HPU:1312Red and Black(dfs)

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14566    Accepted Submission(s): 9023

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<cstdio>
#include<cstring>
char map[1010][1010];
int vis[1010][1010];
int m,n,sx,sy,ans;
void dfs(int x,int y)
{
	if(x<0||x>=m||y<0||y>=n||vis[x][y]||map[x][y]=='#')
	return ;
	vis[x][y]=1;
	ans++;
	dfs(x+1,y);
	dfs(x,y+1);
	dfs(x-1,y);
	dfs(x,y-1);
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(m==0&&n==0)
		break;
		
		memset(map,0,sizeof(map));
		memset(vis,0,sizeof(vis));
		ans=0;
		for(int i=0;i<m;i++)
		{
			scanf("%s",map[i]);
			for(int j=0;j<n;j++)
			{
				if(map[i][j]=='@')
				sx=i,sy=j;
		
			}
		}
		dfs(sx,sy);
		printf("%d
",ans);
	}
	return 0;
}