ProgrammerSought

Original post address http://bbs.csdn.net/topics/380191378

There is a bug in the code written before, that is, the first point must be 1. Ha ha, modified it.

original title

public class Demo1 {

	static int a1[][] = {
	 {0 ,0 ,1 ,1 ,1 ,1 ,1}
	,{1 ,0, 1, 0, 0, 0, 1}
	,{1 ,0, 1, 1, 1, 0, 1}
	,{1 ,0, 1, 1, 1, 0, 1}
	,{1 ,0, 1, 1, 1, 0, 1}
	,{1 ,0 ,0 ,0 ,0 ,0, 1}
	,{1 ,1, 1, 1, 1, 1, 1}};
	static int m = a1.length;//Maximum length
	static int n = a1[0].length;//Maximum width
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		
		Demo1 ha = new Demo1();
		boolean flag2 = false;
		for (int i = 0; i < a1.length; i++) {
			for (int j = 0; j < a1[i].length; j++) {
				if(a1[i][j]==1){//Find the first point
					int[][]	list = {{i,j}};
					a1[i][j]=2;
					ha.check(list);
					flag2 = true;
					break;
				}
			}
			if(flag2){
				break;
			}
		}
// a1 [0] [0] = 2;
//		System.out.println(list3[0]);
		
		boolean flag = true;
		for(int i=0;i<m;i++)
		{
			for(int j=0;j<n;j++)
			{
				System.out.print(a1[i][j]);
				if(a1[i][j]==1)
				{
					flag = false;
				}
			}
			System.out.println();
		}
		if(flag)
		{
			System.out.println("is open");
		}
		else
		{
			System.out.println("is closed");
		}
		
}
	/**
	 * starting from the first point 0,0, find 8 points around him,
	 * If the surrounding 8 points have 1, record it,
	 * Next time, find all the points of these 1 points, and cover them continuously
	 * Add 1 to the covered 1 point, and it becomes 2, 0 points are not covered
	 * When there is no 1 in the surrounding 8 points of a point.
	 * Jump out of the loop, indicating that all coverage has been completed
	 * The last circular matrix, if there are still 1 points,
	 * Prove that this matrix is ​​not connected, otherwise it is connected
	 * @param list
	 */
		public void check(int[][] list)
		{
			int[][] list1 = new int[100][];
			int k=0;
			for(int i=0;i<list.length;i++)
			{
				int[] zuobiao = list[i];
				int x = zuobiao[0];//abscissa
				int y = zuobiao[1];//ordinate
				/*
				 * Start to find 8 points around
				 * coordinate of
				 */
				int[][] eight = new int[8][2];//
				//upper left
				if(x>0 && y>0)
				{
					eight[0][0] = x-1;	
					eight[0][1] = y-1;
					if(a1[x-1][y-1] ==1)
					{
						a1[x-1][y-1] +=1;
						list1[k] = eight[0];
						k++;
					}
					
				}
				//superior
				if(y>0)
				{
					eight[1][0] = x;	
					eight[1][1] = y-1;
					if(a1[x][y-1] ==1)
					{
						a1[x][y-1] +=1;
						list1[k] = eight[1];
						k++;

					}
				}
				//upper right
				if(x<m-1 && y>0)
				{
					eight[2][0] = x+1;
					eight[2][1] = y-1;
					if(a1[x+1][y-1] ==1)
					{
						a1[x+1][y-1] +=1;
						list1[k] = eight[2];
						k++;
					}
					
				}
				//Left
				if(x>0)
				{
					eight[3][0] = x-1;
					eight[3][1] = y;
					if(a1[x-1][y] ==1)
					{
						a1[x-1][y] +=1;
						list1[k] = eight[3];
						k++;
					}
					
				}
				//right
				if(x<m-1)
				{
					eight[4][0] = x+1;
					eight[4][1] = y;
					if(a1[x+1][y] ==1)
					{
						a1[x+1][y] +=1;
						list1[k] = eight[4];
						k++;
					}
					
				}
				//lower left
				if(x>0 && y<n-1)
				{
					eight[5][0] = x-1;
					eight[5][1] = y+1;
					if(a1[x-1][y+1] ==1)
					{
						a1[x-1][y+1] +=1;
						list1[k] = eight[5];
						k++;
					}
					
				}
				//Down
				if(y<n-1)
				{
					eight[6][0] = x;
					eight[6][1] = y+1;
					if(a1[x][y+1] ==1)
					{
						a1[x][y+1] +=1;
						list1[k] = eight[6];
						k++;
					}
					
				}
				// Bottom right
				if(x<m-1 && y<n-1)
				{
					eight[7][0] = x+1;
					eight[7][1] = y+1;
					if(a1[x+1][y+1] ==1)
					{
						a1[x+1][y+1] +=1;
						list1[k] = eight[7];
						k++;
					}
				}
			}
			//Indicates that there is no point in list1, indicating that the coverage is complete
			if(list1[0] == null)
			{
				return;
			}else
			{
				int[][] list2 = null;
				int i=0;
				for(;i<list1.length;i++)
				{
					if(list1[i]==null)
					{
						list2 = new int[i][];
						break;
					}
				}
				for(int j=0;j<i;j++)
				{
					list2[j]= list1[j];
				}
				check(list2);
			}
		}

}