给定二维数组Edges [] [] ,它表示有向无环连通图中一对节点之间的有向边,该有向无环连通图中包含从[1,N]取值的N个节点和代表每个节点权重的数组arr []是找到任何节点及其任何祖先权重之间的最大绝对差。
例子:
Input: N = 5, M = 4, Edges[][2] = {{1, 2}, {2, 3}, {4, 5}, {1, 3}}, arr[] = {13, 8, 3, 15, 18}
Output: 10
Explanation:
From the above graph, it can be observed that the maximum difference between the value of any node and any of its ancestor = 18 (Node 5) – 8 (Node 2) = 10.
Input: N = 4, M = 3, Edges[][2] = {{1, 2}, {2, 4}, {1, 3}}, arr[] = {2, 3, 1, 5}
Output: 3
方法:解决给定问题的想法是对图执行DFS遍历,并填充每个节点到其子节点的最大值和最小值,并找到最大绝对差。
请按照以下步骤解决给定的问题:
- 初始化一个变量,将ans表示为INT_MIN,以存储所需的最大差异。
- 通过执行以下操作,对给定图执行DFS遍历,以找到节点及其任何祖先权重之间的最大绝对差:
- 对于每个源节点,例如src ,更新ans的值以分别存储src的权重与currentMin和currentMax之间的绝对差的最大值。
- 更新为最小currentMin的和源节点SRC的值currentMin的值。
- 将currentMax的值更新为currentMax和源节点src的最大值。
- 现在,递归地遍历src的子节点,并将currentMax和currentMin的值更新为DFS(child,Adj,ans,currentMin,currentMax) 。
- 完成上述步骤后,打印ans的值作为结果的最大差异。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to perform DFS
// Traversal on the given graph
void DFS(int src, vector Adj[],
int& ans, int arr[],
int currentMin, int currentMax)
{
// Update the value of ans
ans = max(
ans, max(abs(
currentMax - arr[src - 1]),
abs(currentMin - arr[src - 1])));
// Update the currentMin and currentMax
currentMin = min(currentMin,
arr[src - 1]);
currentMax = min(currentMax,
arr[src - 1]);
// Traverse the adjacency
// list of the node src
for (auto& child : Adj[src]) {
// Recursively call
// for the child node
DFS(child, Adj, ans, arr,
currentMin, currentMax);
}
}
// Function to calculate maximum absolute
// difference between a node and its ancestor
void getMaximumDifference(int Edges[][2],
int arr[], int N,
int M)
{
// Stores the adjacency list of graph
vector Adj[N + 1];
// Create Adjacency list
for (int i = 0; i < M; i++) {
int u = Edges[i][0];
int v = Edges[i][1];
// Add a directed edge
Adj[u].push_back(v);
}
int ans = 0;
// Perform DFS Traversal
DFS(1, Adj, ans, arr,
arr[0], arr[0]);
// Print the maximum
// absolute difference
cout << ans;
}
// Driver Code
int main()
{
int N = 5, M = 4;
int Edges[][2]
= { { 1, 2 }, { 2, 3 },
{ 4, 5 }, { 1, 3 } };
int arr[] = { 13, 8, 3, 15, 18 };
getMaximumDifference(Edges, arr, N, M);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
static int ans;
// Function to perform DFS
// Traversal on the given graph
static void DFS(int src,
ArrayList > Adj,
int arr[], int currentMin,
int currentMax)
{
// Update the value of ans
ans = Math.max(ans,
Math.max(Math.abs(currentMax - arr[src - 1]),
Math.abs(currentMin - arr[src - 1])));
// Update the currentMin and currentMax
currentMin = Math.min(currentMin, arr[src - 1]);
currentMax = Math.min(currentMax, arr[src - 1]);
// Traverse the adjacency
// list of the node src
for(Integer child : Adj.get(src))
{
// Recursively call
// for the child node
DFS(child, Adj, arr, currentMin, currentMax);
}
}
// Function to calculate maximum absolute
// difference between a node and its ancestor
static void getMaximumDifference(int Edges[][],
int arr[], int N,
int M)
{
ans = 0;
// Stores the adjacency list of graph
ArrayList> Adj = new ArrayList<>();
for(int i = 0; i < N + 1; i++)
Adj.add(new ArrayList<>());
// Create Adjacency list
for(int i = 0; i < M; i++)
{
int u = Edges[i][0];
int v = Edges[i][1];
// Add a directed edge
Adj.get(u).add(v);
}
// Perform DFS Traversal
DFS(1, Adj, arr, arr[0], arr[0]);
// Print the maximum
// absolute difference
System.out.println(ans);
}
// Driver code
public static void main(String[] args)
{
int N = 5, M = 4;
int Edges[][] = { { 1, 2 }, { 2, 3 },
{ 4, 5 }, { 1, 3 } };
int arr[] = { 13, 8, 3, 15, 18 };
getMaximumDifference(Edges, arr, N, M);
}
}
// This code is contributed by offbeat
Python3
# Python3 program for the above approach
ans = 0
# Function to perform DFS
# Traversal on the given graph
def DFS(src, Adj, arr, currentMin, currentMax):
# Update the value of ans
global ans
ans = max(ans, max(abs(currentMax - arr[src - 1]),
abs(currentMin - arr[src - 1])))
# Update the currentMin and currentMax
currentMin = min(currentMin, arr[src - 1])
currentMax = min(currentMax, arr[src - 1])
# Traverse the adjacency
# list of the node src
for child in Adj[src]:
# Recursively call
# for the child node
DFS(child, Adj, arr, currentMin, currentMax)
# Function to calculate maximum absolute
# difference between a node and its ancestor
def getMaximumDifference(Edges, arr, N, M):
global ans
# Stores the adjacency list of graph
Adj = [[] for i in range(N + 1)]
# Create Adjacency list
for i in range(M):
u = Edges[i][0]
v = Edges[i][1]
# Add a directed edge
Adj[u].append(v)
# Perform DFS Traversal
DFS(1, Adj, arr, arr[0], arr[0])
# Print the maximum
# absolute difference
print(ans)
# Driver Code
if __name__ == '__main__':
N = 5
M = 4
Edges = [[1, 2], [2, 3], [4, 5], [1, 3]]
arr = [13, 8, 3, 15, 18]
getMaximumDifference(Edges, arr, N, M)
# This code is contributed by ipg2016107
输出:
10
时间复杂度: O(N + M)
辅助空间: O(N)