HDOJ 1312 Red and Black(単純dfs)

Red and Black
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12655    Accepted Submission(s): 7810
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

ac  ：

   
   
   
   
#include<stdio.h>
#include<string.h>
char map[25][25];
//int v[25][25];
int n,m,num;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int check(int xx,int yy)
{
	if(xx<0||xx>=n||yy<0||yy>=m||map[xx][yy]=='#')
	return 1;
	return 0;
}
void dfs(int x,int y)
{
	int i;
	map[x][y]='#';
	num++;
	for(i=0;i<4;i++)
	{
		int nx=x+dir[i][0];
		int ny=y+dir[i][1];
		if(check(nx,ny))
		continue;
		dfs(nx,ny);
	}
}
int main()
{
	int bx,by,i,j;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		//memset(v,0,sizeof(v));
		if(n==0&&m==0)
		break;
		for(i=0;i<n;i++)
		{
		    scanf("%s",map[i]);
		    for(j=0;j<m;j++)
		    {
		    	if(map[i][j]=='@')
		    	{
		    		bx=i;
		    		by=j;
				}
			}
		}
		num=0;
		dfs(bx,by);
		printf("%d
",num);
	}
	return 0;
}