DFS解决POJ 1979

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

#include<iostream>
using namespace std;
int used[25][25];
char map[25][25];
int d[4][2] = {{-1,0}, {1,0}, {0,-1}, {0,1}};//4个方向的走法下，上，左， 右
int n, m;
int count;//记录走过的总数
void dfs(int i, int j)
{
	count++;
	used[i][j]=1;
	int ii;
	for(ii=0; ii<4; ii++)
	{
		if(i+d[ii][0]>=0 && i+d[ii][0]<n && j+d[ii][1]>=0 && j+d[ii][1]<m && map[i+d[ii][0]][j+d[ii][1]]=='.' && !used[i+d[ii][0]][j+d[ii][1]])
		{
			dfs(i+d[ii][0],j+d[ii][1]);
		}
	}
}
int main()
{
	int start_i, start_j, i, j;
	while(cin>>m>>n && (n+m))
	{
		memset(used, 0, sizeof(used));
		for(i=0; i<n; i++)
		{
			for(j=0; j<m; j++)
			{
				cin>>map[i][j];
				if(map[i][j]=='@')
				{
					start_i = i;
					start_j = j;
				}
			}
		}
		count = 0;
		dfs(start_i, start_j);
		cout<<count<<endl;
	}
	return 0;
}