Red and Black

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

6 9....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0

Sample Output

4559613

# include<cstdio># include<iostream>using namespace std;int w,h,x,y,cnt,vis[30][30],dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};char c[30][30];void dfs(int X,int Y){    for(int i=0;i<4;i++)    {        X+=dx[i];Y+=dy[i];              if(c[X][Y]=='.'&&0<X&&0<Y&&h>=X&&w>=Y&&!vis[X][Y])                {            cnt++;            vis[X][Y]=1;                                                  dfs(X,Y);                                              }        X-=dx[i];Y-=dy[i];                                  }}int main(){    while(cin>>w>>h&&w&&h)    {        int i,j;        for(i=1;i<=h;i++)            for(j=1;j<=w;j++)        {            cin>>c[i][j];                        vis[i][j]=0;                         if(c[i][j]=='@')             {                vis[i][j]=1;                         x=i;                                 y=j;            }        }        cnt=1;                              dfs(x,y);                            cout<<cnt<<endl;    }    return 0;}