Red and Black

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

简单的dfs,注意检索位置要算上一个。

#include<iostream>#include <cstring>#include<string>#include<cmath>using namespace std;char map[22][22];int sum,v[22][22],m,n;void dfs(int a,int b){    sum++;    v[a][b]=1;    if ((map[a+1][b]=='.')&&(v[a+1][b]==0)&&((a+1)<=n)) dfs(a+1,b);    if ((map[a-1][b]=='.')&&(v[a-1][b]==0)&&((a-1)>=1)) dfs(a-1,b);    if ((map[a][b+1]=='.')&&(v[a][b+1]==0)&&((b+1)<=m)) dfs(a,b+1);    if ((map[a][b-1]=='.')&&(v[a][b-1]==0)&&((b-1)>=1)) dfs(a,b-1);}int main(){    int i,j,a,b;    char s[21];    while(cin>>m>>n,m+n>0)    {        sum=0;        memset(v,0,sizeof(v));        for (i=1;i<=n;i++)        {            cin>>s;            for (j=1;j<=m;j++)            {                map[i][j]=s[j-1];                if (s[j-1]=='@')                {                    a=i;                    b=j;                }            }        }        dfs(a,b);        cout<<sum<<endl;    }}