算法：深搜

题意：就是让你找到一共可以移动多少次，每次只能移到黑色格子上，

Problem Description

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

..#.#..

..#.#..

###.###

...@...

###.###

..#.#..

..#.#..

0 0

Sample Output

45

59

6

13

代码：

#include 
    
     #include 
     
      #include 
      
       #include 
       
        using namespace std;char ch[25][25];int k,n,m;void dfs(int x,int y,int &k){	ch[x][y]='#';	if(x-1>=0&&x-1
        
         =0&&y
         
          =0&&y>=0&&y
          
           =0&&y-1
           
            =0&&x
            
             =0&&x>=0&&x
             
              >n>>m&&n&&m) { k=1; for(i=0;i
              
               >ch[i][j]; if(ch[i][j]=='@') { p=i;q=j; } } } dfs(p,q,k); cout<
               
                <