poj-1979-Red and Black

            题目传送门：http://poj.org/problem?id=1979

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

#include <stdio.h>
#include<string.h>
char a[22][22];
int m,n,max,vis[21][21];
int next[4][2]={{0,1},{0,-1},{-1,0},{1,0}};
void dfs(int x,int y)
{
    int i,xi,xj;
    for(i=0;i<4;i++)
    {
        xi=x+next[i][0];
        xj=y+next[i][1];
        if(xi>0 && xj>0 && xi<=m && xj<=n)
        {
            if(a[xi][xj]=='.'&&vis[xi][xj]==0)
            {
                max++;
                vis[xi][xj]=1;
                dfs(xi,xj);
            }
        }
    }
    return ;
}
int main()
{
    while(scanf("%d%d%*c",&n,&m),m!=0&&n!=0)
    {
        int i,j,now_x,now_y;
        memset(vis,0,sizeof(vis));
        max=1;
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=n;j++)
            {
                scanf("%c",&a[i][j]);
                if(a[i][j]=='@')
                {
                    now_x=i,now_y=j;
                }
            }
            getchar();
        }
        vis[now_x][now_y]=1;
        dfs(now_x,now_y);
        printf("%d\n",max);
    }
    return 0;
}