red and black

red and black
Time Limit:1000MS Memory Limit:65536K
Total Submit:22 Accepted:18
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)
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

Source

#include<iostream>
using namespace std;
#define M 25
int n,m;
char map[M][M];
bool flag[M][M];
int startx,starty;
int num;
int dir[4][2]={0,1,0,-1,1,0,-1,0};
int init()
{
	num=1;//including itself
	memset(map,' ',sizeof(map));
	for(int i=0;i<n;i++)
	{
		getchar();
		for(int j=0;j<m;j++)
		{
			scanf("%c",&map[i][j]);
			if(map[i][j]=='@')
			{
				startx=i,starty=j;
			}
		}
	}
	return 1;
}

bool judge(int x,int y)
{
	if(x<n&&y<m&&x>=0&&y>=0&&map[x][y]=='.')
		return true;
	return false;
}

int dfs(int x,int y)//find num
{
	int i,j;
	int tx=x, ty=y;
	map[tx][ty]='#';
	for(i=0;i<4;i++)
	{
		if(judge(tx+dir[i][0],ty+dir[i][1]))
		{
			num++;
			dfs(tx+dir[i][0],ty+dir[i][1]);
		}
	}
	return num;
}
int main()
{
	while(cin>>m>>n&&m&&n)
	{
		init();
		cout<<dfs(startx,starty)<<endl;
	}
}