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)


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

 题意描述：

在W*H的矩形房间里有红色和黑色两种砖块，一个男人站在一块黑色砖块上，男人只能移动上下左右相邻的黑色砖块上，求男人能到达的所有黑色砖块数。

'-'为黑色砖块，'#'红色砖块，'@'男人所站的黑色砖块。

解题思路：

直接深搜，模板题。

程序代码：

#include<stdio.h>
#include<string.h>
char map[30][30];
int num,n,m,book[30][30];
int xx[4][2]={0,-1,
			  -1,0,
			  0,1,
			  1,0};
void dfs(int x,int y)
{
	int k,i,j,tx,ty;
	for(k=0;k<4;k++)
	{
		tx=x+xx[k][0];
		ty=y+xx[k][1];
		if(tx>=0&&tx<n&&ty>=0&&ty<m&&book[tx][ty]==0)
		{
			book[tx][ty]=1;
			num++;
			dfs(tx,ty);
		}
	}
	return;
}
int main()
{
	int i,j,x,y;
	while(scanf("%d%d",&m,&n))
	{
		if(n==0&&m==0)
			break;
		num=0;
		for(i=0;i<n;i++)
			scanf("%s",map[i]);
		for(i=0;i<n;i++)
			for(j=0;j<m;j++)
			{
				if(map[i][j]=='#')
					book[i][j]=1;
				else
					book[i][j]=0;
				if(map[i][j]=='@')
				{
					x=i;
					y=j;
				}
			}
		num++;
		book[x][y]=1;
		dfs(x,y);
		printf("%d\n",num);
	}
	return 0;
 }