POJ 1979 Red and Black （DFS）

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

 1 #include<cstdio>
 2 #include<string.h>
 3 int px[4]={-1,1,0,0};
 4 int py[4]={0,0,-1,1};
 5 int ans;
 6 bool key[300][300];
 7 char a[300][300];
 8 int n,m,x,y;
 9 void f(int x,int y)
10 {
11     int nx,ny;
12     for(int i = 0 ;i < 4 ; i++)
13     {
14         nx=x+px[i];
15         ny=y+py[i];
16         if(nx >= 0 && ny >= 0 && a[nx][ny] == '.' && key[nx][ny] == false && nx < m && ny < n)
17         {
18             ans++;
19             key[nx][ny]=true;
20             f(nx,ny);
21         }
22     }
23 }
24 int main()
25 {
26 
27     while(scanf("%d %d",&n,&m) && n && m)
28     {
29         memset(key,false,sizeof(key));
30         for(int i = 0 ; i < m ; i++)
31         {
32             getchar();
33             for(int j = 0 ; j < n ; j++)
34             {
35                 scanf("%c",&a[i][j]);
36                 if(a[i][j] == '@')
37                 {
38                     x=i;
39                     y=j;
40                 }
41             }
42             
43         }
44         key[x][y]=true;
45         ans=0;
46         f(x,y);
47         printf("%d
",ans+1);
48     }
49 }

——将来的你会感谢现在努力的自己。