Red and Black hd 1312

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

#include<stdio.h>  #include<iostream>  using namespace std;  int n,m,dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}},sum;  char map[25][25];    int DFS(int i,int j)  {      int e;      map[i][j]='#';      sum++;      for(e=0;e<4;e++)      if(i+dir[e][0]>=0&&i+dir[e][0]<n&&j+dir[e][1]>=0&&j+dir[e][1]<m)      if(map[i+dir[e][0]][j+dir[e][1]]!='#')      DFS(i+dir[e][0],j+dir[e][1]);  }    int main()  {      int i,j,pi,pj;      while(scanf("%d%d",&m,&n)>0&&(m||n))      {          for(i=0;i<n;i++)          {              getchar();              for(j=0;j<m;j++)              {                  scanf("%c",&map[i][j]);                  if(map[i][j]=='@')                  {                      pi=i;pj=j;                  }              }          }          sum=0;          DFS(pi,pj);          printf("%d\n",sum);      }  }