Halo :D!

Story: It was 4 AM and I was unable to sleep so I brought you a nice trick and a couple of problems to practice it.

Introduction

(Our DP changes if we change the root of the tree, otherwise it won't make any sense to use this trick)

Let's say we want to find $$$dp_v$$$ for every vertex in a tree, we must be able to update $$$dp_v$$$ using the children of vertex $$$v$$$. Then the trick allows you to move the root from a vertex to one of its adjacent vertices in $$$O(1)$$$.

Implementation

First, calculate DP when the root is some vertex $$$rat$$$. Its obvious that if we change root $$$r$$$ to one of its adjacent vertices, $$$v$$$, then only $$$dp_r$$$ and $$$dp_v$$$ can change, so that we can update $$$dp_r$$$ using its new children and last $$$dp_r$$$, also $$$dp_v$$$ can be updated the same way. It really depends on the DP.

Note:

Don't iterate over children of vertices when moving the root, it will be $$$O(\sum\limits_v\binom{d_v}{2})$$$ and $$$W(n^2)$$$ so just don't. ($$$d_v$$$ is the degree of vertex $$$v$$$)

Now being able to move the root with distance one, we will run a DFS from $$$rat$$$, and move the root with DFS.

See the semi-code below, i hope it helps for better understanding:

void dfs(int nw, int parent){
      int fixDPnow = dp[nw], fixDPv;
      for(v in children of nw){//skip v == parent
            fixDPv = dp[v];
            move(nw, v);//it will move the root from nw to v and updates dp
            dfs(v, nw);
            //following two lines will move the root to nw and update dp
            dp[v] = fixDPv;
            dp[nw] = fixDPnow;
      }
}

int main(){
      calculateDP(rat);//it will calculate the DP when the root is rat
      dfs(rat, rat);//this will move the root to all the vertices
}

Problems

I can't open some of the problems so there is no solution for them, sorry.

You can read the same trick here, also the site has some problems.

Problem1 Solution

Problem2

Problem3 Solution

Problem4

Problem5 Solution

Probelm6 Solution

Please comment problems that can be solved using this trick.

Advanced

Let's say that the problem has online queries, such that each query gives two vertices $$$r$$$ and $$$v$$$ and wants you to calculate $$$f(r, v)$$$, it's equal to $$$dp_v$$$ when the root is $$$r$$$. It can be solved with rerooting + ancestors binary lifting, it will answers each query in $$$O(Q*log_2n)$$$ time($$$O(log_2n)$$$ for binary lifting), and also $$$O(n*log_2n)$$$ precalculation.

So lets see how it works, for every edge($$$v \to u$$$) find $$$dp_v$$$ if the root is $$$u$$$ and $$$dp_u$$$ if the root is $$$v$$$, it will take $$$O(n)$$$ time and $$$O(n)$$$ memory, and also for each vertex $$$v$$$, store $$$f(v, v)$$$. Then if the query is in form $$$v$$$ and $$$v$$$(i.e. it wants to find $$$dp_v$$$ such that the tree is rooted at vertex $$$v$$$ itself) then the answer will be $$$f(v, v)$$$, which is calculated in advance, otherwise the problem is reduced to the following problem:

You are given two vertices $$$r$$$ and $$$u$$$$$$(u \ne r)$$$ what is the second vertex in the unique path from $$$u$$$ to $$$r$$$(it means we want to find out which vertex is adjacent to $$$u$$$ and is the closest to $$$r$$$).

This problem can be solved using binary lifting, if $$$u$$$ is not an ancestor of $$$r$$$, then the answer is $$$u$$$'s parent(the root is vertex $$$rat$$$), otherwise, move $$$r$$$ using binary lifting until it's depth is equal to $$$u$$$'s depth plus 1, it can be done in $$$O(log_2n)$$$, let the answer to the reduced problem be $$$z$$$, then the answer to the whole query is equal to $$$f(z, u)$$$ such that $$$z$$$ is adjacent to $$$u$$$, we have already calculated it.

See the semi-code below, i skipped some parts so i will use $$$moveto(nw,\,u)$$$ to move the root from $$$nw$$$ to $$$u$$$, $$$f(r, u)$$$ for $$$dp_u$$$ when the root is $$$r$$$ and $$$rise(r,\,u)$$$ to find the ancestor of $$$r$$$ with depth equal to $$$depth_u+1$$$ in $$$O(log_2n)$$$.

void dfs(int nw, int parent){
      store dp[nw] as the answer to f(nw, nw);
      int fdpnw = dp[nw], fdpu;
      for(u in children of nw){//skip u == parent
            fdpu = dp[u];

            store dp[u] as the answer to f(nw, u);
            moveto(nw, u);//moving root to u

            store dp[nw] as the answer to f(u, nw);
            dfs(u, nw);//calculate the data for all vertices in the subtree of u

            dp[nw] = fdpnw; dp[u] = fdpu;//moving root to nw
      }
}

int main(){
      calculateDP(1);//calculates dp when root is 1
      dfs(1, 1);//it will move the root to all the vertices and calculates the data needed.
      int q;
      cin >> q;
      while(q--){
            int r, u;
            cin >> r >> u;
            if(r == u){
                  cout << f(r, u) << '\n'; //using the calculated data
                  continue;
            }
            if(u is not an ancestor of r){
                  cout << f(par[u], u) << '\n'; //using the calculated data
            }
            else{
                 r = rise(r, u);//its O(log_2n)
                 cout << f(r, u) << '\n'; //again using the calculated data
            }
      }
}

I tried to add a problem from polygon for the advanced part, the problem is completed but I couldn't bring it to Codeforces, but I give you the link for problem's package, it includes 50 tests, validator and checker, main correct solution and problem statement, you can try running them in polygon.

Here is the link to a group, where I added the mashup which contains the problem:

link

Here is the complete implementation for the topic, $$$dp[u]$$$ is the size of the subtree of $$$u$$$ here.

#include <bits/stdc++.h>
#define fr first
#define sc second

using namespace std;
const int MN = 2e5+7;

int dp[MN], dep[MN], timer, l, n, ans[MN], tin[MN], tout[MN];
vector<int> g[MN];
map<pair<int, int>, int> data;
vector<vector<int>> up;

bool is_ancestor(int u, int v){
    return (tin[u] <= tin[v] && tout[u] >= tout[v]);
}

int rise(int u, int v){
    if(dep[u] == dep[v]+1)return u;
    int ln = dep[u] - dep[v] - 1;
    int c = 0;
    while(ln){
        if(ln & 1)u = up[u][c];
        ln >>= 1;
        c++;
    }
    return u;
}

void cal(int nw, int pr){
    dp[nw] = 1;
    for(auto u : g[nw]){
        if(u == pr)continue;
        cal(u, nw);
        dp[nw] += dp[u];
    }
}

void store(int r, int u, int vl){
    data[{r, u}] = vl;
}

void dfs(int nw, int pr){
    int v = nw;

    tin[v] = ++timer;
    up[v][0] = pr;
    for (int i = 1; (1 << i) <= dep[v]; ++i)
        up[v][i] = up[up[v][i-1]][i-1];

    for(auto u : g[nw]){
        if(u == pr)continue;
        dep[u] = dep[nw]+1;
        int fdpnw = dp[nw], fdpu = dp[u];

        store(nw, u, dp[u]);

        dp[nw] -= dp[u];
        dp[u] += dp[nw];

        store(u, nw, dp[nw]);
        dfs(u, nw);

        dp[nw] = fdpnw;
        dp[u] = fdpu;
    }

    tout[nw] = ++timer;
}

void preprocess(int root) {
    l = ceil(log2(n));
    up.assign(n, vector<int>(l + 1));
    dfs(root, root);
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie();
    cout.tie();
    int q;
    cin >> n >> q;
    for(int i = 1; i < n; i++){
        int u, v;
        cin >> u >> v;
        u--; v--;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    cal(0, 0);
    for(int i = 0; i < n; i++)ans[i] = dp[i];
    preprocess(0);

    while(q--){
        int u, r;
        cin >> r >> u;
        r--; u--;
        if(r == u){
            cout << n << '\n';
            continue;
        }
        if(is_ancestor(u, r)){
            r = rise(r, u);
            cout << data[{r, u}] << '\n';
        }
        else{
            cout << ans[u] << '\n';
        }
    }
}

I used map to store the data, so the solution is a little slower than expected.

I hope it will help you with solving DP-Tree problems.

Thanks for your attention.