Dado un árbol y pesos de Nodes. Los pesos son enteros no negativos. La tarea es encontrar el tamaño máximo de un subárbol de un árbol dado de modo que todos los Nodes tengan pesos pares.
Requisito previo: Unión de conjuntos disjuntos
Ejemplos:
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.
Enfoque: podemos encontrar una solución simplemente ejecutando DFS en el árbol. La solución DFS nos da la respuesta en O(n) . Pero, ¿cómo podemos usar DSU para este problema? Primero iteramos a través de todos los bordes. Si ambos Nodes son pares en pesos, hacemos unión de ellos. Conjunto de Nodes con tamaño máximo es la respuesta. Si usamos union-find con compresión de ruta, entonces la complejidad del tiempo es O(n) .
A continuación se muestra la implementación del enfoque anterior:
C++
// 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;
}
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-Ji
Python3
# 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_56
C#
// 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
Javascript
<script>
// Javascript code to find maximum subtree such
// that all nodes are even in weight
let N = 100010;
/*
'id': stores parent of a node.
'sz': stores size of a DSU tree.
*/
let id = new Array(N);
let sz = new Array(N);
// Function to assign root
function Root(idx)
{
let i = idx;
while(i != id[i])
{
id[i] = id[id[i]];
i = id[i];
}
return i;
}
// Function to find Union
function Union(a, b)
{
let 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
function UnionUtil(e, W, q)
{
for(let i = 0; i < q; i++)
{
// Edge between 'u' and 'v'
let u, v;
u = e[i][0];
v = e[i][1];
// 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
function findMax(n, W)
{
let maxi = 0;
for(let i = 1; i < n; i++)
if (W[i] % 2 == 0)
maxi = Math.max(maxi, sz[i]);
return maxi;
}
// Driver code
/*
Nodes are 0-indexed in this code
So we have to make necessary changes
while taking inputs
*/
// Weights of nodes
let W = [ 1, 2, 6, 4, 2, 0, 3 ];
// Number of nodes in a tree
let n = W.length;
// Initializing every node as
// a tree with single node.
for(let i = 0; i < n; i++)
{
id[i] = i;
sz[i] = 1;
}
let e = [ [ 1, 2 ], [ 1, 3 ],
[ 2, 4 ], [ 2, 5 ],
[ 4, 6 ], [ 6, 7 ] ];
let q = e.length;
UnionUtil(e, W, q);
// Find maximum size of DSU tree.
let maxi = findMax(n, W);
document.write("Maximum size of the subtree with ");
document.write("even weighted nodes = " + maxi);
// This code is contributed by divyesh072019
</script>
Producción:
Maximum size of the subtree with even weighted nodes = 4
Publicación traducida automáticamente
Artículo escrito por fsociety99 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA