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<iostream>#include<cstdio>using namespace std;char a[25][25];int sum,b[25][25],dx[4]={0,0,1,-1},dy[4]={1,-1,0,0},s[410][3],w,h;void search(int ,int );int main(){//freopen("a.txt","r",stdin);int m,n;while(cin>>w>>h&&w&&h){sum=1;for(int j=1;j<=h;j++)  for(int i=1;i<=w;i++)  {cin>>a[j][i];  if(a[j][i]=='@') {m=j;n=i;}  //找到起始位置b[j][i]=0;　　　　　　　　　 //标记每一位置  }search(m,n);printf("%d\n",sum);}return 0;}void search(int m,int n){int head=0,tail=1;b[m][n]=1; s[1][1]=m; s[1][2]=n;while(head<tail){head++;m=s[head][1]; n=s[head][2];for(int i=0;i<4;i++){m+=dy[i]; n+=dx[i];  if(a[m][n]=='.'&&!b[m][n]&&m>0&&n>0&&m<=h&&n<=w)　　//判断下一步是否满足条件{sum++;tail++;s[tail][1]=m;s[tail][2]=n;　　//保存结果b[m][n]=1;}m-=dy[i]; n-=dx[i];　　//还原}}}