Red and Black


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10122    Accepted Submission(s): 6313


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


 



Source


​​Asia 2004, Ehime (Japan), Japan Domestic​​ 


#include<stdio.h>
char s[25][25];
int a[4]={0,0,1,-1};
int b[4]={1,-1,0,0};
int m,n,sum;
void bfs(int x,int y)
{
    int k,v,t;
    s[x][y]='#';
    for(k=0;k<4;k++)
    {
        v=x+a[k];
        t=y+b[k];
        if(s[v][t]=='.'&&v>=1&&v<=n&&t>=0&&t<m)
        {
            bfs(v,t);
            sum++;
        }
    }
}
int main()
{
    int i,j;
    while(scanf("%d%d",&m,&n)&&(m+n))
    {
        for(i=1;i<=n;i++)
        scanf("%s",s[i]);
        for(i=1,sum=1;i<=n;i++)
        {
            for(j=0;j<m;j++)
            {
               if(s[i][j]=='@')
               bfs(i,j);
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}