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

Source

Asia 2004, Ehime (Japan), Japan Domestic

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//参照走“日”算法（马的走法）#include<cstdio>#include<string.h>#include<iostream>using namespace std;int mov[4][2]={0,1,0,-1,1,0,-1,0};/*0  10 -11  0-1 0*/int cnt,sx,sy,vis[21][21],c[21][21],w,h;void dfs(int x,int y){    int i,nx,ny;    for(i=0;i<4;i++)   {     nx=x+mov[i][0],ny=y+mov[i][1];    if(nx>=0&&nx<h&&ny>=0&&ny<w&&c[nx][ny]!=-1&&!vis[nx][ny])//未走的，标志为0，走过的变为1    {   cnt++;//计数        vis[nx][ny]=1;        dfs(nx,ny);    }   }}int main(){    int i,j;    char s[21][21];    while(scanf("%d%d",&w,&h)!=EOF&&w&&h)//w为列，h为行    {          for(i=0;i<h;i++)        {            for(j=0;j<w;j++)            cin>>s[i][j];        }        for(i=0;i<h;i++)        {               for(j=0;j<w;j++)            if(s[i][j]=='@'){c[i][j]=1;sx=i;sy=j;}//当前位置，把字符转换成数字，用C数组存放            else if(s[i][j]=='.')c[i][j]=0;//可以走的            else if(s[i][j]=='#')c[i][j]=-1;        }      memset(vis,0,sizeof(vis));            cnt=1;            vis[sx][sy]=1;            dfs(sx,sy);            printf("%d\n",cnt);    }    return 0;}