给定一棵树和节点的权重。权重是非负整数。任务是找到给定树的子树的最大大小，以使所有节点的权重相等。
先决条件：不交集集合
例子 ：
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.方法：我们只需在树上运行DFS即可找到解决方案。 DFS解决方案在O(n)中给出了答案。但是，我们如何使用DSU解决此问题?我们首先遍历所有边缘。如果两个节点的权重相等，我们将它们合并。答案是具有最大大小的节点集。如果我们将联合查找与路径压缩一起使用，则时间复杂度为O(n) 。
下面是上述方法的实现：
C++
// CPP code to find maximum subtree such
// that all nodes are even in weight
#include
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;
} Java
// Java code to find maximum subtree such
// that all nodes are even in weight
class GFG
{
static int N = 100010;
// Structure for Edge
static class Edge
{
int u, v;
public Edge(int u, int v)
{
this.u = u;
this.v = v;
}
}
/*
'id': stores parent of a node.
'sz': stores size of a DSU tree.
*/
static int []id = new int[N];
static int []sz = new int[N];
// Function to assign root
static 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
static 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
static void UnionUtil(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
static int findMax(int n, int W[])
{
int maxi = 0;
for(int i = 1; i < n; i++)
if(W[i] % 2 == 0)
maxi = Math.max(maxi, sz[i]);
return maxi;
}
// Driver code
public static void main(String[] args)
{
/*
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 = W.length;
// Initializing every node as
// a tree with single node.
for(int i = 0; i < n; i++)
{
id[i] = i;
sz[i] = 1;
}
Edge e[] = {new Edge(1, 2), new Edge(1, 3),
new Edge(2, 4), new Edge(2, 5),
new Edge(4, 6), new Edge(6, 7)};
int q = e.length;
UnionUtil(e, W, q);
// Find maximum size of DSU tree.
int maxi = findMax(n, W);
System.out.printf("Maximum size of the subtree with ");
System.out.printf("even weighted nodes = %d\n", maxi);
}
}
// This code is contributed by Rajput-JiPython3
# Python3 code to find maximum subtree such
# that all nodes are even in weight
N = 100010
# Structure for Edge
class Edge:
def __init__(self, u, v):
self.u = u
self.v = v
'''
'id': stores parent of a node.
'sz': stores size of a DSU tree.
'''
id = [0 for i in range(N)]
sz = [0 for i in range(N)];
# Function to assign root
def Root(idx):
i = idx;
while(i != id[i]):
id[i] = id[id[i]]
i = id[i];
return i;
# Function to find Union
def Union(a, b):
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
def UnionUtil( e, W, q):
for i in range(q):
# Edge between 'u' and 'v'
u = e[i].u
v = e[i].v
# 0-indexed nodes
u -= 1
v -= 1
# If weights of both 'u' and 'v'
# are even then we make union of them.
if(W[u] % 2 == 0 and W[v] % 2 == 0):
Union(u, v);
# Function to find maximum
# size of DSU tree
def findMax(n, W):
maxi = 0
for i in range(1, n):
if(W[i] % 2 == 0):
maxi = max(maxi, sz[i]);
return maxi;
# Driver code
if __name__=='__main__':
'''
Nodes are 0-indexed in this code
So we have to make necessary changes
while taking inputs
'''
# Weights of nodes
W = [1, 2, 6, 4, 2, 0, 3]
# Number of nodes in a tree
n = len(W)
# Initializing every node as
# a tree with single node.
for i in range(n):
id[i] = i
sz[i] = 1;
e = [Edge(1, 2), Edge(1, 3), Edge(2, 4),
Edge(2, 5), Edge(4, 6), Edge(6, 7)]
q = len(e)
UnionUtil(e, W, q);
# Find maximum size of DSU tree.
maxi = findMax(n, W);
print("Maximum size of the subtree with ", end='');
print("even weighted nodes =", maxi);
# This code is contributed by rutvik_56C#
// C# code to find maximum subtree such
// that all nodes are even in weight
using System;
class GFG
{
static int N = 100010;
// Structure for Edge
public class Edge
{
public int u, v;
public Edge(int u, int v)
{
this.u = u;
this.v = v;
}
}
/*
'id': stores parent of a node.
'sz': stores size of a DSU tree.
*/
static int []id = new int[N];
static int []sz = new int[N];
// Function to assign root
static 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
static 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
static void UnionUtil(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
static int findMax(int n, int []W)
{
int maxi = 0;
for(int i = 1; i < n; i++)
if(W[i] % 2 == 0)
maxi = Math.Max(maxi, sz[i]);
return maxi;
}
// Driver code
public static void Main(String[] args)
{
/*
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 = W.Length;
// Initializing every node as
// a tree with single node.
for(int i = 0; i < n; i++)
{
id[i] = i;
sz[i] = 1;
}
Edge []e = {new Edge(1, 2), new Edge(1, 3),
new Edge(2, 4), new Edge(2, 5),
new Edge(4, 6), new Edge(6, 7)};
int q = e.Length;
UnionUtil(e, W, q);
// Find maximum size of DSU tree.
int maxi = findMax(n, W);
Console.Write("Maximum size of the subtree with ");
Console.WriteLine("even weighted nodes = {0}\n", maxi);
}
}
// This code is contributed by Princi Singh输出：
Maximum size of the subtree with even weighted nodes = 4