给定一个由N 个整数组成的数组arr[] ，最初只包含0并且查询形式为{L, R, C} 的Q[][] ，每个查询的任务是更新子数组[L, R ]值C 。打印执行所有查询后生成的最终数组。
例子：
Input: N = 5, Q = {{1, 4, 1}, {3, 5, 2}, {2, 4, 3}}
Output: 1 3 3 3 2
Explanation:
Initially, the array is {0, 0, 0, 0, 0}
Query 1 modifies the array to {1, 1, 1, 1, 0}
Query 2 modifies the array to {1, 1, 2, 2, 2}
Query 3 modifies the array to {1, 3, 3, 3, 2}
Input: N = 3, Q = {{1, 2, 1}, {2, 3, 2}}
Output: 1 2 2
Explanation:
Initially, the array is {0, 0, 0}
Query 1 modifies the array to {1, 1, 0}
Query 2 modifies the array to {1, 2, 2}
方法：思路是使用Disjoint Set Union来解决问题。按照以下步骤解决问题：
- 最初，所有数组元素将被视为单独的集合和自身的父元素，并将存储值为 0 的下一个数组元素。
- 首先，存储查询并以从最后到第一的相反顺序处理查询，因为分配给每个集合的值将是最终的。
- 处理完第一个查询后，具有更改值的元素将指向下一个元素。这样在执行查询时，我们只需要为子数组[l, r] 中的未更新集合赋值。所有其他单元格已经包含它们的最终值。
- 找到最左边的未更新集，并更新它，并使用指针移动到右侧的下一个未更新集。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Maximum possible size of array
#define MAX_NODES 100005
// Stores the parent of each element
int parent[MAX_NODES];
// Stores the final array values
int final_val[MAX_NODES];
// Structure to store queries
struct query {
int l, r, c;
};
// Function to initialize the
// parent of each vertex
void make_set(int v)
{
// Initially parent
// of each node points
// to itself
parent[v] = v;
}
// Function to find the representative
// of the set which contain element v
int find_set(int v)
{
if (v == parent[v])
return v;
// Path compression
return parent[v] = find_set(parent[v]);
}
// Function to assign a
// parent to each element
void Initialize(int n)
{
for (int i = 0; i <= n; i++)
make_set(i + 1);
}
// Function to process the queries
void Process(query Q[], int q)
{
for (int i = q - 1; i >= 0; i--) {
int l = Q[i].l, r = Q[i].r, c = Q[i].c;
for (int v = find_set(l); v <= r;
v = find_set(v)) {
final_val[v] = c;
parent[v] = v + 1;
}
}
}
// Function to print the final array
void PrintAns(int n)
{
for (int i = 1; i <= n; i++) {
cout << final_val[i] << " ";
}
cout << endl;
}
// Driver Code
int main()
{
int n = 5;
// Set all the elements as the
// parent of itself using make_set
Initialize(n);
int q = 3;
query Q[q];
// Store the queries
Q[0].l = 1, Q[0].r = 4, Q[0].c = 1;
Q[1].l = 3, Q[1].r = 5, Q[1].c = 2;
Q[2].l = 2, Q[2].r = 4, Q[2].c = 3;
// Process the queries
Process(Q, q);
// Print the required array
PrintAns(n);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Maximum possible size of array
static final int MAX_NODES = 100005;
// Stores the parent of each element
static int []parent = new int[MAX_NODES];
// Stores the final array values
static int []final_val = new int[MAX_NODES];
// Structure to store queries
static class query
{
int l, r, c;
};
// Function to initialize the
// parent of each vertex
static void make_set(int v)
{
// Initially parent
// of each node points
// to itself
parent[v] = v;
}
// Function to find the representative
// of the set which contain element v
static int find_set(int v)
{
if (v == parent[v])
return v;
// Path compression
return parent[v] = find_set(parent[v]);
}
// Function to assign a
// parent to each element
static void Initialize(int n)
{
for(int i = 0; i <= n; i++)
make_set(i + 1);
}
// Function to process the queries
static void Process(query Q[], int q)
{
for(int i = q - 1; i >= 0; i--)
{
int l = Q[i].l, r = Q[i].r, c = Q[i].c;
for(int v = find_set(l); v <= r;
v = find_set(v))
{
final_val[v] = c;
parent[v] = v + 1;
}
}
}
// Function to print the final array
static void PrintAns(int n)
{
for(int i = 1; i <= n; i++)
{
System.out.print(final_val[i] + " ");
}
System.out.println();
}
// Driver Code
public static void main(String[] args)
{
int n = 5;
// Set all the elements as the
// parent of itself using make_set
Initialize(n);
int q = 3;
query []Q = new query[q];
for(int i = 0; i < Q.length; i++)
Q[i] = new query();
// Store the queries
Q[0].l = 1; Q[0].r = 4; Q[0].c = 1;
Q[1].l = 3; Q[1].r = 5; Q[1].c = 2;
Q[2].l = 2; Q[2].r = 4; Q[2].c = 3;
// Process the queries
Process(Q, q);
// Print the required array
PrintAns(n);
}
}
// This code is contributed by amal kumar choubeyPython3
# Python3 program to implement
# the above approach
MAX_NODES = 100005
# Stores the parent of each element
parent = [0] * MAX_NODES
# Stores the final array values
final_val = [0] * MAX_NODES
# Structure to store queries
# Function to initialize the
# parent of each vertex
def make_set(v):
# Initially parent
# of each node points
# to itself
parent[v] = v
# Function to find the representative
# of the set which contain element v
def find_set(v):
if (v == parent[v]):
return v
# Path compression
parent[v] = find_set(parent[v])
return parent[v]
# Function to assign a
# parent to each element
def Initialize(n):
for i in range(n + 1):
make_set(i + 1)
# Function to process the queries
def Process(Q, q):
for i in range(q - 1, -1, -1):
l = Q[i][0]
r = Q[i][1]
c = Q[i][2]
v = find_set(l)
while v <= r:
final_val[v] = c
parent[v] = v + 1
v = find_set(v)
# Function to prthe final array
def PrintAns(n):
for i in range(1, n + 1):
print(final_val[i], end = " ")
# Driver Code
if __name__ == '__main__':
n = 5
# Set all the elements as the
# parent of itself using make_set
Initialize(n)
q = 3
Q = [[0 for i in range(3)]
for i in range(q)]
# Store the queries
Q[0][0] = 1
Q[0][1] = 4
Q[0][2] = 1
Q[1][0] = 3
Q[1][1] = 5
Q[1][2] = 2
Q[2][0] = 2
Q[2][1] = 4
Q[2][2] = 3
# Process the queries
Process(Q, q)
# Print the required array
PrintAns(n)
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG{
// Maximum possible size of array
static readonly int MAX_NODES = 100005;
// Stores the parent of each element
static int []parent = new int[MAX_NODES];
// Stores the readonly array values
static int []final_val = new int[MAX_NODES];
// Structure to store queries
class query
{
public int l, r, c;
};
// Function to initialize the
// parent of each vertex
static void make_set(int v)
{
// Initially parent
// of each node points
// to itself
parent[v] = v;
}
// Function to find the representative
// of the set which contain element v
static int find_set(int v)
{
if (v == parent[v])
return v;
// Path compression
return parent[v] = find_set(parent[v]);
}
// Function to assign a
// parent to each element
static void Initialize(int n)
{
for(int i = 0; i <= n; i++)
make_set(i + 1);
}
// Function to process the queries
static void Process(query []Q, int q)
{
for(int i = q - 1; i >= 0; i--)
{
int l = Q[i].l, r = Q[i].r, c = Q[i].c;
for(int v = find_set(l); v <= r;
v = find_set(v))
{
final_val[v] = c;
parent[v] = v + 1;
}
}
}
// Function to print the readonly array
static void PrintAns(int n)
{
for(int i = 1; i <= n; i++)
{
Console.Write(final_val[i] + " ");
}
Console.WriteLine();
}
// Driver Code
public static void Main(String[] args)
{
int n = 5;
// Set all the elements as the
// parent of itself using make_set
Initialize(n);
int q = 3;
query []Q = new query[q];
for(int i = 0; i < Q.Length; i++)
Q[i] = new query();
// Store the queries
Q[0].l = 1; Q[0].r = 4; Q[0].c = 1;
Q[1].l = 3; Q[1].r = 5; Q[1].c = 2;
Q[2].l = 2; Q[2].r = 4; Q[2].c = 3;
// Process the queries
Process(Q, q);
// Print the required array
PrintAns(n);
}
}
// This code is contributed by amal kumar choubey输出：
1 3 3 3 2时间复杂度： O(log N)
辅助空间： O(MAX_NODES)
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