Red and Black

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 70   Accepted Submission(s) : 64

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<stdio.h>char s[22][22];int a,b,ans;void f(int x,int y){        if(x<0||x>b-1||y<0||y>a-1)    return ;    if(s[x][y]=='#')    return ;//    if(x>=0&&x<b||y>=0&&y<a)//    {        ans++;    //printf("%d\n",ans);    s[x][y]='#';//    }    f(x,y+1);    f(x,y-1);    f(x+1,y);    f(x-1,y);}int main(){     int i,j,k,x,y;    while(scanf("%d%d",&a,&b),a|b)    {        for(i=0;i<b;i++)        {            getchar();            for(j=0;j<a;j++)        {            scanf("%c",&s[i][j]);            if(s[i][j]=='@')            {                x=i,y=j;            }            }        }        //printf("%d%d",x,y);        ans=0;        f(x,y);        printf("%d\n",ans);            }    return 0;}