Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7720    Accepted Submission(s): 4842

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

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

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#include<iostream>#include<cstdio>int d[4][2]={1,0,-1,0,0,-1,0,1};char a[30][30];int m,n,sx,sy;int count;int bfs(int x,int y){    int X;    int Y,i;    if(x<0||x>=n||y<0||y>=m)    return 0;    if(a[x][y]=='#')    return 0;    else    {    int sum=1;    a[x][y]='#';    for(int i=0; i<4; i++)    {        sum+=bfs(x+d[i][0],y+d[i][1]);    }    return sum;    }}int main(){    int i,j;    while(scanf("%d%d",&m,&n)!=EOF)    {        if((m==0)&&(n==0))        break;        count=1;        for(i=0;i<n;i++)        scanf("%s",&a[i]);        for(i=0;i<n;i++)        {            for(j=0;j<m;j++)            {                if(a[i][j]=='@')                {                    sx=i;                    sy=j;                }            }        }        printf("%d\n",bfs(sx,sy));    }    return 0;}