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;
}