使用给定颜色为树着色的最小步骤

给定一棵树，它有N个最初没有颜色的节点，还有一个大小为N的数组color[] ，表示着色过程发生后每个节点的颜色。任务是使用尽可能少的步骤将树着色为给定的颜色。在每一步中，可以选择一个顶点v和一个颜色x ，然后用颜色x为v的子树（包括 v 本身）中的所有顶点着色。请注意，根是顶点编号 1。
例子：

Input: color[] = { 1, 1, 2, 1, 3, 1} 
 

Output: 4 
Color the sub-tree rooted at node 1 with color 1. 
Then all the vertices have colors 1. 
Now, color the sub-tree rooted at 3 with color 2. 
Finally, color the sub-trees rooted at 5 and 6 with colors 3 and 1 respectively.
Input: color[] = { 1, 2, 3, 2, 2, 3} 
 

Output: 3 
 

方法：在顶点 1 调用一个 DFS函数，最初保持 answer 为零。只要子节点和父节点的颜色不同，就增加答案。
请参阅下面的代码以更好地理解。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// To store the required answer
int ans = 0;
 
// To store the graph
vector gr[100005];
 
// Function to add edges
void Add_Edge(int u, int v)
{
    gr[u].push_back(v);
    gr[v].push_back(u);
}
 
// Dfs function
void dfs(int child, int par, int color[])
{
 
    // When there is difference in colors
    if (color[child] != color[par])
        ans++;
 
    // For all it's child nodes
    for (auto it : gr[child]) {
        if (it == par)
            continue;
        dfs(it, child, color);
    }
}
 
// Driver code
int main()
{
 
    // Here zero is for parent of node 1
    int color[] = { 0, 1, 2, 3, 2, 2, 3 };
 
    // Adding edges in the graph
    Add_Edge(1, 2);
    Add_Edge(1, 3);
    Add_Edge(2, 4);
    Add_Edge(2, 5);
    Add_Edge(3, 6);
 
    // Dfs call
    dfs(1, 0, color);
 
    // Required answer
    cout << ans;
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// To store the required answer
static int ans = 0;
 
// To store the graph
static Vector> gr = new Vector>();
 
// Function to add edges
static void Add_Edge(int u, int v)
{
    gr.get(u).add(v);
    gr.get(v).add(u);
}
 
// Dfs function
static void dfs(int child, int par, int color[])
{
 
    // When there is difference in colors
    if (color[child] != color[par])
        ans++;
 
    // For all it's child nodes
    for (int i = 0; i < gr.get(child).size(); i++)
    {
        if (gr.get(child).get(i) == par)
            continue;
        dfs(gr.get(child).get(i), child, color);
    }
}
 
// Driver code
public static void main(String args[])
{
    for(int i = 0; i <= 10; i++)
    gr.add(new Vector());
 
    // Here zero is for parent of node 1
    int color[] = { 0, 1, 2, 3, 2, 2, 3 };
 
    // Adding edges in the graph
    Add_Edge(1, 2);
    Add_Edge(1, 3);
    Add_Edge(2, 4);
    Add_Edge(2, 5);
    Add_Edge(3, 6);
 
    // Dfs call
    dfs(1, 0, color);
 
    // Required answer
    System.out.println( ans);
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
 
# To store the required answer
ans = 0
 
# To store the graph
gr = [[] for i in range(100005)]
 
# Function to add edges
def Add_Edge(u, v):
    gr[u].append(v)
    gr[v].append(u)
 
# Dfs function
def dfs(child, par, color):
    global ans
 
    # When there is difference in colors
    if (color[child] != color[par]):
        ans += 1
 
    # For all it's child nodes
    for it in gr[child]:
        if (it == par):
            continue
        dfs(it, child, color)
     
# Driver code
 
# Here zero is for parent of node 1
color = [0, 1, 2, 3, 2, 2, 3]
 
# Adding edges in the graph
Add_Edge(1, 2)
Add_Edge(1, 3)
Add_Edge(2, 4)
Add_Edge(2, 5)
Add_Edge(3, 6)
 
# Dfs call
dfs(1, 0, color)
 
# Required answer
print(ans)
 
# This code is contributed
# by mohit kumar

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // To store the required answer
    static int ans = 0;
     
    // To store the graph
    static List> gr = new List>();
     
    // Function to add edges
    static void Add_Edge(int u, int v)
    {
        gr[u].Add(v);
        gr[v].Add(u);
    }
     
    // Dfs function
    static void dfs(int child, int par, int []color)
    {
     
        // When there is difference in colors
        if (color[child] != color[par])
            ans++;
     
        // For all it's child nodes
        for (int i = 0; i < gr[child].Count; i++)
        {
            if (gr[child][i] == par)
                continue;
            dfs(gr[child][i], child, color);
        }
    }
 
    // Driver code
    public static void Main(String []args)
    {
        for(int i = 0; i <= 10; i++)
        gr.Add(new List());
     
        // Here zero is for parent of node 1
        int []color = { 0, 1, 2, 3, 2, 2, 3 };
     
        // Adding edges in the graph
        Add_Edge(1, 2);
        Add_Edge(1, 3);
        Add_Edge(2, 4);
        Add_Edge(2, 5);
        Add_Edge(3, 6);
     
        // Dfs call
        dfs(1, 0, color);
     
        // Required answer
        Console.WriteLine( ans);
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript

// Javascript implementation of the approach
 
// To store the required answer
let ans = 0;
 
// To store the graph
let gr = [];
 
// Function to add edges
function Add_Edge(u,v)
{
    gr[u].push(v);
    gr[v].push(u);
}
 
// Dfs function
function dfs(child,par,color)
{
    // When there is difference in colors
    if (color[child] != color[par])
        ans++;
   
    // For all it's child nodes
    for (let i = 0; i < gr[child].length; i++)
    {
        if (gr[child][i] == par)
            continue;
        dfs(gr[child][i], child, color);
    }
}
 
// Driver code
for(let i = 0; i <= 10; i++)
    gr.push([]);
   
    // Here zero is for parent of node 1
    let color = [ 0, 1, 2, 3, 2, 2, 3 ];
   
    // Adding edges in the graph
    Add_Edge(1, 2);
    Add_Edge(1, 3);
    Add_Edge(2, 4);
    Add_Edge(2, 5);
    Add_Edge(3, 6);
   
    // Dfs call
    dfs(1, 0, color);
   
    // Required answer
    document.write( ans);
     
// This code is contributed by unknown2108

输出：

3

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