C++ 如何在至少有1个邻居等于1的数组中设置长方体

我试图遍历一个数组，找出哪些数组至少有一个直接邻居

我得到的答案是

display
1 1 1 1 1 
1 1 1 1 1 
1 1 1 1 1 
1 1 1 1 1 
1 1 1 1 1 

 display
11111
10001
11111
11111
11111

box :1 :1

box :1 :3

这是完美的。我可以从输出中看到哪些阵列正好有1个相邻阵列[1][1]和阵列[1][3]

---

现在，我该如何将这2个元素（数组[1][1]和数组[1][3]）=1？

如果您只需要“至少有一个邻居是1”，我真的很困惑为什么要尝试匹配整个邻居。你有没有理由不能这么做：

// going through the loop to figure out which ones have at least 1 neighbor
  for (int x = 0; x <5 ; x++)
  {
      for (int y = 0; y <5 ; y++)
      {
           if (  (array[x][y] == 0 )
            && 
            (array [x-1][y] == 1 || array [x+1][y] == 1 || 
             array [x][y-1] == 1 || array [x][y+1] ==1 || 
             array [x+1][y+1]==1 || array [x-1][y-1]==1 || 
             array [x-1][y+1]==1 || array [x+1][y-1] == 1)
            )
            {
                cout << endl;
                cout << "box :" << x << " :" << y << endl;
            } 
      }

//遍历循环以确定哪些循环至少有一个邻居
对于（int x=0；x由于多次使用嵌套的“display”循环，因此最好将其分解为自己的函数：

void Draw(int array[5][5]) {
    for (int x = 0; x < 5; ++x) {
        for (int y = 0; y < 5; ++y) 
            std::cout << array[x][y] << ' ';
        std::cout << '\n';
    }
}

void Draw（int数组[5][5]）{
对于（int x=0；x<5；++x）{
对于（int y=0；y<5；++y）
std：：cout你说的“仅设置那些…=0”是什么意思？它们已经是0了。我的错误。我的意思是1。现在在主函数堆栈的级别上获得另一个包含两个索引的数组，将上面的点存储在其中，然后读取数组中的点，最后将它们设为1。@myohmy，那么你说的是形态学上的“扩展”带3x3内核的过滤器？我对编程不太熟悉。不确定“带3x3内核的形态学“扩展”过滤器”是什么意思Hanks peter。我主要关心的是将那些带2个邻居的过滤器设置为1；我不确定什么是扩展谢谢..我从你的代码中得到了一些类似的东西，但有些东西让我有点不舒服..如果你设置..if（count==2）array[x][y]=1；对于if（count==1）array[x][y]=1；，它会将只有on邻居的单元格更改为1，您将看到3个零中的最后一个仍然是=0。但实际上，中间的零应该保持为0，其余的为1。现在应该可以正常工作。请确保保持
if（count==2）dummyarray[x][y]=0；
未更改。谢谢。当我们想知道他们是否有两个邻居时，计数==2是正确的。如果我们想知道他们是否只有一个邻居呢？这样如果他们只有一个邻居，他们就会变成1；好的。谢谢！我想这也行。我正在考虑用另一种方式构建它，但我会接受它。谢谢！事实上我想我是仍然困惑。我真的不理解你做的“缩短”。你能做同样的事情，但用我编码的方式。我只是想了解发生了什么。
// going through the loop to figure out which ones have at least 1 neighbor
  for (int x = 0; x <5 ; x++)
  {
      for (int y = 0; y <5 ; y++)
      {
           if (  (array[x][y] == 0 )
            && 
            (array [clamp04(x-1)][y] == 1 || array [clamp04(x+1)][y] == 1 || 
             array [x][clamp04(y-1)] == 1 || array [x][clamp04(y+1)] ==1 || 
             array [clamp04(x+1)][clamp04(y+1)]==1 || array [clamp04(x-1)][clamp04(y-1)]==1 || 
             array [clamp04(x-1)][clamp04(y+1)]==1 || array [clamp04(x+1)][clamp04(y-1)] == 1)
            )
            {
                cout << endl;
                cout << "box :" << x << " :" << y << endl;
            } 
      }

void Draw(int array[5][5]) {
    for (int x = 0; x < 5; ++x) {
        for (int y = 0; y < 5; ++y) 
            std::cout << array[x][y] << ' ';
        std::cout << '\n';
    }
}

#include <iostream>

using namespace std;

void Draw(int array[5][5]) {
    cout << endl;
    for (unsigned x = 0; x < 5; ++x) {
        for (unsigned y = 0; y < 5; ++y)
            cout <<  array[x][y] << ' ';
        cout << '\n';
    }
    cout << endl;
}

int main() {
    int num = 0;
    cout << "How many neighbours?: "; //Asks user for the number of neighbours they'd like to check for.
    cin >> num; //Saves the number of neighbours to look for in 'num'

    int array[5][5];
    int dummyarray[5][5]; //A spare array we'll use to keep track of which boxes have the specified number of neighbours.

    for (unsigned x = 0; x < 5; ++x) { 
        for (unsigned y = 0; y < 5; ++y)
            array[x][y] = 1; //Here we fill our original array with 1's
    }

    for (unsigned x = 0; x < 5; ++x) { 
        for (unsigned y = 0; y < 5; ++y)
            dummyarray[x][y] = 1; //Now we fill our dummy array with 1's so it's the same as our original array
    }

    Draw(array); //Here we call the Draw function, which prints our array for us.

    array[1][1] = 0;
    array[1][2] = 0; //Set some values of our original array to 0
    array[1][3] = 0;

    for (unsigned x = 0; x < 5; ++x) { //Here's where the checking starts
        for (unsigned y = 0; y < 5; ++y) {
            if (array[x][y] == 0) { //If we find a 0, start checking for neighbours of that 0 that are also 0's
                int count = -1; //Because this method counts the original box (array[x][y]) when searching for 0's, we anticipate this by removing one from the count, so that we get the correct number of neighbours
                for (int vertical = -1; vertical < 2; ++vertical) { //This loop checks boxes above and below our original box
                    for (int horizontal = -1; horizontal < 2; ++horizontal) { //This loop checks boxes side to side of our original box
                        if (vertical + y >= 0 && vertical + y < 5 && horizontal + x >= 0 && horizontal + x < 5) { //If we're inside the array bounds (ie, if we're not trying to access array[-1][-2] or something similar which would undefined behaviour), continue
                            if (array[x + horizontal][y + vertical] == 0) //If the box we're checking is a 0, add 1 to the number of neighbours our original box has
                                ++count;
                        }
                    }
                }
                cout << "(" << x << ", " << y << "): " << count << endl; 
                if (count == num) dummyarray[x][y] = 0; //If the box we just checked has the number of neighbours we're looking for, mark it in the dummy array.
            }
        }
    }
    for (unsigned x = 0; x < 5; ++x) {
        for (unsigned y = 0; y < 5; ++y)
            array[x][y] = dummyarray[x][y]; //Here we copy all of dummyarray's data to our original array.
    }
    Draw(array); //Finally, we draw our array again with Draw()

    return 0;
}