hdu1312 red and black 【递归】

Red and Black

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

Total Submission(s): 13080 Accepted Submission(s): 8109



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 【代码】[code]#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[22][22];
int n,m,cnt;
void dfs(int x,int y)
{
	if(x<1||x>m||y<1||y>n)
	   return ;
	if(a[x][y]=='#')
	   return ;
	cnt++;
	a[x][y]='#';
	dfs(x+1,y);
	dfs(x-1,y);
	dfs(x,y-1);
	dfs(x,y+1);
}
int main()
{
	int x,y,i,j;
	while(scanf("%d%d",&n,&m),m|n)
	{
		cnt=0;
		for(i=1;i<=m;i++)
		{
			getchar();
			for(j=1;j<=n;j++)
			{
				scanf("%c",&a[i][j]);
				if(a[i][j]=='@')
				  x=i,y=j;
			}
		}
		dfs(x,y);
		printf("%d\n",cnt);
	}
	return 0;
}

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