hdoj--1312--Red and Black（dfs）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 14311    Accepted Submission(s): 8870

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<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char map[500][500];
int n,m,ans,vis[500][500];
void dfs(int x,int y)
{
	if(x<0||x>=m||y<0||y>=n||map[x][y]=='#'||vis[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),n||m)
	{
		int ex,ey;
		memset(map,'\0',sizeof(map));
		for(int i=0;i<m;i++)
		{
			scanf("%s",&map[i]);
			for(int j=0;j<n;j++)
			{
				if(map[i][j]=='@')
				{
					ex=i;ey=j;map[i][j]='.';
				}
			}
		}
		ans=0;
		memset(vis,0,sizeof(vis));
		dfs(ex,ey);
		printf("%d\n",ans);
	}
	return 0;
}