Red and Black（水）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12138    Accepted Submission(s): 7554

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

 

Sample Output

45

59

6

13

 1 #include <iostream>

 2 #include <cstdio>

 3 using namespace std;

 4 char a[21][21];

 5 int m,n;

 6 int count=0;

 7 void dfs(int x,int y)

 8 {

 9     if(a[x][y]=='.'&&x<n&&x>=0&&y<m&&y>=0)

10     {

11         count++;

12         a[x][y]='#';

13         dfs(x+1,y);

14         dfs(x,y+1);

15         dfs(x,y-1);

16         dfs(x-1,y);

17     }

18 }

19 int main()

20 {

21     freopen("in.txt","r",stdin);

22     while(scanf("%d%d",&m,&n)&&m!=0&&n!=0)

23     {

24         count=0;

25         int i,j,x,y;

26         for(i=0;i<n;i++)

27         {

28             for(j=0;j<m;j++)

29             {

30                 cin>>a[i][j];

31                 if(a[i][j]=='@')

32                 {

33                     x=i;y=j;

34                 }

35             }

36         }

37         a[x][y]='.';

38         dfs(x,y);

39         cout<<count<<endl;

40     }

41 }