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

题意：求从字符‘@'开始，能够走过的点的个数，其中‘.'（点)表示空地，可以走，‘#‘表示墙，只能上下左右走

#include <cstdio>#include <cstring>#include <queue>using namespace std;int n,m,ans;char map1[21][21];int vis[21][21];int dx[]={-1,0,1,0};int dy[]={0,1,0,-1};struct A{    int x,y;}f,g;int bfs(int x,int y){    queue<A> Q;    f.x=x;    f.y=y;    ans=1;    vis[x][y]=1;    Q.push(f);    while(!Q.empty())    {        f=Q.front();        Q.pop();        for(int i=0;i<4;i++)        {            g.x=f.x+dx[i];            g.y=f.y+dy[i];            if(g.x>=0&&g.x<n&&g.y>=0&&g.y<m&&map1[g.x][g.y]!='#'&&!vis[g.x][g.y])            {                vis[g.x][g.y]=1;                ans++;                Q.push(g);            }        }    }    return ans;}int main(){    int sx,sy;    while(~scanf("%d %d",&m,&n),n||m)    {        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            getchar();            for(int j=0;j<m;j++)            {                scanf("%c",&map1[i][j]);                if(map1[i][j]=='@')                {                    sx=i;                    sy=j;                }            }        }        ans=bfs(sx,sy);        printf("%d\n",ans);    }    return 0;}