搜索 B - 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

1.以后再遇到类似关于找方向的题目，就定义方向的二维数组，哪一个方向的具体位置没有规定，无先后顺序的要求，只要定义四个方向即可。

2.因为题目提要求我们，它自己也算一个，所以cnt的初始化一定是1，这一点读题的时候一定要注意细心。

#include<iostream>

#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e3;
char mapp[maxn][maxn];
int vis[maxn][maxn];
int ans;
int dir[4][2]= {{0,1},{0,-1},{-1,0},{1,0}};
int l,c;
int check(int x,int y)
{
    if(x<0||y<0||x>=c||y>=l||vis[x][y]||mapp[x][y]!='.')
        return 1;
    else
        return 0;
}
void dfs(int x,int y)
{
    vis[x][y] = 1;
    for(int i=0; i<4; i++)
    {
        int a = x+dir[i][0];
        int b = y+dir[i][1];
        if(!check(a,b))
        {
            ans++;
            dfs(a,b);
        }
    }
}
int main()
{
    while(cin>>l>>c&&(c||l))
    {
        int a,b;
        ans = 1;
        memset(vis,0,sizeof(vis));
        for(int i=0; i<c; i++)
            for(int j=0; j<l; j++)
            {
                cin>>mapp[i][j];
                if(mapp[i][j]=='@')
                {
                    a = i;
                    b = j;
                }
            }
        dfs(a,b);
        cout<<ans<<endl;
    }
    return 0;
}