4.2.5 Red and Black

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

思路：数据范围很小啊，dfs即可，注意数组的清零

#include <cstdio>
#include <cstring>
using namespace std;

bool vis[22][22];
char map[22][22];
int n,m,ax,ay,ans;
const int dx[]={0,0,1,-1};
const int dy[]={1,-1,0,0};

void dfs(int x,int y)
{
for (int k=0;k<4;k++)
{
	int xx=x+dx[k];
	int yy=y+dy[k];
	if (map[xx][yy]=='.' && vis[xx][yy]==false 
		&& xx>=0 && yy>=0 && xx<n && y<m
		)
	{
		ans++;
		vis[xx][yy]=true;
		dfs(xx,yy);
	}
}
}

void init ()
{
  while(scanf("%d%d",&m,&n)!=EOF)
  {
	  ans=0;
	  if (n==0 && m==0) break;
	  memset(vis,false,sizeof(vis));
	  memset(map,'\0',sizeof(map));
	//  printf("TEST!!!\n");
	  for (int i=0;i<n;i++)
	  {
		  scanf("%s",map[i]);
		  for (int j=0;j<m;j++)
			  if (map[i][j]=='@')
			  {
				  ax=i;
				  ay=j;
			  }
	  }
//	  printf("ax:%d ay:%d\n",ax,ay);
	  dfs(ax,ay);
//	  printf("ans::%d\n",ans+1);
	  printf("%d\n",ans+1);
	}
}

int main ()
{
init();
return 0;
}