poj-1979-Red and Black【DFS】

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 30684		Accepted: 16701

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

DFS原理图：

有些小细节要注意：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};
int x,y,W,H,cnt;
bool vis[50][50];
char map[50][50];
void f(int a,int b)
{
	for(int i=0;i<4;i++)
	{
		int nx=a+dx[i];
		int ny=b+dy[i];
		if(map[nx][ny]=='.'&&!vis[nx][ny]&&nx>=0&&nx<H&&ny>=0&&ny<W)
		{
			cnt++;
			vis[nx][ny]=1;
			f(nx,ny);
		}
	}
}
int main()
{
	while(scanf("%d %d",&W,&H)&&(W||H))
	{
		memset(vis,0,sizeof(vis));
		for(int i=0;i<H;i++)   //注意是 H 行，W 列 
		{
			getchar();  // 输入数据每行结束都有一个回车，用 getchar() 吸收 
			for(int j=0;j<W;j++)
			{
				scanf("%c",&map[i][j]);
				if(map[i][j]=='@')
				{
					x=i;	y=j;
					vis[x][y]=1;
				}
			}
		}
		cnt=1;
		f(x,y);
		printf("%d\n",cnt);
	}	
	return 0;
}