POJ 1979 Red and Black （DFS）

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<cstdio>
 #include<string.h>
 int px[]={-,,,};
 int py[]={,,-,};
 int ans;
 bool key[][];
 char a[][];
 int n,m,x,y;
 void f(int x,int y)
 {
     int nx,ny;
     for(int i =  ;i <  ; i++)
     {
         nx=x+px[i];
         ny=y+py[i];
         if(nx >=  && ny >=  && a[nx][ny] == '.' && key[nx][ny] == false && nx < m && ny < n)
         {
             ans++;
             key[nx][ny]=true;
             f(nx,ny);
         }
     }
 }
 int main()
 {

     while(scanf("%d %d",&n,&m) && n && m)
     {
         memset(key,false,sizeof(key));
         for(int i =  ; i < m ; i++)
         {
             getchar();
             for(int j =  ; j < n ; j++)
             {
                 scanf("%c",&a[i][j]);
                 if(a[i][j] == '@')
                 {
                     x=i;
                     y=j;
                 }
             }

         }
         key[x][y]=true;
         ans=;
         f(x,y);
         printf("%d\n",ans+);
     }
 }