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=;
 int vis[][];
 char title[][];
 int fx[]={-,,,},fy[]={,,,-};

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

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

     return ;
 }