poj 1979(递归)

1979:Red and Black

总时间限制:

1000ms

内存限制:

65536kB

描述

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

输入

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)
The end of the input is indicated by a line consisting of two zeros.

输出

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

样例输入

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

样例输出

45
59
6
13

翻译：大致题意如下：

问题描述：有一间长方形的房子，地上铺了红色、黑色两种颜色的正方形瓷砖。你站在其中一块黑色的瓷砖上，只能向相邻的黑色瓷砖移动。请写一个程序，计算你总共能够到达多少块黑色的瓷砖。

 

输入：包括多个数据集合。每个数据集合的第一行是两个整数W和H，分别表示x方向和y方向瓷砖的数量。W和H都不超过20。在接下来的H行中，每行包括W个字符。每个字符表示一块瓷砖的颜色，规则如下
‘.’：黑色的瓷砖
‘#’：白色的瓷砖
‘@’：黑色的瓷砖，并且你站在这块瓷砖上。该字符在每个数据集合中唯一出现一次
当在一行中读入的是两个零时，表示输入结束。
输出：对每个数据集合，分别输出一行，显示你从初始位置出发能到达的瓷砖数(记数时包括初始位置的瓷砖)

 

解决方法：算法：设f(x,y)为从点(x,y)出发能够走过的黑瓷砖总数，则
    f(x,y) = 1+f(x-1,y)+f(x+1,y)
                    +f(x,y-1)+f(x,y+1)
这里需要注意，凡是走过的瓷砖不能够被重复走过。

如何解决这一问题，有个好方法就是我们把走过的砖变为白砖，这样问题句简单了。

 

解决方案： Java

import java.util.Scanner;

/**
 * 黑瓷砖上行走
 * @author tanlvxu
 *
 */
public class Main{
   char rect[][] = new char[22][22] ;
	public static void main(String[] args) {
		
     new Main().test() ;
	}
	public int count(int sCol,int sRow)
	{
		if(rect[sCol][sRow]=='#')
			return 0 ;
		else rect[sCol][sRow] = '#' ;
		return 1+count(sCol-1,sRow)+count(sCol,sRow-1)+count(sCol+1,sRow)+count(sCol,sRow+1);
	}
	
    public void test()
    {
    	Scanner sc = new Scanner(System.in) ;
    	int W = sc.nextInt() ;
    	int H = sc.nextInt() ;
    	while(W!=0)
    	{
    		int i,j,sCol=0,sRow=0 ;
    		//初始化,把数组全部设为白瓷砖
    		for(i=0;i<22;i++)
    			for(j=0;j<22;j++)
    				rect[i][j] = '#';
    		//读入数据
    		for(i=1;i<=H;i++){
    			String s = sc.next() ;
    			for(j=1;j<=W;j++)
    			{    rect[i][j] = s.charAt(j-1) ;
    				if(rect[i][j]=='@')
    				{
    					sCol = i ;
    					sRow = j ;
    					rect[i][j] = '.' ;
    				}    				
    			}
    		}
    		System.out.println(count(sCol,sRow));
    		 W = sc.nextInt() ;
        	 H = sc.nextInt() ;
    		
    	}
    }

 

这个问题和滑雪问题很相似，可以参照滑雪难题来考虑这一类型的问题。