Red and Black

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
 
 
 
#include<stdio.h>
#include<string.h>
using namespace std;
#define max 1000
char a[max][max];
int n,m;
int dg(int x,int y)
{
	if(x>m||x<1||y<1||y>n)
		return 0;
	if(a[x][y]=='#')
		return 0;
	a[x][y]='#';
	return 1+dg(x-1,y)+dg(x+1,y)+dg(x,y-1)+dg(x,y+1);
	
}
int main()
{
	int x,y,count=0;
	while(scanf("%d%d",&n,&m),m&&n)     //m后有个回车键。	
	{
		memset(a,0,sizeof(a));
		for(int i=1;i<=m;i++)
		{
			getchar();         //注意  吸收回车键
			for(int j=1;j<=n;j++)   //一行之后有个回车键；
			{
				scanf("%c",&a[i][j]);
				if(a[i][j]=='@')
				{
					x=i;
					y=j;
				}
				
			}
		}
		count=dg(x,y);
		printf("%d\n",count);
	}
	return 0;
}