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

#include <stdio.h>
#include <cstring>
char a[22][22];
int b[22][22];
int fx[4][2]={1,0,-1,0,0,1,0,-1};
int ans;
void dfs(int x,int y,int p,int q)
{
   b[y][x]=1; /*把@点标记为1，即已走过*/
   for(int i=0;i<4;i++)
   {
       int x1=x+fx[i][0];
       int y1=y+fx[i][1];
       if(x1>=0 && x1<p && y1>=0 && y1<q && b[y1][x1]==0 && a[y1][x1]=='.')
       {
           ans++;
           b[y1][x1]=1; /*把走过的点标记为1*/
           dfs(x1,y1,p,q);
       }
   }
}

int main()
{
   int w,h;
   while(scanf("%d%d",&w,&h) && w && h)
   {
       memset(b,0,sizeof(b));
       memset(a,0,sizeof(a));
       ans=1;
       for(int i=0;i<h;i++)
       scanf("%s",&a[i]);
       for(int i=0;i<h;i++)
       {
           for(int j=0;j<w;j++)
           {
               if(a[i][j]=='@')
               {
               dfs(j,i,w,h); /*j是@点的横坐标，i是@点的纵坐标*/
               printf("%d\n",ans);
               break;
           }
           }
       }
   }
   return 0;
}

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