Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 34862		Accepted: 18856

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) 

The end of the input is indicated by a line consisting of two zeros. 

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

Source

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #define maxn 105using namespace std;int n,m,num,vis[maxn][maxn];int dx[]={
    1,-1,0,0},dy[]={
    0,0,1,-1};char mapn[maxn][maxn];struct node{    int x,y;};void bfs(int a,int b){    queue
           
             q; node p; p.x = a; p.y = b; num++; vis[a][b] = 1; q.push(p); while(!q.empty())//加入队列的都是可以走到的点，所以只需ans++即可 { node now = q.front(); int x = now.x,y = now.y; q.pop(); for(int i=0;i<4;i++) { int xx = x + dx[i]; int yy = y + dy[i]; if(xx<0||yy<0||xx>(n-1)||yy>(m-1)||mapn[xx][yy]=='#'||vis[xx][yy]) continue; num++; node tmp; tmp.x = xx; tmp.y = yy; q.push(tmp); vis[xx][yy] = 1; } }}int main(){ while(cin >> m >> n)//注意这里输入的是m,n { memset(vis,0,sizeof(vis)); num = 0; if(n==0||m==0) break; for(int i=0;i
            
             > mapn[i][j]; for(int i=0;i