Java程序练习-Red and Black

http://poj.grids.cn/practice/1979/
http://poj.grids.cn/practice/3866/

Red and Black
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
参考代码

import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class Main { public static char map[][]; public static int row; public static int col; public static int steps; public static int ds[][]=new int[][]{{1,0},{-1,0},{0,1},{0,-1}}; public static void main(String[] args) throws IOException { BufferedReader cin = new BufferedReader(new InputStreamReader(System.in)); while(true){ String dn[] = cin.readLine().split(" "); col = Integer.parseInt(dn[0]); row = Integer.parseInt(dn[1]); if(0 == col && 0 == row) break; map = new char[row][col]; int si = 0,sj = 0; for(int i = 0;i < row;++ i){ String sline = cin.readLine(); for(int j = 0;j < col;++ j){ map[i][j] = sline.charAt(j); if(map[i][j] == '@'){ si = i; sj = j; } } } steps = 1; movesteps(si,sj); System.out.println(steps); } } private static void movesteps(int si, int sj) { for(int i = 0;i < 4;++ i){ int sx = si+ds[i][0],sy = sj+ds[i][1]; if(check(sx,sy)){ map[sx][sy] = '1'; movesteps(sx,sy); steps ++; } } } private static boolean check(int x, int y) { if(x >= 0&&x < row&&y >= 0&&y < col&&map[x][y] == '.') return true; return false; } }