Red and Black ---路线问题

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) 

 

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

 

 

 

 

#include <stdio.h>
#define N 22
char ch[N][N];
int m ,n;
int fin(int x, int y)
{
    if(x<0 || x>=n || y<0 || y>=m)
        return 0;
    else if(ch[x][y]=='#')
        return 0;
    else
    {
        ch[x][y] = '#';
        return 1+fin(x-1, y)+fin(x+1, y)+fin(x, y-1)+fin(x, y+1);
    }

}
int main()
{
    int i, j;
    int x, y, a;
    while (scanf("%d%d", &m, &n), m!=0 && n!=0)
    {
        for(i = 0; i<n; i++)
        {
            for (j =0; j<m; j++)
            {
                scanf(" %c", &ch[i][j]);
                if (ch[i][j]=='@')
                {
                    x = i;
                    y = j;

                }
            }
        }
        a = fin(x, y);
        printf("%d\n", a);
    }

    return 0;
}