Red and Black

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

百度翻译：有一个长方形的房间，上面铺着方砖。每个瓷砖都是红色或黑色的。一个男人站在一块黑瓷砖上。从一个瓷砖，他可以移动到四个相邻的瓷砖之一。但他不能在红瓷砖上移动，只能在黑瓷砖上移动。

写一个程序来计算黑瓷砖的数量，他可以通过重复上面描述的动作来达到。

思路：先找到人的坐标，如何搜索。注意输入时先输入的是列数，再输入的是行数。

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#define mod 1000000007
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

int m,n;
int ans;//计数器
int fx[4]={1,-1,0,0},fy[4]={0,0,-1,1};
string cz[25];

void dfs(int i,int j)
{
    ans++;
    cz[i][j]='#';
    for(int k=0;k<4;k++)
    {
        int x=i+fx[k];
        int y=j+fy[k];
        if(x>=0&&x<n&&y>=0&&y<m&&cz[x][y]=='.')
        {
            dfs(x,y);
        }
    }
}

int main()
{
    while(cin>>m>>n,m!=0,n!=0)
    {
        int x,y;
        for(int i=0;i<n;i++)
        {
            cin>>cz[i];
            for(int j=0;j<m;j++)
            {
                if(cz[i][j]=='@')//找人
                {
                    x=i;
                    y=j;
                }
            }
        }
        ans=0;
        dfs(x,y);
        printf("%d
",ans);
    }
}