Red and Black POJ

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) 
The end of the input is indicated by a line consisting of two zeros. 

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
BFS

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int lenx,leny;
char a[25][25];
struct lll
{
    int x,y;
}qidian;
int TM=0;
int c[25][25];
int b[4][2]={1,0,-1,0,0,1,0,-1};
void dfs(int y,int x)
{
    if(TM>lenx*leny)
        return ;

    if(y<1||x<1||x>lenx||y>leny)
            return ;

    if(a[y][x]=='#'||c[y][x]!=0)
            return ;
    c[y][x]=++TM;
    for(int i=0;i<4;i++)
    {
        dfs(y+b[i][1],x+b[i][0]);
    }
}
int main()
{
    while(cin>>lenx>>leny)
    {
        if(lenx==0&&leny==0)
            break;
        memset(a,0,sizeof(a));
        memset(c,0,sizeof(c));
        for(int i=1;i<=leny;i++)
        {
            for(int j=1;j<=lenx;j++)
            {
                cin>>a[i][j];
                if(a[i][j]=='@')
                    {
                        qidian.y=i;
                        qidian.x=j;
                    }
            }
        }
        TM=0;
        dfs(qidian.y,qidian.x);
        cout<<TM<<endl;
    }
    return 0;
}