hdu Red and Black

算法：深搜

题意：就是让你找到一共可以移动多少次，每次只能移到黑色格子上，

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

代码：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <iomanip>
using namespace std;
char ch[25][25];
int k,n,m;
void dfs(int x,int y,int &k)
{
	ch[x][y]='#';
	if(x-1>=0&&x-1<m&&y>=0&&y<n&&ch[x-1][y]=='.') 
	{k++;dfs(x-1,y,k);}
	if(x+1<m&&x+1>=0&&y>=0&&y<n&&ch[x+1][y]=='.') 
	{k++;dfs(x+1,y,k);}
	if(y-1>=0&&y-1<n&&x>=0&&x<m&&ch[x][y-1]=='.') 
	{k++;dfs(x,y-1,k);}
	if(y+1<n&&y+1>=0&&x>=0&&x<m&&ch[x][y+1]=='.') 
	{k++;dfs(x,y+1,k);}
	else return ;
} 
int main()
{
	int i,j,q,p;
	while(cin>>n>>m&&n&&m)
	{   k=1;
		for(i=0;i<m;i++)
		{
			for(j=0;j<n;j++)
			{
				cin>>ch[i][j];
				if(ch[i][j]=='@') 
				{
					p=i;q=j;
				}
			}
			
		}
		dfs(p,q,k);
		cout<<k<<endl;
	}
	return 0;
}