hdu-1312 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
   
简单的深搜，如果知道的话，这是很简单的，但是不知道的话就不知道从哪下手，可能还有其他的方法。。。。
代码：
#include<stdio.h>
#include<string.h>
int n,m,ans;
int temp[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
char map[30][30];
void dfs(int x,int y){
	ans++;
	map[x][y]='#';
	for(int k=0;k<4;k++){
		int a=x+temp[k][0];
		int b=y+temp[k][1];
		if(a<m&&b<n&&a>=0&&b>=0&&map[a][b]=='.')
		   dfs(a,b);
	}
	return ;
}
int main(){
	int fx,fy;
	int i,j,k,t;
	while(~scanf("%d%d%*c",&n,&m),n+m!=0){%*c的意思是吸收一个字符串，并将其舍弃，读取字符串的时候会将字符串读取，将换行字符串吸收后舍弃，是的后边的字符串能够读取
		ans=0;
		for(i=0;i<m;i++){
			for(j=0;j<n;j++){
				scanf("%c",&map[i][j]);
				if(map[i][j]=='@'){
					fx=i;
					fy=j;
				}
			}
			getchar();
		}
			dfs(fx,fy);
		printf("%d\n",ans);
	}
	return 0;
}