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

思路：本题是简单的深搜bfs模板

MY  DaiMa：

#include<iostream>
#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
int n,m,t;
char flag[21][21];
struct point
{
    int x,y;
}start;
int bfs()
{
    queue<point>Q;
    point cur,next;
    flag[start.x][start.y]='#';
    Q.push(start);
    while(!Q.empty())
    {
        cur=Q.front();
        Q.pop();
        for(int i=0;i<4;i++)
        {
            next.x=cur.x+dir[i][0];
            next.y=cur.y+dir[i][1];
            if((next.x>=0&&next.x<n)&&(next.y>=0&&next.y<m)&&flag[next.x][next.y]=='.')
            {
                t++;
                flag[next.x][next.y]='#';
                Q.push(next);
            }
        }
    }
    return t;
}
int main()
{
    while(cin>>m>>n&&(n!=0&&m!=0))
    {
        t=1;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                cin >> flag[i][j];
                if(flag[i][j]=='@')
                {
                    start.x=i;
                    start.y=j;
                }
            }
        }
        //t = bfs();
        cout << bfs() <<endl;
    }
    return 0;
}