给定一棵树,以及所有节点的权重。每个查询包含两个整数u和v ,任务是在 u 和 v 之间的简单路径上找到最小和最大权重(两者都包括在内)。
例子:
Input:
Query=[{1, 3}, {2, 4}, {3, 5}]
Output:
-1 5
3 5
-2 5
Explanation:
Weight on path 1 to 3 is [-1, 5, -1]. Hence, the minimum and maximum weight is -1 and 5 respectively.
Weight on path 2 to 4 is [5, 3]. Hence, the minimum and maximum weight is 3 and 5 respectively.
Weight on path 2 to 4 is [-1, 5, -1, -2]. Hence, the minimum and maximum weight is -2 and 5 respectively.
方法:这个想法是使用二进制提升技术在树中使用 LCA。
- 二元提升是一种动态规划方法,我们预先计算数组lca[i][j]其中 i = [1, n], j = [1, log(n)] 和 lca[i][j] 包含 2第j个节点 i 的祖先。
- 为了计算 lca[][] 的值,可以使用以下递归
lca[i][j] = parent[i] if j = 0 and
lca[i][j] = lca[lca[i][j – 1]][j – 1] if j > 0.
- 因为我们将计算 lca[][] 数组,我们还将计算MinWeight[][]和MaxWeight[][]其中 MinWeight[i][j] 包含从节点 i 到其第 2 j个祖先的最小权重, MaxWeight[i][j] 包含从节点 i 到其第 2 j个祖先的最大权重
- 为了计算 MinWeight[][] 和 MaxWeight[] 的值,可以使用以下递归。
[Tex]MaxWeight[i][j] =\begin{cases} max (weight[i], weight[parent[i]]) & \text{ ;if } j=0 \\ max( MaxWeight[i][ j-1], MaxWeight[lca[i][j – 1]][j – 1]) & \text{ ;if } j>0 \end{cases} [/Tex]
- 在预计算之后,我们找到 (u, v) 之间的最小和最大权重,因为我们找到了 (u, v) 的最不共同祖先。
下面是上述方法的实现:
C++
// C++ Program to find the maximum and
// minimum weight between two nodes
// in the given tree using LCA
#include
using namespace std;
#define MAX 1000
#define log 10 // log2(MAX)
// Array to store the level
// of each node
int level[MAX];
int lca[MAX][log];
int minWeight[MAX][log];
int maxWeight[MAX][log];
// Vector to store tree
vector graph[MAX];
// Array to store weight of nodes
int weight[MAX];
void addEdge(int u, int v)
{
graph[u].push_back(v);
graph[v].push_back(u);
}
// Pre-Processing to calculate
// values of lca[][], MinWeight[][]
// and MaxWeight[][]
void dfs(int node, int parent, int h)
{
// Using recursion formula to
// calculate the values
// of lca[][]
lca[node][0] = parent;
// Storing the level of
// each node
level[node] = h;
if (parent != -1)
{
minWeight[node][0]
= min(weight[node], weight[parent]);
maxWeight[node][0]
= max(weight[node], weight[parent]);
}
for (int i = 1; i < log; i++)
{
if (lca[node][i - 1] != -1)
{
// Using recursion formula to
// calculate the values of lca[][],
// MinWeight[][] and MaxWeight[][]
lca[node][i] = lca[lca[node][i - 1]][i - 1];
minWeight[node][i]
= min(minWeight[node][i - 1],
minWeight[lca[node][i - 1]][i - 1]);
maxWeight[node][i]
= max(maxWeight[node][i - 1],
maxWeight[lca[node][i - 1]][i - 1]);
}
}
for (int i : graph[node])
{
if (i == parent)
continue;
dfs(i, node, h + 1);
}
}
// Function to find the minimum and
// maximum weights in the given range
void findMinMaxWeight(int u, int v)
{
int minWei = INT_MAX;
int maxWei = INT_MIN;
// The node which is present
// farthest from the root node
// is taken as v If u is
// farther from root node
// then swap the two
if (level[u] > level[v])
swap(u, v);
// Finding the ancestor of v
// which is at same level as u
for (int i = log - 1; i >= 0; i--)
{
if (lca[v][i] != -1
&& level[lca[v][i]] >= level[u])
{
// Calculating Minimum and
// Maximum Weight of node
// v till its 2^i-th ancestor
minWei = min(minWei, minWeight[v][i]);
maxWei = max(maxWei, maxWeight[v][i]);
v = lca[v][i];
}
}
// If u is the ancestor of v
// then u is the LCA of u and v
if (v == u)
{
cout << minWei << " " << maxWei << endl;
}
else {
// Finding the node closest to the
// root which is not the common
// ancestor of u and v i.e. a node
// x such that x is not the common
// ancestor of u and v but lca[x][0] is
for (int i = log - 1; i >= 0; i--)
{
if (lca[v][i] != lca[u][i])
{
// Calculating the minimum of
// MinWeight of v to its 2^i-th
// ancestor and MinWeight of u
// to its 2^i-th ancestor
minWei = min(minWei, min(minWeight[v][i],
minWeight[u][i]));
// Calculating the maximum of
// MaxWeight of v to its 2^i-th
// ancestor and MaxWeight of u
// to its 2^i-th ancestor
maxWei = max(maxWei, max(maxWeight[v][i],
maxWeight[u][i]));
v = lca[v][i];
u = lca[u][i];
}
}
// Calculating the Minimum of
// first ancestor of u and v
minWei = min(minWei,
min(minWeight[v][0], minWeight[u][0]));
// Calculating the maximum of
// first ancestor of u and v
maxWei = max(maxWei,
max(maxWeight[v][0], maxWeight[u][0]));
cout << minWei << " " << maxWei << endl;
}
}
// Driver Code
int main()
{
// Number of nodes
int n = 5;
// Add edges
addEdge(1, 2);
addEdge(1, 5);
addEdge(2, 4);
addEdge(2, 3);
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// Initialising lca values with -1
// Initialising MinWeight values
// with INT_MAX
// Initialising MaxWeight values
// with INT_MIN
for (int i = 1; i <= n; i++) {
for (int j = 0; j < log; j++) {
lca[i][j] = -1;
minWeight[i][j] = INT_MAX;
maxWeight[i][j] = INT_MIN;
}
}
// Perform DFS
dfs(1, -1, 0);
// Query 1: {1, 3}
findMinMaxWeight(1, 3);
// Query 2: {2, 4}
findMinMaxWeight(2, 4);
// Query 3: {3, 5}
findMinMaxWeight(3, 5);
return 0;
}
Java
// Java Program to find the maximum and
// minimum weight between two nodes
// in the given tree using LCA
import java.util.*;
class GFG{
static final int MAX = 1000;
// Math.log(MAX)
static final int log = 10 ;
// Array to store the
// level of each node
static int []level =
new int[MAX];
static int [][]lca =
new int[MAX][log];
static int [][]minWeight =
new int[MAX][log];
static int [][]maxWeight =
new int[MAX][log];
// Vector to store tree
static Vector []graph =
new Vector[MAX];
// Array to store
// weight of nodes
static int []weight =
new int[MAX];
private static void swap(int x,
int y)
{
int temp = x;
x = y;
y = temp;
}
static void addEdge(int u,
int v)
{
graph[u].add(v);
graph[v].add(u);
}
// Pre-Processing to
// calculate values of
// lca[][], MinWeight[][]
// and MaxWeight[][]
static void dfs(int node,
int parent, int h)
{
// Using recursion formula to
// calculate the values
// of lca[][]
lca[node][0] = parent;
// Storing the level of
// each node
level[node] = h;
if (parent != -1)
{
minWeight[node][0] = Math.min(weight[node],
weight[parent]);
maxWeight[node][0] = Math.max(weight[node],
weight[parent]);
}
for (int i = 1; i < log; i++)
{
if (lca[node][i - 1] != -1)
{
// Using recursion formula to
// calculate the values of lca[][],
// MinWeight[][] and MaxWeight[][]
lca[node][i] =
lca[lca[node][i - 1]][i - 1];
minWeight[node][i] =
Math.min(minWeight[node][i - 1],
minWeight[lca[node][i - 1]][i - 1]);
maxWeight[node][i] =
Math.max(maxWeight[node][i - 1],
maxWeight[lca[node][i - 1]][i - 1]);
}
}
for (int i : graph[node])
{
if (i == parent)
continue;
dfs(i, node, h + 1);
}
}
// Function to find the minimum and
// maximum weights in the given range
static void findMinMaxWeight(int u,
int v)
{
int minWei = Integer.MAX_VALUE;
int maxWei = Integer.MIN_VALUE;
// The node which is present
// farthest from the root node
// is taken as v If u is
// farther from root node
// then swap the two
if (level[u] > level[v])
swap(u, v);
// Finding the ancestor of v
// which is at same level as u
for (int i = log - 1; i >= 0; i--)
{
if (lca[v][i] != -1 &&
level[lca[v][i]] >= level[u])
{
// Calculating Minimum and
// Maximum Weight of node
// v till its 2^i-th ancestor
minWei = Math.min(minWei,
minWeight[v][i]);
maxWei = Math.max(maxWei,
maxWeight[v][i]);
v = lca[v][i];
}
}
// If u is the ancestor of v
// then u is the LCA of u and v
if (v == u)
{
System.out.print(minWei + " " +
maxWei + "\n");
}
else
{
// Finding the node closest to the
// root which is not the common
// ancestor of u and v i.e. a node
// x such that x is not the common
// ancestor of u and v but lca[x][0] is
for (int i = log - 1; i >= 0; i--)
{
if(v == -1)
v++;
if (lca[v][i] != lca[u][i])
{
// Calculating the minimum of
// MinWeight of v to its 2^i-th
// ancestor and MinWeight of u
// to its 2^i-th ancestor
minWei = Math.min(minWei,
Math.min(minWeight[v][i],
minWeight[u][i]));
// Calculating the maximum of
// MaxWeight of v to its 2^i-th
// ancestor and MaxWeight of u
// to its 2^i-th ancestor
maxWei = Math.max(maxWei,
Math.max(maxWeight[v][i],
maxWeight[u][i]));
v = lca[v][i];
u = lca[u][i];
}
}
// Calculating the Minimum of
// first ancestor of u and v
if(u == -1)
u++;
minWei = Math.min(minWei,
Math.min(minWeight[v][0],
minWeight[u][0]));
// Calculating the maximum of
// first ancestor of u and v
maxWei = Math.max(maxWei,
Math.max(maxWeight[v][0],
maxWeight[u][0]));
System.out.print(minWei + " " +
maxWei + "\n");
}
}
// Driver Code
public static void main(String[] args)
{
// Number of nodes
int n = 5;
for (int i = 0; i < graph.length; i++)
graph[i] = new Vector();
// Add edges
addEdge(1, 2);
addEdge(1, 5);
addEdge(2, 4);
addEdge(2, 3);
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// Initialising lca values with -1
// Initialising MinWeight values
// with Integer.MAX_VALUE
// Initialising MaxWeight values
// with Integer.MIN_VALUE
for (int i = 1; i <= n; i++)
{
for (int j = 0; j < log; j++)
{
lca[i][j] = -1;
minWeight[i][j] = Integer.MAX_VALUE;
maxWeight[i][j] = Integer.MIN_VALUE;
}
}
// Perform DFS
dfs(1, -1, 0);
// Query 1: {1, 3}
findMinMaxWeight(1, 3);
// Query 2: {2, 4}
findMinMaxWeight(2, 4);
// Query 3: {3, 5}
findMinMaxWeight(3, 5);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 Program to find the
# maximum and minimum weight
# between two nodes in the
# given tree using LCA
import sys
MAX = 1000
# log2(MAX)
log = 10
# Array to store the level
# of each node
level = [0 for i in range(MAX)];
# Initialising lca values with -1
# Initialising MinWeight values
# with INT_MAX
# Initialising MaxWeight values
# with INT_MIN
lca = [[-1 for j in range(log)]
for i in range(MAX)]
minWeight = [[sys.maxsize for j in range(log)]
for i in range(MAX)]
maxWeight = [[-sys.maxsize for j in range(log)]
for i in range(MAX)]
# Vector to store tree
graph = [[] for i in range(MAX)]
# Array to store weight of nodes
weight = [0 for i in range(MAX)]
def addEdge(u, v):
graph[u].append(v);
graph[v].append(u);
# Pre-Processing to calculate
# values of lca[][], MinWeight[][]
# and MaxWeight[][]
def dfs(node, parent, h):
# Using recursion formula to
# calculate the values
# of lca[][]
lca[node][0] = parent;
# Storing the level of
# each node
level[node] = h;
if (parent != -1):
minWeight[node][0] = (min(weight[node],
weight[parent]));
maxWeight[node][0] = (max(weight[node],
weight[parent]));
for i in range(1, log):
if (lca[node][i - 1] != -1):
# Using recursion formula to
# calculate the values of lca[][],
# MinWeight[][] and MaxWeight[][]
lca[node][i] = lca[lca[node][i - 1]][i - 1];
minWeight[node][i] = min(minWeight[node][i - 1],
minWeight[lca[node][i - 1]][i - 1]);
maxWeight[node][i] = max(maxWeight[node][i - 1],
maxWeight[lca[node][i - 1]][i - 1]);
for i in graph[node]:
if (i == parent):
continue;
dfs(i, node, h + 1);
# Function to find the minimum
# and maximum weights in the
# given range
def findMinMaxWeight(u, v):
minWei = sys.maxsize
maxWei = -sys.maxsize
# The node which is present
# farthest from the root node
# is taken as v If u is
# farther from root node
# then swap the two
if (level[u] > level[v]):
u, v = v, u
# Finding the ancestor of v
# which is at same level as u
for i in range(log - 1, -1, -1):
if (lca[v][i] != -1 and
level[lca[v][i]] >=
level[u]):
# Calculating Minimum and
# Maximum Weight of node
# v till its 2^i-th ancestor
minWei = min(minWei,
minWeight[v][i]);
maxWei = max(maxWei,
maxWeight[v][i]);
v = lca[v][i];
# If u is the ancestor of v
# then u is the LCA of u and v
if (v == u):
print(str(minWei) + ' ' +
str(maxWei))
else:
# Finding the node closest to the
# root which is not the common
# ancestor of u and v i.e. a node
# x such that x is not the common
# ancestor of u and v but lca[x][0] is
for i in range(log - 1, -1, -1):
if (lca[v][i] != lca[u][i]):
# Calculating the minimum of
# MinWeight of v to its 2^i-th
# ancestor and MinWeight of u
# to its 2^i-th ancestor
minWei = (min(minWei,
min(minWeight[v][i],
minWeight[u][i])));
# Calculating the maximum of
# MaxWeight of v to its 2^i-th
# ancestor and MaxWeight of u
# to its 2^i-th ancestor
maxWei = max(maxWei,
max(maxWeight[v][i],
maxWeight[u][i]));
v = lca[v][i];
u = lca[u][i];
# Calculating the Minimum of
# first ancestor of u and v
minWei = min(minWei,
min(minWeight[v][0],
minWeight[u][0]));
# Calculating the maximum of
# first ancestor of u and v
maxWei = max(maxWei,
max(maxWeight[v][0],
maxWeight[u][0]));
print(str(minWei) + ' ' +
str(maxWei))
# Driver code
if __name__ == "__main__":
# Number of nodes
n = 5;
# Add edges
addEdge(1, 2);
addEdge(1, 5);
addEdge(2, 4);
addEdge(2, 3);
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
# Perform DFS
dfs(1, -1, 0);
# Query 1: {1, 3}
findMinMaxWeight(1, 3);
# Query 2: {2, 4}
findMinMaxWeight(2, 4);
# Query 3: {3, 5}
findMinMaxWeight(3, 5);
# This code is contributed by Rutvik_56
C#
// C# Program to find the
// maximum and minimum
// weight between two nodes
// in the given tree using LCA
using System;
using System.Collections.Generic;
class GFG{
static readonly int MAX = 1000;
// Math.Log(MAX)
static readonly int log = 10 ;
// Array to store the
// level of each node
static int []level =
new int[MAX];
static int [,]lca =
new int[MAX, log];
static int [,]minWeight =
new int[MAX, log];
static int [,]maxWeight =
new int[MAX, log];
// List to store tree
static List []graph =
new List[MAX];
// Array to store
// weight of nodes
static int []weight =
new int[MAX];
private static void swap(int x,
int y)
{
int temp = x;
x = y;
y = temp;
}
static void addEdge(int u,
int v)
{
graph[u].Add(v);
graph[v].Add(u);
}
// Pre-Processing to
// calculate values of
// lca[,], MinWeight[,]
// and MaxWeight[,]
static void dfs(int node,
int parent, int h)
{
// Using recursion formula to
// calculate the values
// of lca[,]
lca[node, 0] = parent;
// Storing the level of
// each node
level[node] = h;
if (parent != -1)
{
minWeight[node, 0] = Math.Min(weight[node],
weight[parent]);
maxWeight[node, 0] = Math.Max(weight[node],
weight[parent]);
}
for (int i = 1; i < log; i++)
{
if (lca[node, i - 1] != -1)
{
// Using recursion formula to
// calculate the values of lca[,],
// MinWeight[,] and MaxWeight[,]
lca[node, i] =
lca[lca[node, i - 1],
i - 1];
minWeight[node, i] =
Math.Min(minWeight[node, i - 1],
minWeight[lca[node, i - 1],
i - 1]);
maxWeight[node, i] =
Math.Max(maxWeight[node, i - 1],
maxWeight[lca[node, i - 1],
i - 1]);
}
}
foreach (int i in graph[node])
{
if (i == parent)
continue;
dfs(i, node, h + 1);
}
}
// Function to find the minimum and
// maximum weights in the given range
static void findMinMaxWeight(int u,
int v)
{
int minWei = int.MaxValue;
int maxWei = int.MinValue;
// The node which is present
// farthest from the root node
// is taken as v If u is
// farther from root node
// then swap the two
if (level[u] > level[v])
swap(u, v);
// Finding the ancestor of v
// which is at same level as u
for (int i = log - 1; i >= 0; i--)
{
if (lca[v, i] != -1 &&
level[lca[v, i]] >= level[u])
{
// Calculating Minimum and
// Maximum Weight of node
// v till its 2^i-th ancestor
minWei = Math.Min(minWei,
minWeight[v, i]);
maxWei = Math.Max(maxWei,
maxWeight[v, i]);
v = lca[v, i];
}
}
// If u is the ancestor of v
// then u is the LCA of u and v
if (v == u)
{
Console.Write(minWei + " " +
maxWei + "\n");
}
else
{
// Finding the node closest to the
// root which is not the common
// ancestor of u and v i.e. a node
// x such that x is not the common
// ancestor of u and v but lca[x,0] is
for (int i = log - 1; i >= 0; i--)
{
if(v == -1)
v++;
if (lca[v, i] != lca[u, i])
{
// Calculating the minimum of
// MinWeight of v to its 2^i-th
// ancestor and MinWeight of u
// to its 2^i-th ancestor
minWei = Math.Min(minWei,
Math.Min(minWeight[v, i],
minWeight[u, i]));
// Calculating the maximum of
// MaxWeight of v to its 2^i-th
// ancestor and MaxWeight of u
// to its 2^i-th ancestor
maxWei = Math.Max(maxWei,
Math.Max(maxWeight[v, i],
maxWeight[u, i]));
v = lca[v, i];
u = lca[u, i];
}
}
// Calculating the Minimum of
// first ancestor of u and v
if(u == -1)
u++;
minWei = Math.Min(minWei,
Math.Min(minWeight[v, 0],
minWeight[u, 0]));
// Calculating the maximum of
// first ancestor of u and v
maxWei = Math.Max(maxWei,
Math.Max(maxWeight[v, 0],
maxWeight[u, 0]));
Console.Write(minWei + " " +
maxWei + "\n");
}
}
// Driver Code
public static void Main(String[] args)
{
// Number of nodes
int n = 5;
for (int i = 0; i < graph.Length; i++)
graph[i] = new List();
// Add edges
addEdge(1, 2);
addEdge(1, 5);
addEdge(2, 4);
addEdge(2, 3);
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// Initialising lca values with -1
// Initialising MinWeight values
// with int.MaxValue
// Initialising MaxWeight values
// with int.MinValue
for (int i = 1; i <= n; i++)
{
for (int j = 0; j < log; j++)
{
lca[i, j] = -1;
minWeight[i, j] = int.MaxValue;
maxWeight[i, j] = int.MinValue;
}
}
// Perform DFS
dfs(1, -1, 0);
// Query 1: {1, 3}
findMinMaxWeight(1, 3);
// Query 2: {2, 4}
findMinMaxWeight(2, 4);
// Query 3: {3, 5}
findMinMaxWeight(3, 5);
}
}
// This code is contributed by Rajput-Ji
-1 5
3 5
-2 5
时间复杂度:预处理花费的时间是O(N logN)并且每个查询花费O(logN) 的时间。所以解决方案的整体时间复杂度是O(N logN) 。
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