Day2-K-Red and Black-HDU1312

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. 

InputThe 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) 
OutputFor 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

思路：比较简单的DFS求连通，读入时找到起点即可，代码如下：

const int maxm = 25;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};

int m, n, sx, sy, vis[maxm][maxm], t;
char buf[maxm][maxm];

bool inside(int x,int y) {
    return x >= 0 && x < m && y >= 0 && y < n;
}

void dfs(int x,int y) {
    if(vis[x][y])
        return;
    vis[x][y] = 1;
    for (int i = 0; i < 4; ++i) {
        int nx = x + dx[i], ny = y + dy[i];
        if(inside(nx,ny) && buf[nx][ny] == '.' &&!vis[nx][ny]) {
            ++t;
            dfs(nx, ny);
        }
    }
}


int main() {
    while(scanf("%d%d",&n,&m) && m + n) {
        getchar();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                scanf("%c", &buf[i][j]);
                if(buf[i][j] == '@') {
                    sx = i, sy = j;
                }
            }
            getchar();
        }
        memset(vis,0,sizeof(vis));
        t = 1;
        dfs(sx, sy);
        printf("%d\n", t);
    }
    return 0;
}