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.
InputThe 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)
OutputFor 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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13 @可以通过上下左右通到的.有几个。可以DFS也可以BFS，只需把图遍历即可。
1 #include<iostream>
2 #include<string.h>
3
4 using namespace std;
5
6 char a[22][22];
7 int s;
8
9 void DFS(int x1,int y1)
10 {
11 if(a[x1][y1]==‘.‘||a[x1][y1]==‘@‘)
12 {
13 a[x1][y1]=‘#‘;
14 s++;
15 DFS(x1+1,y1);
16 DFS(x1-1,y1);
17 DFS(x1,y1+1);
18 DFS(x1,y1-1);
19 return;
20 }
21 }
22
23 int main()
24 {
25 int x,y;
26 while(scanf("%d %d",&y,&x),x||y)
27 {
28 int x1,y1;
29 s=0;
30 memset(a,0,sizeof(a));
31 for(int i=1;i<=x;i++)
32 {
33 getchar();
34 for(int j=1;j<=y;j++)
35 {
36 scanf("%c",&a[i][j]);
37 //cout<<i<<" "<<j<<endl;
38 if(a[i][j]==‘@‘)
39 {
40 x1=i;
41 y1=j;
42 }
43 }
44 }
45 DFS(x1,y1);
46 cout<<s<<endl;
47 }
48
49 return 0;
50 }