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.

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.

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).

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

45
59
6
13

题意描述：
输入矩阵的大小W和H（均小于20）
计算并输出从‘@‘位置最多能走多少块‘.‘
解题思路：
输入的时候找到‘@‘的位置，随后对其进行DFS搜索，下面的代码实现的搜索有点模拟广搜的意思。
代码实现：

 1 #include<stdio.h> 
 2 char map[30][30];
 3 int dfs(int x,int y);
 4 int w,h;
 5 int main()
 6 {
 7     int i,j,sx,sy;
 8     while(scanf("%d%d",&w,&h),w+h != 0)
 9     {
10         for(i=1;i<=h;i++)
11         {
12             for(j=1;j<=w;j++){
13                 scanf(" %c",&map[i][j]);
14                 if(map[i][j]==‘@‘)
15                 { sx=i;sy=j; }
16             }
17             getchar();
18         }
19         printf("%d\n",dfs(sx,sy));
20     }
21     return 0;
22 }
23 int dfs(int x,int y)
24 {
25     if(x<1 || x>h || y<1 || y>w)
26         return 0;
27     //如果进入不了dfs函数就是边界问题，注意行数和列数就是x和y的范围 
28     if(map[x][y]==‘#‘)
29         return 0;
30     else
31     {
32         map[x][y]=‘#‘;
33         return 1+dfs(x-1,y)+dfs(x+1,y)+dfs(x,y-1)+dfs(x,y+1);
34     }
35 }