HDU - 1312 Red and Black（DFS）

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>
#include<string.h>
char a[101][101];
int m,n,t;
int xy[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};
void dfs(int x,int y)
{
	
	int i;
    for(i=0;i<4;i++)
    { 
		
		int x1 = x+xy[i][0];
        int y1 = y+xy[i][1];
		if(x1>=0&&x1<n&&y1>=0&&y1<m&&a[x1][y1]=='.')
		{
		a[x1][y1]='*';
		dfs(x1,y1);
	}
    } 
}

int main()
{
	
    while(1)
    {
		memset(a,0, sizeof(a));
		scanf("%d%d",&m,&n);
		if(m+n==0)
			break;
        int i,j,x,y;
		t=0;
		for(i=0; i<n; i++)
		{
			scanf("%s",a[i]);
            for(j=0; j<m; j++)
            {
                if(a[i][j]=='@')   
				{          
                    x=i;
					y=j;
				}

            }
		}
		t++;
		dfs(x,y);
		for(i=0; i<n; i++)
            for(j=0; j<m; j++)	
            	if(a[i][j]=='*')
            		t++; 
      
        printf("%d\n",t);
    }
    return 0;
}