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.
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
..#.#..
..#.#..

.

…@…

.

..#.#..
.3

.#..

0 0
Sample Output
45
59
6
13
大意：就是让你以@为起点，只能走点，#不能走，最多能走到达多少个地方。
思路：bfs和dfs都可以做，下面用dfszuoyixia。

#include<stdio.h>
#include<string.h>
#define maxn 50
char str[maxn][maxn];
int cs[maxn][maxn];
int n,m,ans;

int d[4][2] = {1,0,-1,0,0,1,0,-1};
void dfs(int x,int y)
{
    cs[x][y] = 1;
    ans++;
    for(int i = 0;i < 4; i++)
    {
        int dx = x + d[i][0];
        int dy = y + d[i][1];
        if(dx>=0 && dx<n && dy>=0 && dy<m && !cs[dx][dy] && str[dx][dy]!='#')
        {
            dfs(dx,dy);
        }
    }
}
int main()
{
    while(scanf("%d%d",&m,&n)!=EOF && (n||m))
    {
        memset(cs,0,sizeof(cs));
        for(int i = 0;i < n; i++)
            scanf("%s",str[i]);
        int xf,yf;
        ans = 0;
        for(int i = 0;i < n; i++)
            for(int j = 0;j < m; j++)
            {
                if(str[i][j] == '@'){
                    xf = i,yf = j;
                    break;
                }
            }
        dfs(xf,yf);
        printf("%d\n",ans);
    }
    return 0;
}