Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 41222		Accepted: 22338

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

思路：水题，bfs即可。

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;int map[21][21];int w,h,sx,sy,ans;int dx[4]={
    1,-1,0,0};int dy[4]={
    0,0,1,-1};struct nond{ int x,y; };int main(){    while(scanf("%d%d",&w,&h)&&w!=0&&h!=0){        swap(w,h);        for(int i=1;i<=w;i++)            for(int j=1;j<=h;j++){                char x;cin>>x;                if(x=='.')    map[i][j]=0;                else if(x=='#')    map[i][j]=1;                else if(x=='@'){ sx=i;sy=j;map[i][j]=0; }            }        queue
           
            que;ans=1; nond tmp;tmp.x=sx;tmp.y=sy; que.push(tmp);map[sx][sy]=1; while(!que.empty()){ nond now=que.front(); que.pop(); for(int i=0;i<4;i++){ int cx=dx[i]+now.x; int cy=dy[i]+now.y; if(cx>=1&&cx<=w&&cy>=1&&cy<=h&&!map[cx][cy]){ map[cx][cy]=1; nond mid;mid.x=cx;mid.y=cy; que.push(mid);ans++; //cout<
            
             <<" "<
             
              <