给定尺寸的矩阵N * M和2D阵列的查询[] []与以下形式的每个查询{K,R1,C1,R2,C2},任务是与k添加到所有的C厄尔在子矩阵kting (r1, c1)到(r2, c2)
例子:
Input: A[][] = {{1, 2, 3}, {1, 1, 0}, {4, -2, 2}}, Queries[][] = {{2, 0, 0, 1, 1}, {-1, 1, 0, 2, 2}}
Output:
3 4 3
2 2 -1
3 -3 1
Explanation:
Query 1: Matrix modifies to {{3, 4, 3}, {3, 3, 0}, {4, -2, 2}
Query 2: Matrix modifies to {{3, 4, 3}, {2, 2, -1}, {3, -3, 1}
Input: A[][] = {{1, 2, 3}, { 4, 5, 6}, {7, 8, 9}}, Queries[][] = {{1, 1, 1, 2, 2}, {2, 0, 1, 0, 2}}
Output:
1 4 5
4 6 7
7 9 10
方法:这个想法是基于Difference Array | O(1) 中的范围更新查询。请按照以下步骤解决问题:
- 初始化一个二维差分数组D[][] ,使得D[i][j]存储A[i][j] – A[i][j – 1] (对于0 ≤ i ≤ N 和 0 < j < M ) 或D[i][j] = A[i][j]否则。
- 遍历每一行并计算并存储相邻元素之间的差异。
- 要将子矩阵(r1, c1)更新为(r2, c2) ,请从r1遍历到r2 (假设 r1 < r2 和 c1 < c2)并更新D[i][c1] = D[i][c1] + k和D[i][c2 + 1] = D[i][c2 + 1] – k 。
- 最后,将修改后的数组打印为D[i][j] + A[i][j-1] for j > 0或D[i][j] for j = 0 。
下面是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
#define N 3
#define M 3
// Function to initialize the difference array
void intializeDiff(int D[N][M + 1],
int A[N][M])
{
for (int i = 0; i < N; i++) {
D[i][0] = A[i][0];
D[i][M] = 0;
}
for (int i = 0; i < N; i++) {
for (int j = 1; j < M; j++)
D[i][j] = A[i][j] - A[i][j - 1];
}
}
// Function to add k to the specified
// submatrix (r1, c1) to (r2, c2)
void update(int D[N][M + 1], int k,
int r1, int c1, int r2,
int c2)
{
for (int i = r1; i <= r2; i++) {
D[i][c1] += k;
D[i][c2 + 1] -= k;
}
}
// Function to print the modified array
void printArray(int A[N][M], int D[N][M + 1])
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (j == 0)
A[i][j] = D[i][j];
else
A[i][j] = D[i][j] + A[i][j - 1];
cout << A[i][j] << " ";
}
cout << endl;
}
}
// Function to perform the given queries
void performQueries(int A[N][M],
vector > Queries)
{
// Difference array
int D[N][M + 1];
// Function to initialize
// the difference array
intializeDiff(D, A);
// Count of queries
int Q = Queries.size();
// Perform Queries
for (int i = 0; i < Q; i++) {
update(D, Queries[i][0],
Queries[i][1], Queries[i][2],
Queries[i][3], Queries[i][4]);
}
printArray(A, D);
}
// Driver Code
int main()
{
// Given Matrix
int A[N][M] = { { 1, 2, 3 },
{ 1, 1, 0 },
{ 4, -2, 2 } };
// Given Queries
vector > Queries
= { { 2, 0, 0, 1, 1 },
{ -1, 1, 0, 2, 2 } };
performQueries(A, Queries);
return 0;
}
Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
static int N = 3;
static int M = 3;
// Function to initialize the difference array
static void intializeDiff(int D[][], int A[][])
{
for(int i = 0; i < N; i++)
{
D[i][0] = A[i][0];
D[i][M] = 0;
}
for(int i = 0; i < N; i++)
{
for(int j = 1; j < M; j++)
D[i][j] = A[i][j] - A[i][j - 1];
}
}
// Function to add k to the specified
// submatrix (r1, c1) to (r2, c2)
static void update(int D[][], int k,
int r1, int c1,
int r2, int c2)
{
for(int i = r1; i <= r2; i++)
{
D[i][c1] += k;
D[i][c2 + 1] -= k;
}
}
// Function to print the modified array
static void printArray(int A[][], int D[][])
{
for(int i = 0; i < N; i++)
{
for(int j = 0; j < M; j++)
{
if (j == 0)
A[i][j] = D[i][j];
else
A[i][j] = D[i][j] + A[i][j - 1];
System.out.print(A[i][j] + " ");
}
System.out.println();
}
}
// Function to perform the given queries
static void performQueries(int A[][],
Vector> Queries)
{
// Difference array
int D[][] = new int[N][M + 1];
// Function to initialize
// the difference array
intializeDiff(D, A);
// Count of queries
int Q = Queries.size();
// Perform Queries
for(int i = 0; i < Q; i++)
{
update(D, Queries.get(i).get(0),
Queries.get(i).get(1),
Queries.get(i).get(2),
Queries.get(i).get(3),
Queries.get(i).get(4));
}
printArray(A, D);
}
// Driver Code
public static void main(String[] args)
{
// Given Matrix
int A[][] = { { 1, 2, 3 },
{ 1, 1, 0 },
{ 4, -2, 2 } };
// Given Queries
Vector> Queries = new Vector>();
Vector list1 = new Vector();
list1.add(2);
list1.add(0);
list1.add(0);
list1.add(1);
list1.add(1);
Vector list2 = new Vector();
list2.add(-1);
list2.add(1);
list2.add(0);
list2.add(2);
list2.add(2);
Queries.add(list1);
Queries.add(list2);
performQueries(A, Queries);
}
}
// This code is contributed by divyeshrabadiya07
Python3
# Python3 Program to implement
# the above approach
N = 3
M = 3
# Function to initialize the difference array
def intializeDiff(D, A):
for i in range(N):
D[i][0] = A[i][0];
D[i][M] = 0;
for i in range(N):
for j in range(1, M):
D[i][j] = A[i][j] - A[i][j - 1];
# Function to add k to the specified
# submatrix (r1, c1) to (r2, c2)
def update(D, k, r1, c1, r2, c2):
for i in range(r1, r2 + 1):
D[i][c1] += k;
D[i][c2 + 1] -= k;
# Function to print the modified array
def printArray(A, D):
for i in range(N):
for j in range(M):
if (j == 0):
A[i][j] = D[i][j];
else:
A[i][j] = D[i][j] + A[i][j - 1];
print(A[i][j], end = ' ')
print()
# Function to perform the given queries
def performQueries(A, Queries):
# Difference array
D = [[0 for j in range(M + 1)] for i in range(N)]
# Function to initialize
# the difference array
intializeDiff(D, A);
# Count of queries
Q = len(Queries)
# Perform Queries
for i in range(Q):
update(D, Queries[i][0],
Queries[i][1], Queries[i][2],
Queries[i][3], Queries[i][4]);
printArray(A, D);
# Driver Code
if __name__=='__main__':
# Given Matrix
A = [ [ 1, 2, 3 ],
[ 1, 1, 0 ],
[ 4, -2, 2 ] ];
# Given Queries
Queries = [ [ 2, 0, 0, 1, 1 ],[ -1, 1, 0, 2, 2 ] ];
performQueries(A, Queries);
# This code is contributed by Pratham76
C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
static int N = 3;
static int M = 3;
// Function to initialize the difference array
static void intializeDiff(int[,] D, int[,] A)
{
for(int i = 0; i < N; i++)
{
D[i, 0] = A[i, 0];
D[i, M] = 0;
}
for(int i = 0; i < N; i++)
{
for(int j = 1; j < M; j++)
D[i, j] = A[i, j] - A[i, j - 1];
}
}
// Function to add k to the specified
// submatrix (r1, c1) to (r2, c2)
static void update(int[,] D, int k,
int r1, int c1,
int r2, int c2)
{
for(int i = r1; i <= r2; i++)
{
D[i, c1] += k;
D[i, c2 + 1] -= k;
}
}
// Function to print the modified array
static void printArray(int[,] A, int[,] D)
{
for(int i = 0; i < N; i++)
{
for(int j = 0; j < M; j++)
{
if (j == 0)
A[i, j] = D[i, j];
else
A[i, j] = D[i, j] + A[i, j - 1];
Console.Write(A[i, j] + " ");
}
Console.WriteLine();
}
}
// Function to perform the given queries
static void performQueries(int[,] A,
List> Queries)
{
// Difference array
int[,] D = new int[N, M + 1];
// Function to initialize
// the difference array
intializeDiff(D, A);
// Count of queries
int Q = Queries.Count;
// Perform Queries
for(int i = 0; i < Q; i++)
{
update(D, Queries[i][0],
Queries[i][1], Queries[i][2],
Queries[i][3], Queries[i][4]);
}
printArray(A, D);
}
// Driver Code
static void Main()
{
// Given Matrix
int[,] A = { { 1, 2, 3 },
{ 1, 1, 0 },
{ 4, -2, 2 } };
// Given Queries
List> Queries = new List>();
Queries.Add(new List{ 2, 0, 0, 1, 1 });
Queries.Add(new List{ -1, 1, 0, 2, 2 });
performQueries(A, Queries);
}
}
// This code is contributed by divyesh072019
输出:
3 4 3
2 2 -1
3 -3 1
时间复杂度: O(N * M)
辅助空间: O(N * M)
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