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2013
11-09

POJ 1111 Image Perimeters [解题报告] Java

Image Perimeters

问题描述 :

Technicians in a pathology lab analyze digitized http://poj.org/images of slides. Objects on a slide are selected for analysis by a mouse click on the object. The perimeter of the boundary of an object is one useful measure. Your task is to determine this perimeter for selected objects.

The digitized slides will be represented by a rectangular grid of periods, ‘.’, indicating empty space, and the capital letter ‘X’, indicating part of an object. Simple examples are


XX Grid 1 .XXX Grid 2
XX .XXX
.XXX
...X
..X.
X...

An X in a grid square indicates that the entire grid square, including its boundaries, lies in some object. The X in the center of the grid below is adjacent to the X in any of the 8 positions around it. The grid squares for any two adjacent X’s overlap on an edge or corner, so they are connected.


XXX
XXX Central X and adjacent X's
XXX

An object consists of the grid squares of all X’s that can be linked to one another through a sequence of adjacent X’s. In Grid 1, the whole grid is filled by one object. In Grid 2 there are two objects. One object contains only the lower left grid square. The remaining X’s belong to the other object.

The technician will always click on an X, selecting the object containing that X. The coordinates of the click are recorded. Rows and columns are numbered starting from 1 in the upper left hand corner. The technician could select the object in Grid 1 by clicking on row 2 and column 2. The larger object in Grid 2 could be selected by clicking on row 2, column 3. The click could not be on row 4, column 3.



One useful statistic is the perimeter of the object. Assume each X corresponds to a square one unit on each side. Hence the object in Grid 1 has perimeter 8 (2 on each of four sides). The perimeter for the larger object in Grid 2 is illustrated in the figure at the left. The length is 18.

Objects will not contain any totally enclosed holes, so the leftmost grid patterns shown below could NOT appear. The variations on the right could appear:


Impossible Possible

XXXX XXXX XXXX XXXX
X..X XXXX X... X...
XX.X XXXX XX.X XX.X
XXXX XXXX XXXX XX.X

..... ..... ..... .....
..X.. ..X.. ..X.. ..X..
.X.X. .XXX. .X... .....
..X.. ..X.. ..X.. ..X..
..... ..... ..... .....

输入:

The input will contain one or more grids. Each grid is preceded by a line containing the number of rows and columns in the grid and the row and column of the mouse click. All numbers are in the range 1-20. The rows of the grid follow, starting on the next line, consisting of ‘.’ and ‘X’ characters.

The end of the input is indicated by a line containing four zeros. The numbers on any one line are separated by blanks. The grid rows contain no blanks.

输出:

For each grid in the input, the output contains a single line with the perimeter of the specified object.

样例输入:

2 2 2 2
XX
XX
6 4 2 3
.XXX
.XXX
.XXX
...X
..X.
X...
5 6 1 3
.XXXX.
X....X
..XX.X
.X...X
..XXX.
7 7 2 6
XXXXXXX
XX...XX
X..X..X
X..X...
X..X..X
X.....X
XXXXXXX
7 7 4 4
XXXXXXX
XX...XX
X..X..X
X..X...
X..X..X
X.....X
XXXXXXX
0 0 0 0

样例输出:

8
18
40
48
8

解题代码:

//* @author: [email protected]
import java.io.*;
class Main
{
	static int cnt;
	static char[][] map;
	static int[][] bool;
	public static void main(String[] args) throws IOException
	{
		InputStreamReader is=new InputStreamReader(System.in);
		BufferedReader in=new BufferedReader(is);
		while(true)
		{
			cnt=0;
			String[] ss=in.readLine().split(" ");
			int row=Integer.parseInt(ss[0]);
			int col=Integer.parseInt(ss[1]);
			int x=Integer.parseInt(ss[2]);
			int y=Integer.parseInt(ss[3]);
			if(row==0&&col==0)break;
			map=new char[row+2][col+2];
			bool=new int[row+2][col+2];
			for(int i=1;i<=row;i++)
			{
				for(int j=1;j<=col;j++)
					map[i][j]=(char)in.read();
				in.readLine();
			}
			f(x,y);
			System.out.println(cnt);
		}
	}
	static void f(int x,int y)
	{
		if(bool[x][y]==1)return;
		bool[x][y]=1;
		if (map[x-1][y]!='X')cnt++;
		if (map[x+1][y]!='X')cnt++;
		if (map[x][y-1]!='X')cnt++;
		if (map[x][y+1]!='X')cnt++;
		for(int i=x-1;i< x+2;i++)
			for(int j=y-1;j< y+2;j++)
				if(map[i][j]=='X')f(i,j);	
	}
}

  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

  2. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?

  3. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮