Red and Black

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.

Input:

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. 

Output:

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).       

Sample Input:

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


Sample Output

45
59
6
13

这道题目要求求出可以走的地方（包括起点），如果用BFS做的话，可以先把起点入队列，并标记该点已经计数。然后，不断向外扩张，如果该点满足条件的话，那么就把该点放入队列中，再标记和计数...如此类推，直到找出所有满足条件的点。
代码如下：

import java.util.Scanner;

public class Main {
    static class Location{
        int x;
        int y;
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int W = in.nextInt();
        int H = in.nextInt();
        while(W != 0 || H != 0) {
            String[] str = new String[H];
            for(int i = 0; i < H; i++) {
                str[i] = in.next();
            }
            char[][] map = new char[H][W];
            for(int r = 0; r < H; r++) {
                for(int c = 0; c < W; c++) {
                    map[r][c] = str[r].charAt(c);
                }
            }
            boolean[][] visited = new boolean[H][W];
            int result = BFS(map, visited, H, W);
            System.out.println(result);
            W = in.nextInt();
            H = in.nextInt();
        }
    }
    
    public static int BFS(char[][] map, boolean[][] visited, int H, int W) {
        int result = 0;
        Location[] queue = new Location[H*W+1];
        int head = 0, tail = 0;
        for(int r = 0; r < H; r++) {
            for(int c = 0; c < W; c++) {
                if(map[r][c] == '@') {
                    Location start = new Location();
                    start.y = r;
                    start.x = c;
                    queue[tail++] = start;
                    visited[r][c] = true;
                    result++;
                }
            }
        }
        int[] dx = {1, 0, -1, 0};
        int[] dy = {0, -1, 0, 1};
        while(head < tail) {
            Location curNode = queue[head++];
            for(int d = 0; d < 4; d++) {
                Location next = new Location();
                next.x = curNode.x + dx[d];
                next.y = curNode.y + dy[d];
                if(next.x >= 0 && next.x < W && next.y >= 0 && next.y < H) {
                    if(map[next.y][next.x] == '.' && !visited[next.y][next.x]) {
                        queue[tail++] = next;
                        visited[next.y][next.x] = true;
                        result++;
                    }
                }
            }
            
        }
        return result;
    }

DFS的做法：

import java.util.Scanner;

public class Main {
    static int result;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int W = in.nextInt();
        int H = in.nextInt();
        while(W != 0 || H != 0) {
            String[] str = new String[H];
            for(int i = 0; i < H; i++) {
                str[i] = in.next();
            }
            char[][] map = new char[H +1][W + 1];
            int start_x = 0, start_y = 0;
            boolean[][] visited = new boolean[H + 1][W + 1];
            for(int r = 0; r < H; r++) {
                for(int c = 0; c < W; c++) {
                    map[r][c] = str[r].charAt(c);
                    if(map[r][c] == '@') {
                        start_x = c;
                        start_y = r;
                        visited[start_y][start_x] = true;
                    }
                }
            }
            result = 0;
            DFS(map, visited, start_x, start_y, W, H);
            System.out.println(result);
            W = in.nextInt();
            H = in.nextInt();
        }
    }
    
    public static void DFS(char[][] map, boolean[][] visited, int cur_x,int cur_y, int W, int H) {
        if(map[cur_y][cur_x] == '#') {
            return;
        }
        result++;
        int[] dx = {1, 0, -1, 0};
        int[] dy = {0, -1, 0, 1};
        for(int i = 0; i < 4; i++) {
            int x = cur_x + dx[i];
            int y = cur_y + dy[i];
            if(x >= 0 && x < W && y >= 0 &&y < H) {
                if(!visited[y][x]) {
                    visited[y][x] = true;
                    DFS(map, visited, x, y, W, H);
                    //visited[y][x] = false;
                }
            }
        }
    }

}