Red and Black

Problem Description

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

//此题根据马的走法改写
#include<iostream>
#include<cstdio>
using namespace std;
int mov[4][2]={0,1,0,-1,1,0,-1,0},a[22][22];
int cnt,sx,sy,vis[22][22],w,h;

void  dfs(int x,int y)
{  int i,nx,ny;
 for(i=0;i<4;i++)
 { 
	 nx=x+mov[i][0];
     ny=y+mov[i][1];

     if(nx>=0 && nx<h && ny>=0&&ny<w&&!vis[nx][ny]&&  a[nx][ny]!=-1) 
        {  cnt++;  
           vis[nx][ny]=1;
           dfs(nx,ny);
          
        }
 }
}
int main()
{
  int i,j,flag=1; 
  char b[22][22];
  while(scanf("%d%d",&w,&h)!=EOF)
  {  if(w==0&&h==0)  break;
     flag=1;
    for(i=0;i<h;i++)
		for(j=0;j<w;j++)
		cin>>b[i][j];

  for(i=0;i<h;i++)
	   for(j=0;j<w;j++)
		   if(b[i][j]=='@')  { a[i][j]=1; flag=0; sx=i;  sy=j;}
		   else if(b[i][j]=='.')  a[i][j]=0;
		   else if(b[i][j]=='#')  a[i][j]=-1;
                 
	 memset(vis,0,sizeof(vis));
	        
        if(flag) printf("0\n"); 
	
		else 
		{   cnt=1;  
            vis[sx][sy]=1; 
			dfs(sx,sy);  
           printf("%d\n",cnt);  
		}
	 } 
return 0;
}