Disjoint Set Union on trees | Set 1

Given a tree and weights of nodes. Weights are non-negative integers. Task is to find maximum size of a subtree of a given tree such that all nodes are even in weights.

Prerequisite : Disjoint Set Union

Examples :

Input : Number of nodes = 7
        Weights of nodes = 1 2 6 4 2 0 3
        Edges = (1, 2), (1, 3), (2, 4), 
                (2, 5), (4, 6), (6, 7)
Output : Maximum size of the subtree 
with even weighted nodes = 4 
Explanation : 
Subtree of nodes {2, 4, 5, 6} gives the maximum size.

Input : Number of nodes = 6
        Weights of nodes = 2 4 0 2 2 6
        Edges = (1, 2), (2, 3), (3, 4), 
                (4, 5), (1, 6)
Output : Maximum size of the subtree
with even weighted nodes = 6
Explanation : 
The given tree gives the maximum size.

Approach :We can find solution by simply running DFS on tree. DFS solution gives us answer in O(n). But, how can we use DSU for this problem? We first iterate through all edges. If both nodes are even in weights, we make union of them. Set of nodes with maximum size is the answer. If we use union-find with path compression then time complexity is O(n).

Below is the implementation of above approach :
[sourcecode language=”CPP”]
// CPP code to find maximum subtree such
// that all nodes are even in weight
#include<bits/stdc++.h>

using namespace std;

#define N 100010

// Structure for Edge
struct Edge
{
int u, v;
};

/*
‘id’: stores parent of a node.
‘sz’: stores size of a DSU tree.
*/
int id[N], sz[N];

// Function to assign root
int Root(int idx)
{
int i = idx;

while(i != id[i])
id[i] = id[id[i]], i = id[i];

return i;
}

// Function to find Union
void Union(int a, int b)
{
int i = Root(a), j = Root(b);

if (i != j)
{
if(sz[i] >= sz[j])
{
id[j] = i, sz[i] += sz[j];
sz[j] = 0;
}
else
{
id[i] = j, sz[j] += sz[i];
sz[i] = 0;
}
}
}

// Utility function for Union
void UnionUtil(struct Edge e[], int W[], int q)
{

for(int i = 0; i < q; i++)
{
// Edge between ‘u’ and ‘v’
int u, v;
u = e[i].u, v = e[i].v;

// 0-indexed nodes
u–, v–;

// If weights of both ‘u’ and ‘v’
// are even then we make union of them.
if(W[u] % 2 == 0 && W[v] % 2 == 0)
Union(u,v);
}
}

// Function to find maximum
// size of DSU tree
int findMax(int n, int W[])
{
int maxi = 0;
for(int i = 1; i <= n; i++)
if(W[i] % 2 == 0)
maxi = max(maxi, sz[i]);

return maxi;
}

// Driver code
int main()
{
/*
Nodes are 0-indexed in this code
So we have to make necessary changes
while taking inputs
*/

// Weights of nodes
int W[] = {1, 2, 6, 4, 2, 0, 3};

// Number of nodes in a tree
int n = sizeof(W) / sizeof(W[0]);

// Initializing every node as
// a tree with single node.
for(int i = 0; i < n; i++)
id[i] = i, sz[i] = 1;

Edge e[] = {{1, 2}, {1, 3}, {2, 4},
{2, 5}, {4, 6}, {6, 7}};

int q = sizeof(e) / sizeof(e[0]);

UnionUtil(e, W, q);

// Find maximum size of DSU tree.
int maxi = findMax(n, W);

printf("Maximum size of the subtree with ");
printf("even weighted nodes = %d\n", maxi);

return 0;
}
[/sourcecode]

Output:

Maximum size of the subtree with even weighted nodes = 4


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