poj_1979(dfs)

Red and Black

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>
 using namespace std;
 int n, m;
 int num;
 char map[][];
 bool vis[][];
 int dir[][] = {,,,,-,,,-};

 void dfs(int x, int y){
     for(int i = ; i < ; i++){
         int xx = x + dir[i][];
         int yy = y + dir[i][];
         if(xx >=  && xx <= m && yy >=  && yy <= n && vis[xx][yy] ==  && map[xx][yy] == '.'){
             vis[xx][yy] = ;
             num++;
             dfs(xx,yy);
         }
     }
 }
 int main(){
     while(cin >> n >> m){
         if(n ==  && m == )
             break;
         int x, y;
         memset(vis,,sizeof(vis));
         for(int i = ; i <= m; i++){
             for(int j = ; j <= n; j++){
                 cin >> map[i][j];
                 if(map[i][j] == '@')
                     x = i, y = j;
             }
         }
         num = ;
         vis[x][y] = ;
         dfs(x,y);
         cout << num << endl;
     }
     return ;
 }