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).

45

59

6

13

解题思路：
深搜的方法解决，题目意思就是从@开始找.并与@连通，碰到#等于碰到了墙，题目很简单，@可以向四个方向上、下、左、右走，所以 用四个坐标标记出来，然后，再一一遍历，递归调用寻找，用一个30*30的数组标识此点有没有走过，避免走重复

程序代码：

#include <cstdio>

#include <cstring>

using namespace std;

int n,m,cot;

char map[][];

int to[][] = {{,},{,},{-,},{,-}};

void dfs(int i,int j)

{

    cot++;

    map[i][j] = '#';

    for(int k = ; k<; k++)

    {

        int x = i+to[k][];

        int y = j+to[k][];

        if(x<n && y<m && x>= && y>= && map[x][y] == '.')

            dfs(x,y);

    }

    return;

}

int main()

{

    int i,j,fi,fj;

    while(~scanf("%d%d%*c",&m,&n)&&m&&n)

    {

        for(i = ; i<n; i++)

        {

            for(j = ; j<m; j++)

            {

                scanf("%c",&map[i][j]);

                if(map[i][j] == '@')

                {

                    fi = i;

                    fj = j;

                }

            }

            getchar();

        }

         cot= ;

        dfs(fi,fj);

        printf("%d\n",cot);

    }

    return ;

}