Question

我正在尝试通过一个数组并找出哪些数组至少有一个直接邻居..

我得到的答案是

这是完美的。我能够从输出中看到哪些正好有1个相邻数组[1] [1]和数组[1] [3]。

现在我该如何设置这两个（数组[1] [1]和数组[1] [3]）= 1？

Answer 1

由于您使用嵌套的＆＃39;显示＆＃39;多次循环，将其分解为自己的函数是理想的：

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';
    }
}

同样，你找到邻居＆＃39;使用额外的嵌套循环可以大大缩短检查。

这也将使您更轻松地解决问题：

#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;
}

Answer 2

我真的很困惑，如果你需要的是＆＃34;至少有一个邻居是1＆＃34;你为什么要尝试匹配整个社区？你有什么理由可以做到：

// 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;
            } 
      }

哦，你应该在边境处理案件。

int clamp04(int coord){return coord<0?0:(coord>4)?4:coord;}

然后像

一样使用它

// 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;
            } 
      }

另外，作为旁注，如果您将这些值设置为1，则您执行的操作是一种称为"dilation"的形态滤镜操作

如何在数组中设置至少1个邻居等于1的框

2 个答案: