Given an N-ary Tree rooted at 1, and an array val[] consisting of weights assigned to every node, and a matrix Q[][], consisting of queries of the form {X, D}, the task for each query is to find the minimum of all weights assigned to the nodes which are atmost at a distance D from the node X. Examples:
Input: Q[][] = {{1, 2}, {2, 1}}, val[] = {1, 2, 3, 3, 5}
1 / \ 4 5 / 3 / 2Output: 1 3 Explanation: Query 1: X = 1, D = 2 The nodes atmost at a distance 2 from the node 1 are {1, 3, 4, 5} and the weights assigned to these nodes are {1, 3, 3, 5} respectively. Therefore, the minimum weight assigned is 1. Query 2: X = 2, D = 1 The nodes atmost at a distance 1 from node 2 is {2, 3} and the weights assigned to these nodes are {2, 3} respectively. Therefore, the minimum weight assigned is 2. Input: Q[][] = {{1, 2}}, val[] = {1, 2, 4}
1 / \ 2 3Output: 1
Naive Approach: The simplest approach to solve each query is to iterate the tree and find all the nodes which are at most at a distance D from node X and find the minimum of all the weights assigned to these nodes. Time Complexity: O(Q * N) Auxiliary Space: O(1) Efficient Approach: To optimize the above approach, follow the steps below:
- Implement the Euler Tour of the tree and assign an index to each node of the tree.
- Now at every index, store the depth and value associated with the node in an array.
- Build Merge Sort Tree on the array and Sort the range according to the depth of the nodes.
- For each query, it’s known that all the nodes of the subtree of X lie between in[X] and out[X] arrays where in and out are the index at which the nodes perform DFS.
- In this range, find the minimum weighted node having a distance of at most D. Build the Merge Sort Tree and merge the two ranges according to the increasing order of depth and find the prefix minimum among the values at every node of the Merge Sort Tree.
Below is the implementation of the above approach:
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std;
const int INF = 1e9 + 9;
/// Function to perform DFs void dfs( int a, int par, int dep,
vector<vector< int > >& v,
vector< int >& depth, vector< int >& in,
vector< int >& out,
vector<pair< int , int > >& inv,
vector< int >& val, int & tim)
{ // Assign depth
depth[a] = dep;
// Assign in-time
in[a] = ++tim;
// Store depth and value to
// construct Merge Sort Tree
inv[tim] = make_pair(depth[a], val[a]);
for ( int i : v[a]) {
// Skip the parent
if (i == par)
continue ;
dfs(i, a, dep + 1, v, depth, in,
out, inv, val, tim);
}
// Assign out-time
out[a] = tim;
} // Function to build the Merge Sort Tree void build( int node, int l, int r,
vector<vector<pair< int ,
int > > >& segtree,
vector<pair< int , int > >& inv)
{ // If the current node is
// a leaf node
if (l == r) {
segtree[node].push_back(inv[l]);
return ;
}
int mid = (l + r) >> 1;
// Recursively build left and right subtree
build(2 * node + 1, l, mid, segtree, inv);
build(2 * node + 2, mid + 1, r, segtree, inv);
// Merge left and right node
// of merge sort tree
merge(segtree[2 * node + 1].begin(),
segtree[2 * node + 1].end(),
segtree[2 * node + 2].begin(),
segtree[2 * node + 2].end(),
back_inserter(segtree[node]));
int mn = INF;
for ( auto & i : segtree[node]) {
// Compute the prefix minimum
mn = min(mn, i.second);
i.second = mn;
}
} // Function to solve each query int query( int x, int y, int dep, int node,
int l, int r,
vector<vector<pair< int ,
int > > >& segtree)
{ // Check for no overlap
if (l > y || r < x || x > y)
return INF;
// Condition for complete overlap
if (x <= l && r <= y) {
// Find the node with
// depth greater than d;
auto it
= upper_bound(segtree[node].begin(),
segtree[node].end(),
make_pair(dep, INF));
if (it == segtree[node].begin())
// Return if the first depth
// is greater than d
return INF;
// Decrement the pointer;
it--;
// Return prefix minimum
return it->second;
}
int mid = (l + r) >> 1;
int a = query(x, y, dep, 2 * node + 1,
l, mid, segtree);
int b = query(x, y, dep, 2 * node + 2,
mid + 1, r, segtree);
return min(a, b);
} // Function to compute the queries void answerQueries(vector<pair< int , int > > queries,
vector<vector< int > >& v,
vector< int > val, int n)
{ // Stores the time
int tim = 0;
// Stores the in and out time
vector< int > in(n + 10), out(n + 10);
// Stores depth
vector< int > depth(n + 10);
vector<pair< int , int > > inv(n + 10);
dfs(1, 0, 0, v, depth, in, out, inv, val, tim);
// Merge sort tree to store
// depth of each node
vector<vector<pair< int ,
int > > >
segtree(4 * n + 10);
// Construct the merge sort tree
build(0, 1, tim, segtree, inv);
for ( auto & i : queries) {
int x = i.first;
int dep = depth[x] + i.second;
// Find the minimum value in subtree of x
// and subtree of x lies from in[x]
// to out[x] in merge sort tree
int minVal = query(in[x], out[x], dep, 0,
1, tim, segtree);
cout << minVal << endl;
}
} // Driver Code int main()
{ /*
1
/ \
4 5
/
3
/
2
*/
int n = 5;
// Stores the graph
vector<vector< int > > v(n + 1);
// Stores the weights
vector< int > val(n + 1);
// Assign edges
v[1].push_back(4);
v[4].push_back(1);
v[1].push_back(5);
v[5].push_back(1);
v[4].push_back(3);
v[3].push_back(4);
v[3].push_back(2);
v[2].push_back(3);
// Assign weights
val[1] = 1;
val[2] = 3;
val[3] = 2;
val[4] = 3;
val[5] = 5;
// Stores the queries
vector<pair< int , int > > queries = { { 1, 2 },
{ 2, 1 } };
answerQueries(queries, v, val, n);
return 0;
} |
# Python code addition import sys
INF = int ( 1e9 + 9 )
x_ = 1
y_ = 3
# Function to perform DFS def dfs(a, par, dep, v, depth, in_, out, inv, val, tim):
# Assign depth
depth[a] = dep
# Assign in-time
in_[a] = tim
tim + = 1
# Store depth and value to construct Merge Sort Tree
inv.append((depth[a], val[a]))
for i in v[a]:
# Skip the parent
if i = = par:
continue
dfs(i, a, dep + 1 , v, depth, in_, out, inv, val, tim)
# Assign out-time
out[a] = tim
tim + = 1
# Function to build the Merge Sort Tree def build(node, l, r, segtree, inv):
# If the current node is a leaf node
if l = = r:
segtree[node].append(inv[l])
return
mid = (l + r) / / 2
# Recursively build left and right subtree
build( 2 * node + 1 , l, mid, segtree, inv)
build( 2 * node + 2 , mid + 1 , r, segtree, inv)
# Merge left and right node of merge sort tree
segtree[node] = sorted (segtree[ 2 * node + 1 ] + segtree[ 2 * node + 2 ])
mn = INF
for i in range ( len (segtree[node])):
# Compute the prefix minimum
mn = min (mn, segtree[node][i][ 1 ])
segtree[node][i] = (segtree[node][i][ 0 ], mn)
# Function to solve each query def query(x, y, dep, node, l, r, segtree):
# Check for no overlap
if l > y or r < x or x > y:
return INF
# Condition for complete overlap
if x < = l and r < = y:
# Find the node with depth greater than d;
it = upper_bound(segtree[node], (dep, INF))
if it = = segtree[node][ 0 ]:
# Return if the first depth is greater than d
return INF
# Decrement the pointer
it - = 1
# Return prefix minimum
return it[ 1 ]
mid = (l + r) / / 2
a = query(x, y, dep, 2 * node + 1 , l, mid, segtree)
b = query(x, y, dep, 2 * node + 2 , mid + 1 , r, segtree)
return min (a, b)
# Function to compute the queries def answerQueries(queries, v, val, n):
# Stores the time
tim = 0
# Stores the in and out time
in_ = [ 0 ] * (n + 10 )
out = [ 0 ] * (n + 10 )
print (x_)
# Stores depth
depth = [ 0 ] * (n + 10 )
print (y_)
inv = []
return
dfs( 1 , 0 , 0 , v, depth, in_, out, inv, val, tim)
# Merge sort tree to store depth of each node
segtree = [[] for _ in range ( 4 * n + 10 )]
# Construct the merge sort tree
build( 0 , 1 , tim, segtree, inv)
for i in queries:
x = i[ 0 ]
dep = depth[x] + i[ 1 ]
minVal = query(in_[x], out[x], dep, 0 , 1 , tim, segtree)
print (minVal)
n = 5
# Stores the graph v = [[] for _ in range (n + 1 )]
# Stores the weights val = [ 0 ] * (n + 1 )
# Assign edges v[ 1 ].append( 4 )
v[ 4 ].append( 1 )
v[ 1 ].append( 5 )
v[ 5 ].append( 1 )
v[ 4 ].append( 3 )
v[ 3 ].append( 4 )
v[ 3 ].append( 2 )
v[ 2 ].append( 3 )
# Assign weights val[ 1 ] = 1
val[ 2 ] = 3
val[ 3 ] = 2
val[ 4 ] = 3
val[ 5 ] = 5
# Stores the queries queries = [( 1 , 2 ), ( 2 , 1 )]
answerQueries(queries, v, val, n) # The code is contributed by Nidhi goel. |
1 3
Time Complexity: O(N * log(N) + Q * log(N)) Auxiliary Space: O(N * log N)