Heavy Light White Falcon


Problem Statement :


Our lazy white falcon finally decided to learn heavy-light decomposition. Her teacher gave an assignment for her to practice this new technique. Please help her by solving this problem.

You are given a tree with N nodes and each node's value is initially 0. The problem asks you to operate the following two types of queries:

"1 u x" assign x to the value of the node .
"2 u v" print the maximum value of the nodes on the unique path between u and v.


Input Format

First line consists of two integers seperated by a space: N and Q.
Following N - 1  lines consisting of two integers denotes the undirectional edges of the tree.
Following Q lines consist of the queries you are asked to operate.

Constraints

1  <=   N , Q , x  <=  50000

It is guaranteed that input denotes a connected tree with N nodes. Nodes are enumerated with 0-based indexing.


Output Format

For each second type of query print single integer in a single line, denoting the asked maximum value.



Solution :



title-img


                            Solution in C :

In   C++   :








#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 50010;

class node {
    
    public :
        int l, r, mx;
        node *left, *right;
    
        void update(int idx, int val) {
            if(l >= r) {
                mx = val;
                return;
            }        
            int mid = (l + r) / 2;
            (idx <= mid ? left : right)->update(idx, val);
            mx = max(left->mx, right->mx);
        }
    
        int query(int a, int b) {
            if(b < l or r < a) return 0;
            if(a <= l and r <= b) return mx;
            return max(left->query(a, b), right->query(a, b));
        }
    
        node(int _l, int _r) : 
            l(_l), r(_r), mx(0), left(NULL), right(NULL) {}
};

node* init(int l, int r) {
    node *p = new node(l, r);
    if(l < r) {
        int mid = (l + r) / 2;
        p->left = init(l, mid);
        p->right = init(mid+1, r);
    }
    return p;
}

vector<int> adj[N];
int n, q;

node* head[N];
vector<int> Path[N];
int sz[N], H[N], P[N], G[N], pos[N];

void dfs_init(int u, int p, int h) {
    P[u] = p;
    H[u] = h;
    sz[u] = 1;
    for(int v : adj[u]) {
        if(v == p) {
            continue;
        }
        dfs_init(v, u, h+1);
        sz[u] += sz[v];
    }
}
void dfs_HLD(int u) {
    Path[u].push_back(u);
    for(int i = 0;i < Path[u].size();i++) {
        int v = Path[u][i];
        G[v] = u;
        pos[v] = i;
        for(int vv : adj[v]) {
            if(vv == P[v]) continue;
            if(2*sz[vv] >= sz[v]) {
                Path[u].push_back(vv);
            }else {
                dfs_HLD(vv);
            }
        }
    }
    head[u] = init(0, Path[u].size() - 1);
}
int query(int u, int v) {
    int ans = 0;
    while(G[u] != G[v]) {
        if(H[G[u]] < H[G[v]]) {
            swap(u, v);
        }
        ans = max(ans, head[G[u]]->query(0, pos[u]));
        u = P[G[u]];
    }
    if(pos[u] > pos[v]) {
        swap(u, v);
    }
    ans = max(ans, head[G[u]]->query(pos[u], pos[v]));
    return ans;
}
int main() {
    
    ios::sync_with_stdio(false);
    cin >> n >> q;
    for(int i = 0;i < n-1;i++) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    
    dfs_init(0, 0, 0);
    dfs_HLD(0);
    
    for(int i = 0;i < q;i++) {
        int type;
        cin >> type;
        if(type == 1) {
            int u, x;
            cin >> u >> x;
            head[G[u]]->update(pos[u], x);
        }else {
            int u, v;
            cin >> u >> v;
            cout << query(u, v) << "\n";
        }
    }
    
    return 0;
}









In   Java  :







import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

 
    // array to store values for each node(vertices)
    static int[] v;
    // array to store connections (edges)
    static ArrayList<Integer>[] e;
    
    // some node with single connection will be tree root
    static int root; 
    // direction to reach root for each node
    static int[] dirToRoot;
    // distance to root
    static int[] distToRoot;
    
    // maximum node value 
    static int totalMax = 0;
    
    static void markRoot(){
        for(int i=0; i<v.length; i++){
            if(e[i].size()==1){
                root = i;
                break;
            }
        }
    }
    
 static void markPathToRoot(int node, int dist, boolean[] visited){
        distToRoot[node] = dist;
        for(Integer item : e[node]){
            int index = item.intValue();
            if(visited[index]==false){
                dirToRoot[index] = node;
                visited[index] = true;
                markPathToRoot(index, dist+1, visited);
            }
        }
    }
    
    static int findMax2(int start, int finish){
        int sDistance = distToRoot[start];
        int fDistance = distToRoot[finish];

        int sIndex = start;
        int fIndex = finish;
        int max = Math.max(v[sIndex], v[fIndex]);
        
        // decrease distance from the one that is more far from root
        while(sDistance>fDistance && max<totalMax){
            sIndex = dirToRoot[sIndex];
            max = max >= v[sIndex] ? max : v[sIndex];
            sDistance--;
        }
        while(fDistance>sDistance && max<totalMax){
            fIndex = dirToRoot[fIndex];
            max = max >= v[fIndex] ? max : v[fIndex];
            fDistance--;
        }

        // run both of them
        while(sIndex!=fIndex){
            fIndex = dirToRoot[fIndex];
            sIndex = dirToRoot[sIndex];
            max = max >= v[fIndex] ? max : v[fIndex];
            max = max >= v[sIndex] ? max : v[sIndex];
            if(sIndex==root || max==totalMax)
                break;
        }
        return Math.max(max, v[sIndex]);
    }
    
    // calculate distance to root node from each node
    static void resetRoot(){
        // direction to the root of tree for each node
        dirToRoot = new int[v.length];
        distToRoot= new int[v.length];
        
        // mark node with only one edge as root
        markRoot();
//        System.out.println("root="+root);
        
        dirToRoot[root] = root;
        boolean[] visited = new boolean[v.length];
        visited[root] = true;
        markPathToRoot(root, 0, visited);
    }
    
    public static void main(String[] args) {
        Reader sc = new Reader();
        sc.init(System.in);
        int N = sc.nextInt();
        int Q = sc.nextInt();
        
        // array to store values for each node(vertices)
        v = new int[N];
        // array to store connections (edges)
        e = new ArrayList[N];

        for(int i=0; i<N; i++)
            e[i] = new ArrayList<Integer>(2);
        
        for(int i=0; i<N-1; i++){
            int v1 = sc.nextInt();
            int v2 = sc.nextInt();
            // add to both because undirectional
            e[v1].add(v2);
            e[v2].add(v1);
        }
        
        resetRoot();
        
        // read queries
        while(Q-->0){
            int type = sc.nextInt();
            if(type==1){
                int node    = sc.nextInt();
                int value   = sc.nextInt();
                v[node] = value;
                totalMax = value > totalMax ? value : totalMax;
            }else{
                int start   = sc.nextInt();
                int finish  = sc.nextInt();
                //search for solution
                System.out.println( findMax2(start, finish) );
            }
        }
    }
    
// code from internet    
/** Class for buffered reading int and double values */
static class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
     reader = new BufferedReader( new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }

    /** get next word */
    static String next() {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            try{
                tokenizer = new StringTokenizer( reader.readLine() );
            }catch(Exception e){}
        }
        return tokenizer.nextToken();
    }

    static int nextInt() {
        return Integer.parseInt( next() );
    }
	
    static double nextDouble() {
        return Double.parseDouble( next() );
    }
}
}








In   Python3  :







def segtree_init(ary):
   ary = list(ary)
   seg = [ary]
   while len(ary) > 1:
      if len(ary) & 1: ary.append(0)
      ary = [max(ary[i],ary[i+1]) for i in range(0,len(ary),2)]
      seg.append(ary)
   return seg
def segtree_set(seg, i, x):
   ary = seg[0]
   ary[i] = x
   for j in range(1, len(seg)):
      x = max(ary[i], ary[i^1])
      ary = seg[j]
      i >>= 1
      ary[i] = x
def segtree_max(seg, lo, hi):
   m = 0
   j = 0
   while lo < hi:
      ary = seg[j]
      if lo & 1:
         x = ary[lo]
         if x > m: m = x
         lo += 1
      if hi & 1:
         hi -= 1
         x = ary[hi]
         if x > m: m = x
      lo >>= 1
      hi >>= 1
      j += 1
   return m
class heavy_light_node:
   def __init__(self, segtree):
      self.parent = None
      self.pos = -1
      self.segtree = segtree
def build_tree(i, edges, location):
   children = []
   members = [i]
   ed = edges[i]
   while ed:
      for j in range(1,len(ed)):
         child = build_tree(ed[j], edges, location)
         child.pos = len(members) - 1
         children.append(child)
      i = ed[0]
      members.append(i)
      ed = edges[i]
   node = heavy_light_node(segtree_init(0 for j in members))
   for child in children:
      child.parent = node
   for j in range(len(members)):
      location[members[j]] = (node, j)
   return node
def read_tree(N):
   edges = [[] for i in range(N)]
   for i in range(N-1):
      x, y = map(int, input().split())
      edges[x].append(y)
      edges[y].append(x)
   size = [0] * N
   active = [0]
   while active:
      i = active[-1]
      if size[i] == 0:
         size[i] = 1
         for j in edges[i]:
            edges[j].remove(i)
            active.append(j)
      else:
         active.pop()
         edges[i].sort(key=lambda j: -size[j])
         size[i] = 1 + sum(size[j] for j in edges[i])
   location = [None] * N
   build_tree(0, edges, location)
   return location
def root_path(i, location):
   loc = location[i]
   path = [ loc ]
   loc = loc[0]
   while loc.parent != None:
      path.append((loc.parent, loc.pos))
      loc = loc.parent
   path.reverse()
   return path
def max_weight(x, y):
   px = root_path(x, location)
   py = root_path(y, location)
   m = 1
   stop = min(len(px), len(py))
   while m < stop and px[m][0] == py[m][0]: m += 1
   loc, a = px[m-1]
   b = py[m-1][1]
   if a > b: a, b = b, a
   w = segtree_max(loc.segtree, a, b+1)
   for j in range(m, len(px)):
      loc, i = px[j]
      x = segtree_max(loc.segtree, 0, i+1)
      if x > w: w = x
   for j in range(m, len(py)):
      loc, i = py[j]
      x = segtree_max(loc.segtree, 0, i+1)
      if x > w: w = x
   return w
N, Q = map(int, input().split())
location = read_tree(N)
for i in range(Q):
   t, x, y = map(int, input().split())
   if t == 1:
      loc, i = location[x]
      segtree_set(loc.segtree, i, y)
   elif t == 2:
      print(max_weight(x, y))
                        








View More Similar Problems

Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

View Solution →

Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

View Solution →

Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from

View Solution →

Kitty's Calculations on a Tree

Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a

View Solution →

Is This a Binary Search Tree?

For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a

View Solution →

Square-Ten Tree

The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the

View Solution →