树上的动态编程套装1

动态编程(DP)是一种通过将问题分解为遵循最佳子结构的重叠子问题来解决问题的技术。使用DP时会遇到各种问题，例如子集总和，背包，零钱等。DP也可以应用于树上以解决一些特定问题。
先决条件： DFS
给定一棵具有N个节点和N-1个边的树，无需重新访问任何节点即可计算从根到任何叶子的节点值的最大和。

上面给出的是N = 14个节点和N-1 = 13个边的树的示意图。节点1、2、3、4….14的节点值分别为3、2、1、10、1、3、9、1、5、3、4、5、9和8。
下图显示了从根到叶的所有路径：

所有路径都用不同的颜色标记：
路径1(红色，3-2-1-4)：所有节点值的总和= 10
路径2(橙色，3-2-1-5)：所有节点值的总和= 11
路径3(黄色，3-2-3)：所有节点值的总和= 8
路径4(绿色，3-1-9-9)：所有节点值的总和= 22
路径5(紫色，3-1-9-8)：所有节点值的总和= 21
路径6(粉红色，3-10-1)：所有节点值的总和= 14
路径7(蓝色，3-10-5)：所有节点值的总和= 18
路径8(棕色，3-10-3)：所有节点值的总和= 16
答案是22，因为路径4在从根到叶的路径中具有节点值的最大和。

在这种情况下，贪婪的方法失败了。从根开始，贪婪地从第一个级别获取3个，从下一个级别获取10个，从第三个级别获取5个。如果遵循贪婪方法，则结果为路径7，因此此处不应用贪婪方法。
使用树上的动态编程可以解决该问题。从叶子开始记忆，然后将最大数量的叶子添加到每个子树的根。在最后一步，将存在根及其下的子树，将节点处的值与子树的最大值相加，将得到从根到任何叶子的节点值的最大和。

上图显示了如何从叶子开始，并将子树的最大叶子添加到其根。向上移动并重复相同的过程，即存储每个子树叶子的最大值并将其添加到其根中。在此示例中，对节点11和12的最大值进行计数，然后将其添加到节点5(在此子树中，5是根，而11和12是其叶子)。同样，对节点13和14的最大值进行计数，然后将其添加到节点7。对每个子树重复上述步骤，直到到达节点。

令DP _i为i与i向下移动的任何叶子之间的路径中节点值的最大总和。使用DFS遍历遍历树。存储子树的所有叶子的最大值，并将其添加到子树的根。最后，DP ₁将具有从根到任何叶子的最大节点值总和，而无需重新访问任何节点。
下面是上述想法的实现：

C++

// C++ code to find the maximum path sum
#include 
using namespace std;
 
int dp[100];
 
// function for dfs traversal and to store the
// maximum value in dp[] for every node till the leaves
void dfs(int a[], vector v[], int u, int parent)
{
    // initially dp[u] is always a[u]
    dp[u] = a[u - 1];
 
    // stores the maximum value from nodes
    int maximum = 0;
 
    // traverse the tree
    for (int child : v[u]) {
 
        // if child is parent, then we continue
        // without recursing further
        if (child == parent)
            continue;
 
        // call dfs for further traversal
        dfs(a, v, child, u);
 
        // store the maximum of previous visited node
        // and present visited node
        maximum = max(maximum, dp[child]);
    }
 
    // add the maximum value returned to the parent node
    dp[u] += maximum;
}
 
// function that returns the maximum value
int maximumValue(int a[], vector v[])
{
    dfs(a, v, 1, 0);
    return dp[1];
}
 
// Driver Code
int main()
{
    // number of nodes
    int n = 14;
 
    // adjacency list
    vector v[n + 1];
 
    // create undirected edges
    // initialize the tree given in the diagram
    v[1].push_back(2), v[2].push_back(1);
    v[1].push_back(3), v[3].push_back(1);
    v[1].push_back(4), v[4].push_back(1);
    v[2].push_back(5), v[5].push_back(2);
    v[2].push_back(6), v[6].push_back(2);
    v[3].push_back(7), v[7].push_back(3);
    v[4].push_back(8), v[8].push_back(4);
    v[4].push_back(9), v[9].push_back(4);
    v[4].push_back(10), v[10].push_back(4);
    v[5].push_back(11), v[11].push_back(5);
    v[5].push_back(12), v[12].push_back(5);
    v[7].push_back(13), v[13].push_back(7);
    v[7].push_back(14), v[14].push_back(7);
 
    // values of node 1, 2, 3....14
    int a[] = { 3, 2, 1, 10, 1, 3, 9, 1, 5, 3, 4, 5, 9, 8 };
 
    // function call
    cout << maximumValue(a, v);
 
    return 0;
}

Java

// Java code to find the maximum path sum
import java.util.Vector;
 
class GFG
{
    static int[] dp = new int[100];
 
    // function for dfs traversal and to
    // store the maximum value in dp[]
    // for every node till the leaves
    static void dfs(int[] a, Vector[] v,
                    int u, int parent)
    {
 
        // initially dp[u] is always a[u]
        dp[u] = a[u - 1];
 
        // stores the maximum value from nodes
        int maximum = 0;
 
        // traverse the tree
        for (int child : v[u])
        {
 
            // if child is parent, then we continue
            // without recursing further
            if (child == parent)
                continue;
 
            // call dfs for further traversal
            dfs(a, v, child, u);
 
            // store the maximum of previous visited
            // node and present visited node
            maximum = Math.max(maximum, dp[child]);
        }
 
        // add the maximum value returned
        // to the parent node
        dp[u] += maximum;
    }
 
    // function that returns the maximum value
    static int maximumValue(int[] a,
                            Vector[] v)
    {
        dfs(a, v, 1, 0);
        return dp[1];
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        // Driver Code
        int n = 14;
 
        // adjacency list
        @SuppressWarnings("unchecked")
        Vector[] v = new Vector[n + 1];
 
        for (int i = 0; i < v.length; i++)
            v[i] = new Vector<>();
 
        // create undirected edges
        // initialize the tree given in the diagram
        v[1].add(2); v[2].add(1);
        v[1].add(3); v[3].add(1);
        v[1].add(4); v[4].add(1);
        v[2].add(5); v[5].add(2);
        v[2].add(6); v[6].add(2);
        v[3].add(7); v[7].add(3);
        v[4].add(8); v[8].add(4);
        v[4].add(9); v[9].add(4);
        v[4].add(10); v[10].add(4);
        v[5].add(11); v[11].add(5);
        v[5].add(12); v[12].add(5);
        v[7].add(13); v[13].add(7);
        v[7].add(14); v[14].add(7);
 
        // values of node 1, 2, 3....14
        int a[] = { 3, 2, 1, 10, 1, 3, 9,
                    1, 5, 3, 4, 5, 9, 8 };
 
        // function call
        System.out.println(maximumValue(a, v));
    }
}
 
// This code is contributed by
// sanjeev2552

Python3

# Python3 code to find the maximum path sum
dp = [0]*100
 
# Function for dfs traversal and
# to store the maximum value in
# dp[] for every node till the leaves
def dfs(a, v, u, parent):
     
    # Initially dp[u] is always a[u]
    dp[u] = a[u - 1]
     
    # Stores the maximum value from nodes
    maximum = 0
     
    # Traverse the tree
    for child in v[u]:
         
        # If child is parent, then we continue
        # without recursing further
        if child == parent:
            continue
         
        # Call dfs for further traversal
        dfs(a, v, child, u)
         
        # Store the maximum of previous visited 
        # node and present visited node
        maximum = max(maximum, dp[child])
         
    # Add the maximum value
    # returned to the parent node
    dp[u] += maximum
 
 
# Function that returns the maximum value
def maximumValue(a, v):
    dfs(a, v, 1, 0)
    return dp[1]
 
# Driver Code
def main():
     
    # Number of nodes
    n = 14
     
    # Adjacency list as a dictionary
    v = {}
    for i in range(n + 1):
        v[i] = []
         
    # Create undirected edges
    # initialize the tree given in the diagram
    v[1].append(2), v[2].append(1)
    v[1].append(3), v[3].append(1)
    v[1].append(4), v[4].append(1)
    v[2].append(5), v[5].append(2)
    v[2].append(6), v[6].append(2)
    v[3].append(7), v[7].append(3)
    v[4].append(8), v[8].append(4)
    v[4].append(9), v[9].append(4)
    v[4].append(10), v[10].append(4)
    v[5].append(11), v[11].append(5)
    v[5].append(12), v[12].append(5)
    v[7].append(13), v[13].append(7)
    v[7].append(14), v[14].append(7)
     
    # Values of node 1, 2, 3....14
    a = [ 3, 2, 1, 10, 1, 3, 9,
          1, 5, 3, 4, 5, 9, 8 ]
     
    # Function call
    print(maximumValue(a, v))
main()
 
# This code is contributed by stutipathak31jan

C#

// C# code to find the maximum path sum
using System;
using System.Collections.Generic;
 
class GFG
{
    static int[] dp = new int[100];
 
    // function for dfs traversal and to
    // store the maximum value in []dp
    // for every node till the leaves
    static void dfs(int[] a, List[] v,
                    int u, int parent)
    {
 
        // initially dp[u] is always a[u]
        dp[u] = a[u - 1];
 
        // stores the maximum value from nodes
        int maximum = 0;
 
        // traverse the tree
        foreach(int child in v[u])
        {
 
            // if child is parent, then we continue
            // without recursing further
            if (child == parent)
                continue;
 
            // call dfs for further traversal
            dfs(a, v, child, u);
 
            // store the maximum of previous visited
            // node and present visited node
            maximum = Math.Max(maximum, dp[child]);
        }
 
        // add the maximum value returned
        // to the parent node
        dp[u] += maximum;
    }
 
    // function that returns the maximum value
    static int maximumValue(int[] a,
                            List[] v)
    {
        dfs(a, v, 1, 0);
        return dp[1];
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
 
        // Driver Code
        int n = 14;
 
        // adjacency list
 
        List[] v = new List[n + 1];
 
        for (int i = 0; i < v.Length; i++)
            v[i] = new List();
 
        // create undirected edges
        // initialize the tree given in the diagram
        v[1].Add(2); v[2].Add(1);
        v[1].Add(3); v[3].Add(1);
        v[1].Add(4); v[4].Add(1);
        v[2].Add(5); v[5].Add(2);
        v[2].Add(6); v[6].Add(2);
        v[3].Add(7); v[7].Add(3);
        v[4].Add(8); v[8].Add(4);
        v[4].Add(9); v[9].Add(4);
        v[4].Add(10); v[10].Add(4);
        v[5].Add(11); v[11].Add(5);
        v[5].Add(12); v[12].Add(5);
        v[7].Add(13); v[13].Add(7);
        v[7].Add(14); v[14].Add(7);
 
        // values of node 1, 2, 3....14
        int []a = { 3, 2, 1, 10, 1, 3, 9,
                    1, 5, 3, 4, 5, 9, 8 };
 
        // function call
        Console.WriteLine(maximumValue(a, v));
    }
}
 
// This code is contributed by PrinciRaj1992

输出

22

时间复杂度： O(N)，其中N是节点数。