从矩阵找出距离K内的元件 [英] Finding elements within distance k from a matrix

问题描述

给定一个* n的矩阵和值k，我们如何找到所有的邻居每一个元素？ 例如：在 4 * 4 基质，具有 K = 2 说矩阵是：

Given a n*n matrix and a value k, how do we find all the neighbors for each element? for example: in a 4*4 matrix, with k=2 say matrix is :

[ 1  2  3  4
  5  6  7  8
  9 10 11 12
 13 14 15 16]

在那里这些值的位置的索引，邻居 1为1,2,3,5,6,9 。该值 3,6和9 来只是因为K = 2，并不会是那里，如果K值是=​​ 1。

where these values are the indexes of the location, the neighbors for 1 are 1,2,3,5,6,9 . The values 3,6 and 9 come only because k =2 and wouldnt be there if k was = 1.

同样的6邻居将 1 2 3 5 6 7 8 9 10 11 14

能否请你帮我写AC code在C ++来实现这一点。

Can you please help me to write a c code to implement this in c++.

这是冯·诺依曼邻居的问题，请能有人实现它在C ++中。谢谢

It is the problem of von Neumann neighborhood, please can some one implement it in c++. Thanks

推荐答案

这应该做的伎俩对于k = 1。做小的改动，使之对于所有的k工作

This should do the trick for k=1. Make minor changes to make it work for all k

int width = 4;
int height = 4;
int k = 1;
int value = 2;

bool hasRight = (value % width != 0);
bool hasLeft = (value % width != 1);
bool hasTop = (value > 4);
bool hasBottom = (value < (height * width - width));

cout << value;  // Always itself
if(hasRight == true) {
 cout << value+1 << " ";  // Right
 if(hasTop == true) {
  cout << value-width << " " << value-width+1 << " "; // Top and Top-right
 }
 if(hasBottom == true) {
  cout << value+width << " " << value+width+1; // Bottom and Bottom-right
 }
}

if(hasLeft == true) {
 cout << value-1 << " ";  // Left
 if(hasTop == true) {
  cout << value-width-1 << " ";  // Top-left
 }
 if(hasBottom == true) {
  cout << value+width-1 << " ";  // Bottom-left
 }
}

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