在 M 个范围内应用增量操作后打印矩阵
原文:https://www . geesforgeks . org/print-matrix-applied-increment-operations-in-m-ranges/
给定尺寸为 N * N 的二维矩阵 mat[][] ，矩阵的所有元素最初都是 0 。需要对矩阵进行多个查询(M 个范围)，其中每个查询由四个整数组成 X1 、 Y1 、 X2 和 Y2 ，任务是将 1 添加到mat【X1】【Y1】和mat【X2】【Y2】之间的所有单元格中(包括二者)，并最终打印更新后矩阵的内容。 例:
输入: N = 2，q[][] = { { 0，0，1，1 }，{ 0，0，0，1 } } 输出: 2 2 1 1 第一次查询后:mat[][] = { {1，1 }，{1，1} } 第二次查询后:mat[][] = { {2，2}，{1，1} } 输入: N = 5，q[][]= } 3，4 } } 输出: 1 1 1 1 1 1 1 2 1 2 2 2 2 2 2 2 2 2 2 2 0 0 0 0
方式:对于每个查询 (X1，Y1) 表示子矩阵的左上角单元格， (X2，Y2) 表示子矩阵的右下角单元格。对于每个左上角单元格，在左上角元素中添加 1 ，并从右下角单元格旁边的元素(如果有)中减去 1 。 然后保持原始(现已修改)矩阵中所有元素的累计总和，每次相加时，当前总和就是当前位置的元素(已更新)。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
// Function to update and print the
// matrix after performing queries
void updateMatrix(int n, int q[3][4])
{
int i, j;
int mat[n][n];
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
mat[i][j] = 0;
for (i = 0; i < 3; i++)
{
int X1 = q[i][0];
int Y1 = q[i][1];
int X2 = q[i][2];
int Y2 = q[i][3];
// Add 1 to the first element of
// the sub-matrix
mat[X1][Y1]++;
// If there is an element after the
// last element of the sub-matrix
// then decrement it by 1
if (Y2 + 1 < n)
mat[X2][Y2 + 1]--;
else if (X2 + 1 < n)
mat[X2 + 1][0]--;
}
// Calculate the running sum
int sum = 0;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
sum += mat[i][j];
// Print the updated element
cout << sum << " ";
}
// Next line
cout << endl;
}
}
// Driver code
int main()
{
// Size of the matrix
int n = 5;
// Queries
int q[3][4] = {{ 0, 0, 1, 2 },
{ 1, 2, 3, 4 },
{ 1, 4, 3, 4 }};
updateMatrix(n, q);
return 0;
}
// This code is contributed by chandan_jnu
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
public class GFG {
// Function to update and print the matrix
// after performing queries
static void updateMatrix(int n, int q[][], int mat[][])
{
int i, j;
for (i = 0; i < q.length; i++) {
int X1 = q[i][0];
int Y1 = q[i][1];
int X2 = q[i][2];
int Y2 = q[i][3];
// Add 1 to the first element of the sub-matrix
mat[X1][Y1]++;
// If there is an element after the last element
// of the sub-matrix then decrement it by 1
if (Y2 + 1 < n)
mat[X2][Y2 + 1]--;
else if (X2 + 1 < n)
mat[X2 + 1][0]--;
}
// Calculate the running sum
int sum = 0;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
sum += mat[i][j];
// Print the updated element
System.out.print(sum + " ");
}
// Next line
System.out.println();
}
}
// Driver code
public static void main(String[] args)
{
// Size of the matrix
int n = 5;
int mat[][] = new int[n][n];
// Queries
int q[][] = { { 0, 0, 1, 2 },
{ 1, 2, 3, 4 },
{ 1, 4, 3, 4 } };
updateMatrix(n, q, mat);
}
}
Python 3
# Python 3 implementation of the approach
# Function to update and print the matrix
# after performing queries
def updateMatrix(n, q, mat):
for i in range(0, len(q)):
X1 = q[i][0];
Y1 = q[i][1];
X2 = q[i][2];
Y2 = q[i][3];
# Add 1 to the first element of
# the sub-matrix
mat[X1][Y1] = mat[X1][Y1] + 1;
# If there is an element after the
# last element of the sub-matrix
# then decrement it by 1
if (Y2 + 1 < n):
mat[X2][Y2 + 1] = mat[X2][Y2 + 1] - 1;
elif (X2 + 1 < n):
mat[X2 + 1][0] = mat[X2 + 1][0] - 1;
# Calculate the running sum
sum = 0;
for i in range(0, n):
for j in range(0, n):
sum =sum + mat[i][j];
# Print the updated element
print(sum, end = ' ');
# Next line
print(" ");
# Driver code
# Size of the matrix
n = 5;
mat = [[0 for i in range(n)]
for i in range(n)];
# Queries
q = [[ 0, 0, 1, 2 ],
[ 1, 2, 3, 4 ],
[ 1, 4, 3, 4 ]];
updateMatrix(n, q, mat);
# This code is contributed
# by Shivi_Aggarwal
C
// C# implementation of the above approach
using System;
public class GFG {
// Function to update and print the matrix
// after performing queries
static void updateMatrix(int n, int [,]q, int [,]mat)
{
int i, j;
for (i = 0; i < q.GetLength(0); i++) {
int X1 = q[i,0];
int Y1 = q[i,1];
int X2 = q[i,2];
int Y2 = q[i,3];
// Add 1 to the first element of the sub-matrix
mat[X1,Y1]++;
// If there is an element after the last element
// of the sub-matrix then decrement it by 1
if (Y2 + 1 < n)
mat[X2,Y2 + 1]--;
else if (X2 + 1 < n)
mat[X2 + 1,0]--;
}
// Calculate the running sum
int sum = 0;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
sum += mat[i,j];
// Print the updated element
Console.Write(sum + " ");
}
// Next line
Console.WriteLine();
}
}
// Driver code
public static void Main()
{
// Size of the matrix
int n = 5;
int [,]mat = new int[n,n];
// Queries
int [,]q = { { 0, 0, 1, 2 },
{ 1, 2, 3, 4 },
{ 1, 4, 3, 4 } };
updateMatrix(n, q, mat);
}
// This code is contributed by Ryuga
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to update and print the
// matrix after performing queries
function updateMatrix($n, $q, $mat)
{
for ($i = 0; $i < sizeof($q); $i++)
{
$X1 = $q[$i][0];
$Y1 = $q[$i][1];
$X2 = $q[$i][2];
$Y2 = $q[$i][3];
// Add 1 to the first element of
// the sub-matrix
$mat[$X1][$Y1]++;
// If there is an element after the last
// element of the sub-matrix then decrement
// it by 1
if ($Y2 + 1 < $n)
$mat[$X2][$Y2 + 1]--;
else if ($X2 + 1 < $n)
$mat[$X2 + 1][0]--;
}
// Calculate the running sum
$sum = 0;
for ($i = 0; $i < $n; $i++)
{
for ($j = 0; $j < $n; $j++)
{
$sum += $mat[$i][$j];
// Print the updated element
echo($sum . " ");
}
// Next line
echo("\n");
}
}
// Driver code
// Size of the matrix
$n = 5;
$mat = array_fill(0, $n,
array_fill(0, $n, 0));
// Queries
$q = array(array( 0, 0, 1, 2 ),
array( 1, 2, 3, 4 ),
array( 1, 4, 3, 4 ));
updateMatrix($n, $q, $mat);
// This code is contributed by chandan_jnu
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to update and print the matrix
// after performing queries
function updateMatrix(n, q, mat)
{
let i, j;
for (i = 0; i < q.length; i++) {
let X1 = q[i][0];
let Y1 = q[i][1];
let X2 = q[i][2];
let Y2 = q[i][3];
// Add 1 to the first element of the sub-matrix
mat[X1][Y1]++;
// If there is an element after the last element
// of the sub-matrix then decrement it by 1
if (Y2 + 1 < n)
mat[X2][Y2 + 1]--;
else if (X2 + 1 < n)
mat[X2 + 1][0]--;
}
// Calculate the running sum
let sum = 0;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
sum += mat[i][j];
// Print the updated element
document.write(sum + " ");
}
// Next line
document.write("</br>");
}
}
// Size of the matrix
let n = 5;
let mat = new Array(n);
for(let i = 0; i < n; i++)
{
mat[i] = new Array(n);
for(let j = 0; j < n; j++)
{
mat[i][j] = 0;
}
}
// Queries
let q = [ [ 0, 0, 1, 2 ],
[ 1, 2, 3, 4 ],
[ 1, 4, 3, 4 ] ];
updateMatrix(n, q, mat);
// This code is contributed by divyeshrabadiya07.
</script>
Output:
1 1 1 1 1
1 1 2 1 2
2 2 2 2 2
2 2 2 2 2
0 0 0 0 0
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