are you talking about? The one with inverse-Ackermann function?

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        11 месяцев назад, # ^ |
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        No. Dsu with size compare. Like this :

        int find(int x){
            return par[x] == x ? x : find(par[x]);
        }
        void merge(int v, int u){
            v = find(v), u = find(u);
            if(v == u)  return ;
            if(size[v] < size[u])  swap(v, u);
            par[u] = v;
            size[v] += size[u];
        }
        
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15 месяцев назад, # |
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In easy to code but O(nlog^ 2), I cant't understand why do we store the size of subtrees of vertices in array sz and use it as the criteria for choosing the big child, I think we should store in the array "sz" the number of distinct colors in the subtree of any node v, because that is what we actually iterate on when transferring the map from v to u, why is this wrong?

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    15 месяцев назад, # ^ |
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    Hi !

    It isn't wrong! Both of your method and mine have the same worst case. But your average is better.

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15 месяцев назад, # |
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Ahh, thanks gabrielsimoes, for anyone struggling to understand: n*log^2n is about answering queries OFFLINE right during the dfs. After the dfs has finished the cnt[v] will no longer be a valid map for vertices that were chosen as bigChild.

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14 месяцев назад, # |
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http://codeforces.com/problemset/problem/291/E 291E - Tree-String Problem Arpa This problem can also be done by dsu on trees. Calculate hash value for each suffix value of target string. Then for each suffix of an edge if it is a valid prefix of the target string we would just need the frequency of the hash value of the remaining suffix of the target string in its subtree which can be maintained by this technique. The case when the entire string occurs in an edge can be dealt with separately.

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12 месяцев назад, # |
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Actually, in China, we call this method as "Heuristic Merge" which always merge the smaller to the bigger. Not hard to understand each vertex will be visited in O(log n) times because when we visited a vertex, then the size of tree which the vertex is in doubled.

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11 месяцев назад, # |
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Hey Arpa,

In your my invented style I'm unable to understand that why in third loop are you not checking for u not being parent of v. Why are you only checking for just u not being the big child.

Thanks in Advance

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11 месяцев назад, # |
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In the easy to code O(nlogn) method vec[v] stores all the vertices in the the subtree rooted at v . How will this fit into memory if we are not deleting the vec of child after merging it with the parent

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    11 месяцев назад, # ^ |
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    Used memory is always less than or equal to time complexity, so when time complexity is , used memory is less than or equal to . In this case, used memory is . Although if you delete useless vec's the memory become .

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      11 месяцев назад, # ^ |
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      Thanks for the reply . I think this problem can also be solved using your approach. (The Grass Type HackerEarth)

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        11 месяцев назад, # ^ |
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        I'll add this problem to the post if I find it related, I'm thankful anyway.

        Edit : Note that this is not my approach, but I'm the first man who publishes a tutorial about this (not sure), Sack has been used in INOI, IOI and ACM several times, so it isn't a new thing, invented be me.

        Edit : Added.

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11 месяцев назад, # |
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Hi Arpa, I can not understand, why is this approach called dsu on tree? This approach has a nice trick to reduce complexity by saving data about "big child". I can't see any special similarity with general dsu approach. In general dsu problems, we merge 2 subset into 1 set by linked list approach. But, in your tutorial there is no "merge" function. Am I missing something?

Also I see that, in your 600E's solution 14554536 you used merge functon. I can't understand, could you please explain that code?

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    11 месяцев назад, # ^ |
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    In fact we are merging information of small children with big child. Think more.

    In that code, mrg function merges information in u into v.

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11 месяцев назад, # |
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Hi Arpa! Thanks for making this tutorial.

I just want to make sure my understanding is correct: this merging smaller maps into larger ones takes logarithmic time because when a vertex is merged, the new map it is in at least twice its size. Hence, merging can only happen log(n) times for each of the n vertices, leading to a total runtime of O(nlogn)?

Thanks!

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    11 месяцев назад, # ^ |
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    Yes, but note that if you use map, it's .

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      11 месяцев назад, # ^ |
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      If you use unordered_map, does it become O(n·logn), then?

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        11 месяцев назад, # ^ |
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        Unordered_map is theoretically O(n) per query. But you can suppose that it's O(1) per query in code.

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10 месяцев назад, # |
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This 758E.Read this comment on how to use it.

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8 месяцев назад, # |
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Why do we need to iterate through the children of v after add(v,p,-1)in the naive approach?

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    8 месяцев назад, # ^ |
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    dfs() solves the problem for all the nodes, not just one. So, after you've gotten the answer for v, it will calculate the answer for its children.

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8 месяцев назад, # |
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101 Hack 47 Summing Tree was solved using this technique by satyaki3794 Submission

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8 месяцев назад, # |
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also 778C - Peterson Polyglot is solvable with a similar tecnique: is that DSU on tree?

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6 месяцев назад, # |
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Can anyone give me a link to "Shaazzz contest round #4 (season 2016-2017) problem 3" or tell me where can I find it? Thanks.

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5 месяцев назад, # |
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I can AC easily Problem 375D by the 3th way ,but WA by the 4th way.... WA on the test 4.. why..

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5 месяцев назад, # |
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APIO 2016 Fireworks uses this, but is a much harder problem.

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4 месяца назад, # |
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Arpa, in the Easy to code but O(nlog2n) section code you have written a commented line that is : //now (*cnt)[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.. But I think it would be //now (*cnt[v])[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.. Will (*cnt)[c] changed with (*cnt[v])[c] ?

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3 месяца назад, # |
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You can solve APIO 2012 Dispatching with this technique too.

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3 недели назад, # |
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IOI 2011 — Race What is the idea of DSU on tree for this problem? I know of a solution based on Centroid Decomposition.