二维矩阵中一个元素的 8 个邻域的问题
给定二维矩阵和整数“K”，任务是在“K”次迭代后预测矩阵，如下所示:
- Element 1 in the current matrix will remain 1 in the next iteration only if it is surrounded by multiple 1s, where 0 < = range 1a < = A < = range 1b.
- The element 0 in the current matrix will only become 1 in the next iteration if it is surrounded by the number of b of 1, where 0<= range 0a<= B<= range 0b.
我们用一个例子来理解一下:
约束: 1<= K<= 100000 0<=范围 1a、范围 1b、范围 0a、范围 0b < = 8
- 在上述单元格(0，0)的图像中，单元格在第一次迭代中为“0”，但由于它仅被一个包含“1”的相邻单元格包围，因此不在范围[range0a，range0b]内。因此它将继续保持“0”。
- 对于第二次迭代，单元格(0，0)为 0，但这次它被两个包含“1”的单元格包围，两个单元格位于范围[range0a，range0b]内。因此，它在下一次(第二次)迭代中变为“1”。
示例:
输入:范围 1a = 2 范围 1b = 2 范围 0a = 2 范围 0b = 3 K = 1 输出: 0 1 1 0 0 1 1 1 0 0 1 0 0 1 0 输入:范围 1a = 2 范围 1b = 2 范围 0a = 2 范围 0b = 3
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Dimension of Array
#define N 4
void predictMatrix(int arr[N][N],
int range1a,
int range1b,
int range0a,
int range0b,
int K,
int b[N][N])
{
// Count of 1s
int c = 0;
while (K--) {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
c = 0;
// Counting all neighbouring 1s
if (i > 0 && arr[i - 1][j] == 1)
c++;
if (j > 0 && arr[i][j - 1] == 1)
c++;
if (i > 0 && j > 0
&& arr[i - 1][j - 1] == 1)
c++;
if (i < N - 1 && arr[i + 1][j] == 1)
c++;
if (j < N - 1 && arr[i][j + 1] == 1)
c++;
if (i < N - 1 && j < N - 1
&& arr[i + 1][j + 1] == 1)
c++;
if (i < N - 1 && j > 0
&& arr[i + 1][j - 1] == 1)
c++;
if (i > 0 && j < N - 1
&& arr[i - 1][j + 1] == 1)
c++;
// Comparing the number of
// neighbouring 1s with
// given ranges
if (arr[i][j] == 1) {
if (c >= range1a && c <= range1b)
b[i][j] = 1;
else
b[i][j] = 0;
}
if (arr[i][j] == 0) {
if (c >= range0a && c <= range0b)
b[i][j] = 1;
else
b[i][j] = 0;
}
}
}
// Copying changes to
// the main matrix
for (int k = 0; k < N; k++)
for (int m = 0; m < N; m++)
arr[k][m] = b[k][m];
}
}
// Driver code
int main()
{
int arr[N][N] = { 0, 0, 0, 0,
0, 1, 1, 0,
0, 0, 1, 0,
0, 1, 0, 1 };
int range1a = 2, range1b = 2;
int range0a = 2, range0b = 3;
int K = 3, b[N][N] = { 0 };
// Function call to calculate
// the resultant matrix
// after 'K' iterations.
predictMatrix(arr, range1a, range1b,
range0a, range0b, K, b);
// Printing Result
for (int i = 0; i < N; i++) {
cout << endl;
for (int j = 0; j < N; j++)
cout << b[i][j] << " ";
}
return 0;
}
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{
// Dimension of Array
final static int N = 4 ;
static void predictMatrix(int arr[][],
int range1a,
int range1b,
int range0a,
int range0b,
int K,
int b[][])
{
// Count of 1s
int c = 0;
while (K != 0) {
K--;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
c = 0;
// Counting all neighbouring 1s
if (i > 0 && arr[i - 1][j] == 1)
c++;
if (j > 0 && arr[i][j - 1] == 1)
c++;
if (i > 0 && j > 0
&& arr[i - 1][j - 1] == 1)
c++;
if (i < N - 1 && arr[i + 1][j] == 1)
c++;
if (j < N - 1 && arr[i][j + 1] == 1)
c++;
if (i < N - 1 && j < N - 1
&& arr[i + 1][j + 1] == 1)
c++;
if (i < N - 1 && j > 0
&& arr[i + 1][j - 1] == 1)
c++;
if (i > 0 && j < N - 1
&& arr[i - 1][j + 1] == 1)
c++;
// Comparing the number of
// neighbouring 1s with
// given ranges
if (arr[i][j] == 1) {
if (c >= range1a && c <= range1b)
b[i][j] = 1;
else
b[i][j] = 0;
}
if (arr[i][j] == 0) {
if (c >= range0a && c <= range0b)
b[i][j] = 1;
else
b[i][j] = 0;
}
}
}
// Copying changes to
// the main matrix
for (int k = 0; k < N; k++)
for (int m = 0; m < N; m++)
arr[k][m] = b[k][m];
}
}
// Driver code
public static void main(String []args)
{
int arr[][] = { {0, 0, 0, 0},
{0, 1, 1, 0},
{0, 0, 1, 0},
{0, 1, 0, 1 } };
int range1a = 2, range1b = 2;
int range0a = 2, range0b = 3;
int K = 3;
int b[][] = new int[N][N] ;
// Function call to calculate
// the resultant matrix
// after 'K' iterations.
predictMatrix(arr, range1a, range1b,
range0a, range0b, K, b);
// Printing Result
for (int i = 0; i < N; i++) {
System.out.println();
for (int j = 0; j < N; j++)
System.out.print(b[i][j]+ " ");
}
}
// This Code is contributed by Ryuga
}
Python 3
# Python3 implementation of the approach
# Dimension of Array
N = 4
def predictMatrix(arr, range1a, range1b,
range0a, range0b, K, b):
# Count of 1s
c = 0
while (K):
for i in range(N) :
for j in range(N):
c = 0
# Counting all neighbouring 1s
if (i > 0 and arr[i - 1][j] == 1):
c += 1
if (j > 0 and arr[i][j - 1] == 1):
c += 1
if (i > 0 and j > 0 and
arr[i - 1][j - 1] == 1):
c += 1
if (i < N - 1 and arr[i + 1][j] == 1):
c += 1
if (j < N - 1 and arr[i][j + 1] == 1):
c += 1
if (i < N - 1 and j < N - 1
and arr[i + 1][j + 1] == 1):
c += 1
if (i < N - 1 and j > 0
and arr[i + 1][j - 1] == 1):
c += 1
if (i > 0 and j < N - 1
and arr[i - 1][j + 1] == 1):
c += 1
# Comparing the number of neighbouring
# 1s with given ranges
if (arr[i][j] == 1) :
if (c >= range1a and c <= range1b):
b[i][j] = 1
else:
b[i][j] = 0
if (arr[i][j] == 0):
if (c >= range0a and c <= range0b):
b[i][j] = 1
else:
b[i][j] = 0
K -= 1
# Copying changes to the main matrix
for k in range(N):
for m in range( N):
arr[k][m] = b[k][m]
# Driver code
if __name__ == "__main__":
arr = [[0, 0, 0, 0],
[0, 1, 1, 0],
[0, 0, 1, 0],
[0, 1, 0, 1]]
range1a = 2
range1b = 2
range0a = 2
range0b = 3
K = 3
b = [[0 for x in range(N)]
for y in range(N)]
# Function call to calculate
# the resultant matrix
# after 'K' iterations.
predictMatrix(arr, range1a, range1b,
range0a, range0b, K, b)
# Printing Result
for i in range( N):
print()
for j in range(N):
print(b[i][j], end = " ")
# This code is contributed
# by ChitraNayal
C
// C# implementation of the approach
using System;
class GFG
{
// Dimension of Array
readonly static int N = 4 ;
static void predictMatrix(int [,]arr, int range1a,
int range1b, int range0a,
int range0b, int K, int [,]b)
{
// Count of 1s
int c = 0;
while (K != 0)
{
K--;
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
c = 0;
// Counting all neighbouring 1s
if (i > 0 && arr[i - 1, j] == 1)
c++;
if (j > 0 && arr[i, j - 1] == 1)
c++;
if (i > 0 && j > 0
&& arr[i - 1, j - 1] == 1)
c++;
if (i < N - 1 && arr[i + 1, j] == 1)
c++;
if (j < N - 1 && arr[i, j + 1] == 1)
c++;
if (i < N - 1 && j < N - 1 &&
arr[i + 1, j + 1] == 1)
c++;
if (i < N - 1 && j > 0 &&
arr[i + 1, j - 1] == 1)
c++;
if (i > 0 && j < N - 1 &&
arr[i - 1, j + 1] == 1)
c++;
// Comparing the number of
// neighbouring 1s with
// given ranges
if (arr[i,j] == 1)
{
if (c >= range1a && c <= range1b)
b[i, j] = 1;
else
b[i, j] = 0;
}
if (arr[i,j] == 0)
{
if (c >= range0a && c <= range0b)
b[i, j] = 1;
else
b[i, j] = 0;
}
}
}
// Copying changes to the main matrix
for (int k = 0; k < N; k++)
for (int m = 0; m < N; m++)
arr[k, m] = b[k, m];
}
}
// Driver code
public static void Main()
{
int [,]arr = { {0, 0, 0, 0},
{0, 1, 1, 0},
{0, 0, 1, 0},
{0, 1, 0, 1 } };
int range1a = 2, range1b = 2;
int range0a = 2, range0b = 3;
int K = 3;
int [,]b = new int[N, N];
// Function call to calculate
// the resultant matrix
// after 'K' iterations.
predictMatrix(arr, range1a, range1b,
range0a, range0b, K, b);
// Printing Result
for (int i = 0; i < N; i++)
{
Console.WriteLine();
for (int j = 0; j < N; j++)
Console.Write(b[i, j] + " ");
}
}
}
// This code is contributed by 29AjayKumar
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Dimension of Array
#define N 4
function predictMatrix($arr, $range1a, $range1b,
$range0a, $range0b, $K, $b)
{
$N = 4;
// Count of 1s
$c = 0;
while ($K--)
{
for ($i = 0; $i < $N; $i++)
{
for ($j = 0; $j < $N; $j++)
{
$c = 0;
// Counting all neighbouring 1s
if ($i > 0 && $arr[$i - 1][$j] == 1)
$c++;
if ($j > 0 && $arr[$i][$j - 1] == 1)
$c++;
if ($i > 0 && $j > 0 && $arr[$i - 1][$j - 1] == 1)
$c++;
if ($i < $N - 1 && $arr[$i + 1][$j] == 1)
$c++;
if ($j < $N - 1 && $arr[$i][$j + 1] == 1)
$c++;
if ($i < $N - 1 && $j < $N - 1 &&
$arr[$i + 1][$j + 1] == 1)
$c++;
if ($i < $N - 1 && $j > 0 && $arr[$i + 1][$j - 1] == 1)
$c++;
if ($i > 0 && $j < $N - 1 && $arr[$i - 1][$j + 1] == 1)
$c++;
// Comparing the number of
// neighbouring 1s with
// given ranges
if ($arr[$i][$j] == 1)
{
if ($c >= $range1a && $c <= $range1b)
$b[$i][$j] = 1;
else
$b[$i][$j] = 0;
}
if ($arr[$i][$j] == 0) {
if ($c >= $range0a && $c <= $range0b)
$b[$i][$j] = 1;
else
$b[$i][$j] = 0;
}
}
}
// Copying changes to
// the main matrix
for ($k = 0; $k < $N; $k++)
for ($m = 0; $m < $N; $m++)
$arr[$k][$m] = $b[$k][$m];
}
return $b;
}
// Driver code
$N = 4;
$arr= array(array(0, 0, 0, 0),
array(0, 1, 1, 0),
array(0, 0, 1, 0),
array(0, 1, 0, 1));
$range1a = 2; $range1b = 2;
$range0a = 2; $range0b = 3;
$K = 3; $b = array(array(0));
// Function call to calculate
// the resultant matrix
// after 'K' iterations.
$b1 = predictMatrix($arr, $range1a, $range1b,
$range0a, $range0b, $K, $b);
// Printing Result
for ($i = 0; $i < $N; $i++)
{
echo "\n";
for ($j = 0; $j < $N; $j++)
echo $b1[$i][$j] . " ";
}
// This code is contributed by Akanksha Rai
java 描述语言
<script>
// Javascript implementation of the approach
// Dimension of Array
let N = 4 ;
function predictMatrix(arr,range1a,range1b,range0a,range0b,K,b)
{
// Count of 1s
let c = 0;
while (K != 0) {
K--;
for (let i = 0; i < N; i++) {
for (let j = 0; j < N; j++) {
c = 0;
// Counting all neighbouring 1s
if (i > 0 && arr[i - 1][j] == 1)
c++;
if (j > 0 && arr[i][j - 1] == 1)
c++;
if (i > 0 && j > 0
&& arr[i - 1][j - 1] == 1)
c++;
if (i < N - 1 && arr[i + 1][j] == 1)
c++;
if (j < N - 1 && arr[i][j + 1] == 1)
c++;
if (i < N - 1 && j < N - 1
&& arr[i + 1][j + 1] == 1)
c++;
if (i < N - 1 && j > 0
&& arr[i + 1][j - 1] == 1)
c++;
if (i > 0 && j < N - 1
&& arr[i - 1][j + 1] == 1)
c++;
// Comparing the number of
// neighbouring 1s with
// given ranges
if (arr[i][j] == 1) {
if (c >= range1a && c <= range1b)
b[i][j] = 1;
else
b[i][j] = 0;
}
if (arr[i][j] == 0) {
if (c >= range0a && c <= range0b)
b[i][j] = 1;
else
b[i][j] = 0;
}
}
}
// Copying changes to
// the main matrix
for (let k = 0; k < N; k++)
for (let m = 0; m < N; m++)
arr[k][m] = b[k][m];
}
}
// Driver code
let arr = [[0, 0, 0, 0],
[0, 1, 1, 0],
[0, 0, 1, 0],
[0, 1, 0, 1]];
let range1a = 2, range1b = 2;
let range0a = 2, range0b = 3;
let K = 3;
let b = new Array(N) ;
for(let i=0;i<N;i++)
{
b[i]=new Array(N);
for(let j=0;j<N;j++)
{
b[i][j]=0;
}
}
// Function call to calculate
// the resultant matrix
// after 'K' iterations.
predictMatrix(arr, range1a, range1b,
range0a, range0b, K, b);
// Printing Result
for (let i = 0; i < N; i++) {
document.write("<br>");
for (let j = 0; j < N; j++)
document.write(b[i][j]+ " ");
}
// This code is contributed by avanitrachhadiya2155
</script>
Output:
0 1 0 0
0 1 0 0
1 1 1 0
0 0 1 0
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