2色树中色差最小的子树

具有 N 个节点和 N-1 条边的树为其节点提供了 2 种不同的颜色。
找到色差最小的子树，即 abs(1-color nodes – 2-colour nodes) 最小。
例子：

Input : 
Edges : 1 2
        1 3
        2 4
        3 5
Colours : 1 1 2 2 1 [1-based indexing where 
                    index denotes the node]
Output : 2
Explanation : The sub-tree {1-2} and {1-2-3-5}
have color difference of 2. Sub-tree {1-2} has two
1-colour nodes and zero 2-colour nodes. So, color 
difference is 2. Sub-tree {1-2-3-5} has three 1-colour
nodes and one 2-colour nodes. So color diff = 2.

方法 1：可以通过检查树的每个节点的每个可能的子树来解决问题。这将花费指数时间，因为我们将检查每个节点的子树。
方法 2：(高效)如果我们观察，我们正在多次求解树的一部分。这会产生重复出现的子问题。我们可以使用动态规划方法在一次遍历中获得最小的色差。为了让事情更简单，我们可以将颜色值设置为 1 和 -1。现在，如果我们有一个两个颜色节点相等的子树，我们的颜色总和将为 0。为了获得最小差异，我们应该有最大的负和或最大的正和。

• 案例 1当我们需要一个具有最大 sum 的子树时：如果它的值 > 0，我们就取一个节点，即 sum(parent) += max(0, sum(child))
• 案例 2当我们需要一个具有最小总和(或最大负总和)的子树时：如果它的值 < 0，我们就取一个节点，即 sum(parent) += min(0, sum(child))

为了得到最小和，我们可以交换节点的颜色，即-1变成1，反之亦然。
下面是实现：

C++

// CPP code to find the sub-tree with minimum color
// difference in a 2-coloured tree
#include 
using namespace std;
 
// Tree traversal to compute minimum difference
void dfs(int node, int parent, vector tree[],
                    int colour[], int answer[])
{
    // Initial min difference is the color of node
    answer[node] = colour[node];
 
    // Traversing its children
    for (auto u : tree[node]) {
 
        // Not traversing the parent
        if (u == parent)
            continue;
 
        dfs(u, node, tree, colour, answer);
 
        // If the child is adding positively to
        // difference, we include it in the answer
        // Otherwise, we leave the sub-tree and
        // include 0 (nothing) in the answer
        answer[node] += max(answer[u], 0);
    }
}
 
int maxDiff(vector tree[], int colour[], int N)
{
       int answer[N + 1];
       memset(answer, 0, sizeof(answer));
 
    // DFS for colour difference : 1colour - 2colour
    dfs(1, 0, tree, colour, answer);
 
    // Minimum colour difference is maximum answer value
    int high = 0;
    for (int i = 1; i <= N; i++) {
        high = max(high, answer[i]);
 
        // Clearing the current value
        // to check for colour2 as well
        answer[i] = 0;
    }
 
    // Interchanging the colours
    for (int i = 1; i <= N; i++) {
        if (colour[i] == -1)
            colour[i] = 1;
        else
            colour[i] = -1;
    }
 
    // DFS for colour difference : 2colour - 1colour
    dfs(1, 0, tree, colour, answer);
 
    // Checking if colour2 makes the minimum colour
    // difference
    for (int i = 1; i < N; i++)
        high = max(high, answer[i]);
         
    return high;
}
 
// Driver code
int main()
{
    // Nodes
    int N = 5;
 
    // Adjacency list representation
    vector tree[N + 1];
 
    // Edges
    tree[1].push_back(2);
    tree[2].push_back(1);
 
    tree[1].push_back(3);
    tree[3].push_back(1);
 
    tree[2].push_back(4);
    tree[4].push_back(2);
 
    tree[3].push_back(5);
    tree[5].push_back(3);
 
    // Index represent the colour of that node
    // There is no Node 0, so we start from
    // index 1 to N
    int colour[] = { 0, 1, 1, -1, -1, 1 };
 
    // Printing the result
    cout << maxDiff(tree,  colour,  N);
     
    return 0;
}

Java

// Java code to find the sub-tree
// with minimum color difference
// in a 2-coloured tree
import java.util.*;
class GFG
{
 
// Tree traversal to compute minimum difference
static void dfs(int node, int parent,
                Vector tree[], 
                int colour[], int answer[])
{
    // Initial min difference is
    // the color of node
    answer[node] = colour[node];
 
    // Traversing its children
    for (Integer u : tree[node])
    {
 
        // Not traversing the parent
        if (u == parent)
            continue;
 
        dfs(u, node, tree, colour, answer);
 
        // If the child is adding positively to
        // difference, we include it in the answer
        // Otherwise, we leave the sub-tree and
        // include 0 (nothing) in the answer
        answer[node] += Math.max(answer[u], 0);
    }
}
 
static int maxDiff(Vector tree[],
                   int colour[], int N)
{
    int []answer = new int[N + 1];
 
    // DFS for colour difference : 1colour - 2colour
    dfs(1, 0, tree, colour, answer);
 
    // Minimum colour difference is
    // maximum answer value
    int high = 0;
    for (int i = 1; i <= N; i++)
    {
        high = Math.max(high, answer[i]);
 
        // Clearing the current value
        // to check for colour2 as well
        answer[i] = 0;
    }
 
    // Interchanging the colours
    for (int i = 1; i <= N; i++)
    {
        if (colour[i] == -1)
            colour[i] = 1;
        else
            colour[i] = -1;
    }
 
    // DFS for colour difference : 2colour - 1colour
    dfs(1, 0, tree, colour, answer);
 
    // Checking if colour2 makes the
    // minimum colour difference
    for (int i = 1; i < N; i++)
        high = Math.max(high, answer[i]);
         
    return high;
}
 
// Driver code
public static void main(String []args)
{
     
    // Nodes
    int N = 5;
 
    // Adjacency list representation
    Vector tree[] = new Vector[N + 1];
    for(int i = 0; i < N + 1; i++)
        tree[i] = new Vector();
 
    // Edges
    tree[1].add(2);
    tree[2].add(1);
 
    tree[1].add(3);
    tree[3].add(1);
 
    tree[2].add(4);
    tree[4].add(2);
 
    tree[3].add(5);
    tree[5].add(3);
 
    // Index represent the colour of that node
    // There is no Node 0, so we start from
    // index 1 to N
    int colour[] = { 0, 1, 1, -1, -1, 1 };
 
    // Printing the result
    System.out.println(maxDiff(tree, colour, N));
}
}
 
// This code is contributed by 29AjayKumar

C#

// C# code to find the sub-tree
// with minimum color difference
// in a 2-coloured tree
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Tree traversal to compute minimum difference
static void dfs(int node, int parent,
                List []tree,
                int []colour, int []answer)
{
    // Initial min difference is
    // the color of node
    answer[node] = colour[node];
 
    // Traversing its children
    foreach (int u in tree[node])
    {
 
        // Not traversing the parent
        if (u == parent)
            continue;
 
        dfs(u, node, tree, colour, answer);
 
        // If the child is Adding positively to
        // difference, we include it in the answer
        // Otherwise, we leave the sub-tree and
        // include 0 (nothing) in the answer
        answer[node] += Math.Max(answer[u], 0);
    }
}
 
static int maxDiff(List []tree,
                         int []colour, int N)
{
    int []answer = new int[N + 1];
 
    // DFS for colour difference : 1colour - 2colour
    dfs(1, 0, tree, colour, answer);
 
    // Minimum colour difference is
    // maximum answer value
    int high = 0;
    for (int i = 1; i <= N; i++)
    {
        high = Math.Max(high, answer[i]);
 
        // Clearing the current value
        // to check for colour2 as well
        answer[i] = 0;
    }
 
    // Interchanging the colours
    for (int i = 1; i <= N; i++)
    {
        if (colour[i] == -1)
            colour[i] = 1;
        else
            colour[i] = -1;
    }
 
    // DFS for colour difference : 2colour - 1colour
    dfs(1, 0, tree, colour, answer);
 
    // Checking if colour2 makes the
    // minimum colour difference
    for (int i = 1; i < N; i++)
        high = Math.Max(high, answer[i]);
         
    return high;
}
 
// Driver code
public static void Main(String []args)
{
     
    // Nodes
    int N = 5;
 
    // Adjacency list representation
    List []tree = new List[N + 1];
    for(int i = 0; i < N + 1; i++)
        tree[i] = new List();
 
    // Edges
    tree[1].Add(2);
    tree[2].Add(1);
 
    tree[1].Add(3);
    tree[3].Add(1);
 
    tree[2].Add(4);
    tree[4].Add(2);
 
    tree[3].Add(5);
    tree[5].Add(3);
 
    // Index represent the colour of that node
    // There is no Node 0, so we start from
    // index 1 to N
    int []colour = { 0, 1, 1, -1, -1, 1 };
 
    // Printing the result
    Console.WriteLine(maxDiff(tree, colour, N));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// JavaScript code to find the sub-tree
// with minimum color difference
// in a 2-coloured tree
 
// Tree traversal to compute minimum difference
function dfs(node, parent, tree, colour, answer)
{
    // Initial min difference is
    // the color of node
    answer[node] = colour[node];
 
    // Traversing its children
    for(var u of tree[node])
    {
 
        // Not traversing the parent
        if (u == parent)
            continue;
 
        dfs(u, node, tree, colour, answer);
 
        // If the child is Adding positively to
        // difference, we include it in the answer
        // Otherwise, we leave the sub-tree and
        // include 0 (nothing) in the answer
        answer[node] += Math.max(answer[u], 0);
    }
}
 
function maxDiff(tree, colour, N)
{
    var answer = Array(N+1).fill(0);
 
    // DFS for colour difference : 1colour - 2colour
    dfs(1, 0, tree, colour, answer);
 
    // Minimum colour difference is
    // maximum answer value
    var high = 0;
    for (var i = 1; i <= N; i++)
    {
        high = Math.max(high, answer[i]);
 
        // Clearing the current value
        // to check for colour2 as well
        answer[i] = 0;
    }
 
    // Interchanging the colours
    for (var i = 1; i <= N; i++)
    {
        if (colour[i] == -1)
            colour[i] = 1;
        else
            colour[i] = -1;
    }
 
    // DFS for colour difference : 2colour - 1colour
    dfs(1, 0, tree, colour, answer);
 
    // Checking if colour2 makes the
    // minimum colour difference
    for (var i = 1; i < N; i++)
        high = Math.max(high, answer[i]);
         
    return high;
}
 
// Driver code
// Nodes
var N = 5;
// Adjacency list representation
var tree = Array.from(Array(N+1), ()=>Array());
 
// Edges
tree[1].push(2);
tree[2].push(1);
tree[1].push(3);
tree[3].push(1);
tree[2].push(4);
tree[4].push(2);
tree[3].push(5);
tree[5].push(3);
// Index represent the colour of that node
// There is no Node 0, so we start from
// index 1 to N
var colour = [0, 1, 1, -1, -1, 1];
// Printing the result
document.write(maxDiff(tree, colour, N));
 
 

输出：

2

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