poj-1979-Red and Black

            题目传送门：http://poj.org/problem?id=1979

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

#include <stdio.h>#include<string.h>char a[22][22];int m,n,max,vis[21][21];int next[4][2]={{0,1},{0,-1},{-1,0},{1,0}};void dfs(int x,int y){    int i,xi,xj;    for(i=0;i<4;i++)    {        xi=x+next[i][0];        xj=y+next[i][1];        if(xi>0 && xj>0 && xi<=m && xj<=n)        {            if(a[xi][xj]=='.'&&vis[xi][xj]==0)            {                max++;                vis[xi][xj]=1;                dfs(xi,xj);            }        }    }    return ;}int main(){    while(scanf("%d%d%*c",&n,&m),m!=0&&n!=0)    {        int i,j,now_x,now_y;        memset(vis,0,sizeof(vis));        max=1;        for(i=1;i<=m;i++)        {            for(j=1;j<=n;j++)            {                scanf("%c",&a[i][j]);                if(a[i][j]=='@')                {                    now_x=i,now_y=j;                }            }            getchar();        }        vis[now_x][now_y]=1;        dfs(now_x,now_y);        printf("%d\n",max);    }    return 0;}