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[25][25];
bool vis[25][25];
int dir[4][2] = {1,0,0,1,-1,0,0,-1};

void dfs(int x, int y){
    for(int i = 0; i < 4; i++){
        int xx = x + dir[i][0];
        int yy = y + dir[i][1];
        if(xx >= 1 && xx <= m && yy >= 1 && yy <= n && vis[xx][yy] == 0 && map[xx][yy] == '.'){
            vis[xx][yy] = 1;
            num++;
            dfs(xx,yy);
        }
    }
}
int main(){
    while(cin >> n >> m){
        if(n == 0 && m == 0)
            break;
        int x, y;
        memset(vis,0,sizeof(vis));
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                cin >> map[i][j];
                if(map[i][j] == '@')
                    x = i, y = j;
            }
        }
        num = 1;
        vis[x][y] = 1;
        dfs(x,y);
        cout << num << endl;
    }
    return 0;
}