Perform the given queries on the rooted tree
Given a rooted tree and not necessarily binary. The tree contains N nodes, labeled 1 to N. You are given the tree in the form of an array A[1..N] of size N. A[i] denotes label of the parent of node labeled i. For clarity, you may assume that the tree satisfies the following conditions.
- The root of the tree is labeled 1. Hence A[1] is set to 0.
- The parent of node T will always have a label less than T.
The task is to perform the following operations according to the type of query given.
- ADD, X, Y: add Y to the value at node X.
- ADDUP, X, Y: add Y to the value at node X. Then, add Y to the value at A[X] (i.e. the parent of X). The, add Y to the value at A[A[X]] (i.e. the parent of A[X]).. and so on, till you add Y to the value at root.
After you have performed all the given operations, you are asked to answer several queries of the following type
- VAL, X: print the value at node X.
- VALTREE, X: print the sum of values at all nodes in the subtree rooted at X (including X).
Source: Directi Interview | Set 13
Examples:
Input:
N = 7, M = 4, Q = 5
0 1 2 2 2 1 2
ADD 6 76
ADDUP 1 49
ADD 4 48
ADDUP 2 59
VALTREE 1
VALTREE 5
VAL 5
VALTREE 2
VAL 2
Output:
291
0
0
107
59
Input:
N = 5, M = 5, Q = 3
0 1 1 1 3
ADD 1 10
ADD 2 20
ADD 3 30
ADD 4 40
ADDUP 5 50
VAL 3
VALTREE 3
VALTREE 1
Output:
80
130
250
Explanation: This problem is a slight variation of dfs. In this, we have stored the node’s original value and addup value in the vector of the pair. We did 2 times dfs.
- dfs1 for offline queries i.e to calculate the addup sum for each node.
- dfs2 to store the subtree sum in an array.
Now all the queries can be answered in a constant time.
Graph before dfs1
Graph after dfs1
Below is the required implementation:
C++
// C++ implementation to perform// above operations and queries#include <bits/stdc++.h>using namespace std;/*Code Parametersp->for every node first value is it's original valueand second value is it's addup valuesubtree_sum[]-> to store the subtree_sum at every nodevisit-> for dfs1visit2->for dfs2*/vector<pair<int, int> > p;vector<int> adj[10000];int subtree_sum[10000], visit[10000], visit2[10000];int dfs1(int root){ // for leaf node if (adj[root].size() == 0) { // if leaf node then add the addup // sum to it's original value p[root].first += p[root].second; return 0; } int sum = 0; for (int i = 0; i < adj[root].size(); i++) { if (visit[adj[root][i]] == 0) { dfs1(adj[root][i]); // add the addup sum of all the adjacent // neighbors to the current node p[root].second += p[adj[root][i]].second; visit[adj[root][i]] = 1; } } // process the root node p[root].first += p[root].second; return 0;}int dfs2(int root){ if (adj[root].size() == 0) { // for the leaf node subtree_sum // will be it's own value subtree_sum[root] = p[root].first; return p[root].first; } int sum = p[root].first; for (int i = 0; i < adj[root].size(); i++) { if (visit2[adj[root][i]] == 0) { sum += dfs2(adj[root][i]); visit2[adj[root][i]] = 1; } } // calculate the subtree_sum // for the particular root node subtree_sum[root] = sum; return sum;}// Driver codeint main(){ int nodes = 7, m = 4, qu = 5, b; int a[] = { 0, 1, 2, 2, 2, 1, 2 }; // for root node p.push_back(make_pair(0, 0)); for (int i = 0; i < nodes; i++) { if (a[i] != 0) adj[a[i]].push_back(i + 1); // for every node p.push_back(make_pair(0, 0)); } vector<pair<string, pair<int, int> > > v; v.push_back(make_pair("ADD", make_pair(6, 76))); v.push_back(make_pair("ADDUP", make_pair(1, 49))); v.push_back(make_pair("ADD", make_pair(4, 48))); v.push_back(make_pair("ADDUP", make_pair(2, 59))); for (int i = 0; i < m; i++) { string s = v[i].first; int a = v[i].second.first; int b = v[i].second.second; if (s == "ADD") // adding to it's own value p[a].first += b; else // adding to it's addup value p[a].second += b; } // to process the offline queries dfs1(1); // to store the subtree sum for every root node dfs2(1); vector<pair<string, int> > q; q.push_back(make_pair("VALTREE", 1)); q.push_back(make_pair("VALTREE", 5)); q.push_back(make_pair("VAL", 5)); q.push_back(make_pair("VALTREE", 2)); q.push_back(make_pair("VAL", 2)); for (int i = 0; i < qu; i++) { string s = q[i].first; int a = q[i].second; if (s == "VAL") cout << p[a].first << "\n"; else cout << subtree_sum[a] << "\n"; }} |
Python3
# Python3 implementation to perform# above operations and queriesp = []adj = [0] * 10000for i in range(10000): adj[i] = []subtree_sum, visit, visit2 = [0] * 10000, [0] * 10000, [0] * 10000# Code Parameters# p->for every node first value is it's original value# and second value is it's addup value# subtree_sum[]-> to store the subtree_sum at every node# visit-> for dfs1# visit2->for dfs2def dfs1(root: int) -> int: # for leaf node if len(adj[root]) == 0: # if leaf node then add the addup # sum to it's original value p[root][0] += p[root][1] return 0 summ = 0 for i in range(len(adj[root])): if visit[adj[root][i]] == 0: dfs1(adj[root][i]) # add the addup sum of all the adjacent # neighbors to the current node p[root][1] += p[adj[root][i]][1] visit[adj[root][i]] = 1 # process the root node p[root][0] += p[root][1] return 0def dfs2(root: int) -> int: if len(adj[root]) == 0: # for the leaf node subtree_sum # will be it's own value subtree_sum[root] = p[root][0] return p[root][0] summ = p[root][0] for i in range(len(adj[root])): if visit2[adj[root][i]] == 0: summ += dfs2(adj[root][i]) visit2[adj[root][i]] = 1 # calculate the subtree_sum # for the particular root node subtree_sum[root] = summ return summ# Driver Codeif __name__ == "__main__": nodes, m, qu = 7, 4, 5 a = [0, 1, 2, 2, 2, 1, 2] # for root node p.append([0, 0]) for i in range(nodes): if a[i] != 0: adj[a[i]].append(i + 1) # for every node p.append([0, 0]) v = [] v.append(("ADD", [6, 76])) v.append(("ADDUP", [1, 49])) v.append(("ADD", [4, 48])) v.append(("ADDUP", [2, 59])) for i in range(m): s = v[i][0] a = v[i][1][0] b = v[i][1][1] if s == "ADD": # adding to it's own value p[a][0] += b else: # adding to it's addup value p[a][1] += b # to process the offline queries dfs1(1) # to store the subtree sum for every root node dfs2(1) q = [] q.append(["VALTREE", 1]) q.append(["VALTREE", 5]) q.append(["VAL", 5]) q.append(["VALTREE", 2]) q.append(["VAL", 2]) for i in range(qu): s = q[i][0] a = q[i][1] if s == "VAL": print(p[a][0]) else: print(subtree_sum[a])# This code is contributed by# sanjeev2552 |
291 0 0 107 59
Time Complexity: O(1) per query, O(N) for preprocessing is taken by dfs1() and dfs2() function.
Auxiliary Space: O(N)
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