给定一棵N-Ary树，其中包含N个节点和一个数组权重[] ，该权重表示可以为正数或负数的节点的权重，每个节点的任务是通过包括当前节点的一系列节点来打印可能的最大和。 。
例子：
Input: N = 7
weight[] = [-8, 9, 7, -4, 5, -10, -6]
N-Ary tree:
-8
/ \
9 7
/ \ /
-4 5 -10
/
-6
Output: 13 14 13 10 14 3 4
Explanations:
Node -8: [-8 + 9 + 7 + 5] = 13
Node 9: [9 + 5] = 14
Node 3: [7 + (-8) + 9 + 5] = 13
Node 4: [-4 + 9 + 5] = 10
Node: [5 + 9] = 14
Node 6: [-10 + 7 + (-8) + 9 + 5] = 3
Node 7: [-6 + (-4) + 9 + 5] = 4
Input: N = 6
weight[] = [2, -7, -5, 8, 4, -10]
N-Ary tree:
2
/ \
-7 -5
/ \ \
8 4 -10
Output: 7 7 2 8 7 -8方法：可通过应用两个DFS来使用Dp on Trees技术来解决此问题。
- 应用第一个DFS可以通过将每个节点及其后继节点按顺序包括在其中来存储每个节点可能的最大和。将最大可能的和存储在dp1 []中。大批。
- 可以通过以下方式获得第一个DFS中每个节点的最大可能值：
dp1[node] += maximum(0, dp1[child1], dp1[child2], …)
- 通过在第二个Dfs中将它们的祖先也包含在一个序列中，来执行第二个Dfs以更新dp1 []中每个节点的最大和。每个节点中dp2 []中存储的最大值是必需的答案。
- 可以通过以下方式获得第二个DFS中每个节点的最大可能值：
dp2[node] = dp1[node] + maximum(0, maxSumAncestors)
Best answer can be obtained by including or excluding the maximum sum possible for its ancestors
where maxSumAncestors = dp2[parent] – maximum(0, dp1[node]), i.e. including or excluding contribution of the maximum sum of the current node stored in dp1[]
请参阅图片说明以更好地理解：
下面是上述方法的实现：
C++
// C++ program to calculate the maximum
// sum possible for every node by including
// it in a segment of the N-Ary Tree
#include
using namespace std;
// Stores the maximum
// sum possible for every node
// by including them in a segment
// with their successors
int dp1[100005];
// Stores the maximum
// sum possible for every node
// by including them in a segment
// with their ancestors
int dp2[100005];
// Store the maximum sum
// for every node by
// including it in a
// segment with its successors
void dfs1(int u, int par,
vector g[],
int weight[])
{
dp1[u] = weight[u];
for (auto c: g[u]) {
if (c != par) {
dfs1(c, u, g, weight);
dp1[u] += max(0, dp1);
}
}
}
// Update the maximum sums
// for each node by including
// them in a sequence with
// their ancestors
void dfs2(int u, int par,
vector g[],
int weight[])
{
// Condition to check,
// if current node is not root
if (par != 0) {
int maxSumAncestors = dp2[par]
- max(0, dp1[u]);
dp2[u] = dp1[u] + max(0,
maxSumAncestors);
}
for (auto c: g[u]) {
if (c != par) {
dfs2(c, u, g, weight);
}
}
}
// Add edges
void addEdge(int u, int v, vector g[])
{
g[u].push_back(v);
g[v].push_back(u);
}
// Function to find the maximum
// answer for each node
void maxSumSegments(vector g[],
int weight[],
int n)
{
// Compute the maximum sums
// with successors
dfs1(1, 0, g, weight);
// Store the computed maximums
for (int i = 1; i <= n; i++) {
dp2[i] = dp1[i];
}
// Update the maximum sums
// by including their
// ancestors
dfs2(1, 0, g, weight);
}
// Print the desired result
void printAns(int n)
{
for (int i = 1; i <= n; i++) {
cout << dp2[i] << " ";
}
}
// Driver Program
int main()
{
// Number of nodes
int n = 6;
int u, v;
// graph
vector g[100005];
// Add edges
addEdge(1, 2, g);
addEdge(1, 3, g);
addEdge(2, 4, g);
addEdge(2, 5, g);
addEdge(3, 6, g);
addEdge(4, 7, g);
// Weight of each node
int weight[n + 1];
weight[1] = -8;
weight[2] = 9;
weight[3] = 7;
weight[4] = -4;
weight[5] = 5;
weight[6] = -10;
weight[7] = -6;
// Compute the max sum
// of segments for each
// node
maxSumSegments(g, weight, n);
// Print the answer
// for every node
printAns(n);
return 0;
} Python3
# Python3 program to calculate the maximum
# sum possible for every node by including
# it in a segment of the N-Ary Tree
# Stores the maximum
# sum possible for every node
# by including them in a segment
# with their successors
dp1 = [0 for i in range(100005)]
# Stores the maximum sum possible
# for every node by including them
# in a segment with their ancestors
dp2 = [0 for i in range(100005)]
# Store the maximum sum for every
# node by including it in a
# segment with its successors
def dfs1(u, par, g, weight):
dp1[u] = weight[u]
for c in g[u]:
if (c != par):
dfs1(c, u, g, weight)
dp1[u] += max(0, dp1)
# Update the maximum sums
# for each node by including
# them in a sequence with
# their ancestors
def dfs2(u, par, g, weight):
# Condition to check,
# if current node is not root
if (par != 0):
maxSumAncestors = dp2[par] - max(0, dp1[u])
dp2[u] = dp1[u] + max(0, maxSumAncestors)
for c in g[u]:
if (c != par):
dfs2(c, u, g, weight)
# Add edges
def addEdge(u, v, g):
g[u].append(v)
g[v].append(u)
# Function to find the maximum
# answer for each node
def maxSumSegments(g, weight, n):
# Compute the maximum sums
# with successors
dfs1(1, 0, g, weight)
# Store the computed maximums
for i in range(1, n + 1):
dp2[i] = dp1[i]
# Update the maximum sums
# by including their
# ancestors
dfs2(1, 0, g, weight)
# Print the desired result
def printAns(n):
for i in range(1, n):
print(dp2[i], end = ' ')
# Driver code
if __name__=='__main__':
# Number of nodes
n = 7
u = 0
v = 0
# Graph
g = [[] for i in range(100005)]
# Add edges
addEdge(1, 2, g)
addEdge(1, 3, g)
addEdge(2, 4, g)
addEdge(2, 5, g)
addEdge(3, 6, g)
addEdge(4, 7, g)
# Weight of each node
weight=[0 for i in range(n + 1)]
weight[1] = -8
weight[2] = 9
weight[3] = 7
weight[4] = -4
weight[5] = 5
weight[6] = -10
weight[7] = -6
# Compute the max sum
# of segments for each
# node
maxSumSegments(g, weight, n)
# Print the answer
# for every node
printAns(n)
# This code is contributed by pratham76输出：
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