原题目：

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)

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

4559613 题目大意： 给出一个w列和h行的方阵，方阵中有'.','#'和'@'。其中'.'是可以移动的地方，'#'是红色的砖，是不可以移动的地方，'@'是最初人的地点就（图的入口），人可以上下左右移动，以0 0 为结束。 要求输出人可以移动的砖的个数。该题利用搜索。 注意： 先输入w，然后输入h，但是w是列数，h是行数。 代码：

#include
    
     #include
     
       using namespace std; void dfs(int x,int y); int w,h,tmp=0; char ma[100][100]; int dx[4]={ 0,0,1,-1}; int dy[4]={ 1,-1,0,0}; int main(){ while(cin>>h>>w){ if(w==0&&h==0) break; memset(ma,'0',sizeof(ma)); int ax,ay; for(int i=0;i
      
       >ma[i][j]; if(ma[i][j]=='@'){ ax=i; ay=j; } } } tmp=0; dfs(ax,ay); cout<
       
        <
        
         =0&&nx
         
          =0&&ny
         
        
       
      
     
    

Select Code

#include
    
     #include
     
       using namespace std; void dfs(int x,int y); int w,h,tmp=0; char ma[100][100]; int dx[4]={ 0,0,1,-1}; int dy[4]={ 1,-1,0,0}; int main(){ while(cin>>h>>w){ if(w==0&&h==0) break; memset(ma,'0',sizeof(ma)); int ax,ay; for(int i=0;i
      
       >ma[i][j]; if(ma[i][j]=='@'){ ax=i; ay=j; } } } tmp=0; dfs(ax,ay); cout<
       
        <
        
         =0&&nx
         
          =0&&ny