Red and Black---POJ - 1979

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m;
int ans=0;
int vis[30][30];
char title[30][30];
int fx[4]={-1,0,1,0},fy[4]={0,1,0,-1};

void dfs(int x,int y){
    ans++;
    vis[x][y]=1;
    for(int i=0; i<4; i++ ){
        int nx=x+fx[i];
        int ny=y+fy[i];
        if(nx>=0&&nx<n&&ny>=0&&ny<m&&title[nx][ny]=='.'&&!vis[nx][ny]){
            dfs(nx,ny);
        }
    }
}

int main(){
    while(~scanf("%d%d",&m,&n)&&n&&m){
//        getchar();
        ans=0;
        memset(vis,0,sizeof(vis));
        for( int i=0; i<n; i++ ){
            cin>>title[i];
        }
        for( int i=0; i<n; i++ ){
            for(int j=0; j<m; j++ ){
                if(title[i][j]=='@'&&!vis[i][j]){
                    dfs(i,j);
                    break;
                }
            }
        }
        printf("%d\n",ans);
    }

    return 0;
}