如何在二维数组中找到索引为 0 或 1 的二维形状的周长和面积？

Question

给定一个大小为 20x20 的二维数组，其值类似于二维形状，例如正方形或矩形：

public static int[][] rectangle= {
        {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
        {0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
        {0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0},
        {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}
        
};

我想在执行以下操作的算法中找到它的周长和面积：

从中心点 (point[10][10]) 开始寻找最近的 (1) 点。
以最近的（1）点为起点，遍历所有剩余的1来统计周长.
计算圆周围的零点个数，计算面积

下面是我现在所在的位置； “猜测”方法计算最近的 (1) 点并执行“计数”方法，然后计算周长。

public static void guess() {

    boolean found = false;
    
    if(!found) {
    int  y = 10;
    for(int x = 10; x <= 20; x++) { 
            if(rectangle[x][y]==1) {
                rectangle[x][y] = 2;
                found = true;
                break;
            }else if(rectangle[x][y++]==1) {
                    rectangle[x][y] = 2;
                    found = true;
                    break;
                }
            }
     }
    
    
    if(!found) {
     int  y = 10;
     for(int x = 10; x >= 0; x--) { 
            if(rectangle[x][y]==1) {
                rectangle[x][y] = 2;
                found = true;
                 break;
            }else if(rectangle[x][y--]==1) {
                    rectangle[x][y] = 2;
                    found = true;
                     break;
                }
            }
     }
    
     if(!found) {
            int  x  = 10;
            for(int y = 10; y <= 20; y++) { 
                    if(rectangle[x][y]==1) {
                        rectangle[x][y] = 2;
                        found = true;
                        break;
                    }else if(rectangle[x][y++]==1) {
                            rectangle[x][y] = 2;
                            found = true;
                            break;
                        }
                    }
             }
            
            if(!found) {
             int  x = 10;
             for(int y = 10; y >= 0; y--) { 
                    if(rectangle[x][y]==1) {
                        rectangle[x][y] = 2;
                        found = true;
                         break;
                    }else if(rectangle[x][y--]==1) {
                            rectangle[x][y] = 2;
                            found = true;
                             break;
                        }
                    }
             }
    
    for(int i = 0; i < 20; i++) {
        for(int j = 0; j < 20; j++) {
            if(rectangle[i][j] == 2) {
                Count(i, j);
                break;
            }
        }
    }
    
}

public static void Count(int x, int y) {
    
    public int circumf;

    int tx = x;
    int ty = y;
    
    for(int c = 40; c >=0; c--) {
        if((c/2)-1<0 || x>=20 || x<0 || y>=20 || y<0) break;
        
        if(rectangle[x][(int) (c/2)-1]==1 || rectangle[(int) (c/2)-1][y]==1 
                || rectangle[x++][y]==1 || rectangle[x][y++]==1
                || rectangle[x--][(int) (c/2)-1]==1 || rectangle[(int) (c/2)-1][y--]==1 || rectangle[x--][y]==1 || rectangle[x][y--]==1) {
            circumf++;
        }
        
    }
    
    x = tx;
    y = ty;
    
    for(int c = 0; c <=40; c++) {
        if((c/2)>=20 || x>=20 || x<0 || y>=20 || y<0) break;
        
        if(rectangle[x][(int) (c/2)]==1 || rectangle[(int) (c/2)][y]==1 
                || rectangle[x++][y]==1 || rectangle[x][y++]==1
                || rectangle[x--][(int) (c/2)]==1 || rectangle[(int) (c/2)] 
       [y--]==1 || rectangle[x--][y]==1 || rectangle[x][y--]==1) {
            circumf++;
        }
    
        
    }
    
    System.out.print(circumf);

现在，guess 方法可以正确计算最近的点，但是 count 方法不能正确计算上例中接近 70 的周长。

关于面积计算的算法，我还没搞清楚

上面的代码并不是我所知道的最出色或最有条理的代码，但我们将不胜感激任何帮助！

Answer 1

假设您在索引 [i][j] 找到最近的一个。由于您的对象是矩形，因此您需要检查交叉图案中最接近的值。

例如你在矩形左边缘的中间一个：

0 1 0
0 1 0
0 1 0

你检查 1 的“继续”找到边缘方向。有两种可能的结果

[x-1][y]=[x][y]=[x+1][y] or [x][y-1]=[x][y]=[x][y+1]==1

所以，你发现边缘是垂直的。继续迭代这条垂直线，直到条件 [x][y-1]==1 为 false。然后对水平线做同样的事情。

现在对于 0 的计数，您可以在执行上述操作时存储所有角的索引 i、j check.If您知道它们的位置并且数组的大小是固定的您可以计算 0喜欢：

0's area = [(DL_corner_index - 1) - (UL_corner_index + 1)] x [(UR_corner_index - 1) - (UL_corner_index + 1)]

int rectangle[10][10] ={
   {0,0,0,0,0,0,0,0,0,0},
   {0,1,1,1,1,1,1,1,1,0},
   {0,1,0,0,0,0,0,0,1,0},
   {0,1,0,0,0,0,0,0,1,0},
   {0,1,0,0,0,0,0,0,1,0},
   {0,1,0,0,0,0,0,0,1,0},
   {0,1,0,0,0,0,0,0,1,0},
   {0,1,0,0,0,0,0,0,1,0},
   {0,1,1,1,1,1,1,1,1,0},
   {0,0,0,0,0,0,0,0,0,0}
};

//calculating area of zeros with known corner coordinates
//you have stored the coordinates in variables like this
int UL_X=1,UR_X=8;
int UL_Y=1,DL_Y=8;
int area = ((UR_X-1) - UL_X)*((DL_Y-1)-UL_Y);

cout<<area; //36
}

如何在二维数组中找到索引为 0 或 1 的二维形状的周长和面积？

How to find circumference and area of a 2D shape whose indices are given as 0 or 1 in a 2D array?

java

arrays

algorithm

2d

path-finding