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：

45

59

6

13

解题思路：

如果说，你写搜索的题写的挺多了话，那么这道题可能你看到样例，就能猜到这是一道要用DFS的题，题目的大致意思就是说给你一个图，‘@’代表这个人初始的位置；‘#’代表红色瓷砖，表示不能走；‘.’代表黑色瓷砖，可以走。最后要计算的是遍历走完整幅图，一共可以经过多少块黑色瓷砖。

程序代码：  

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int w,h,ans;
char map[25][25];
int num1[4]={-1,1,0,0};
int num2[4]={0,0,-1,1};
void dfs(int x,int y)
{
  map[x][y]='#';
  ans++;
  for(int i=0;i<4;i++)
  {
    int tx=x+num1[i];
    int ty=y+num2[i];
    if(tx>=0&&tx<h&&ty>=0&&ty<w&&map[tx][ty]=='.')
      dfs(tx,ty);
  }
  return ;
}
int main()
{
  while(cin>>w>>h)
  {
    if(w==0&&h==0)
      break;
    ans=0;
    int x,y;
    for(int i=0;i<h;i++)
      for(int j=0;j<w;j++)
        cin>>map[i][j];
    for(int i=0;i<h;i++)
    {
      for(int j=0;j<w;j++)
      {
        if(map[i][j]=='@')
        {
          x=i;
          y=j;
        }
      }
    }
    dfs(x,y);
    cout<<ans<<endl;
  }
  return 0;
}