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
4559
6
13
深搜實作方案：
1 #include<stdio.h>
2 #include<string.h>
3 #include<algorithm>
4 #include<iostream>
5 using namespace std;
6 char map[110][110];
7 int vis[110][100];
8 int dir[8][2]={{0,1},{0,-1},{1,0},{-1,0}};
9 int n,m,num;
10 void DFS(int x,int y)
11 {
12 int a,b,i;
13 vis[x][y]=1;
14 num++;
15 for(i=0;i<8;i++)
16 {
17 a=x+dir[i][0];
18 b=y+dir[i][1];
19 if(a>=0&&a<m&&b>=0&&b<n&&vis[a][b]==0&&map[a][b]=='.')
20 {
21 DFS(a,b);
22 }
23 }
24 }
25 int main()
26 {
27 int i,j,x,y;
28 while(scanf("%d%d",&n,&m)!=EOF)
29 {
30 getchar();
31 if(m==0&&n==0)
32 break;
33 memset(map,0,sizeof(map));
34 memset(vis,0,sizeof(vis));
35 for(i=0; i<m; i++)
36 {
37 scanf("%s",map[i]);
38 }
39 for(i=0; i<m; i++)
40 {
41 for(j=0; j<n; j++)
42 {
43 if(map[i][j]=='@')///隻有一個人
44 {
45 x=i;
46 y=j;
47 }
48 }
49 }
50 num=0;
51 DFS(x,y);
52 printf("%d\n",num);
53 }
54 return 0;
55 }