poj1979 Red and Black

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 35903		Accepted: 19444

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) 
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

45
59
6

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int a,b,n,m;
char c[22][22];
int dfs(int a,int b)
{
    if(c[a][b]=='#'||a>=m||b>=n||a<0||b<0)
        return 0;
    else
    {
        c[a][b]='#';
        return 1+dfs(a,b+1)+dfs(a,b-1)+dfs(a-1,b)+dfs(a+1,b);
    }
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m)&&n&&m)
    {
        for(i=0;i<m;i++)
        {
            getchar();
            for(j=0;j<n;j++)
            {
                scanf("%c",&c[i][j]);
                if(c[i][j]=='@')
                {
                    a=i;b=j;
                }
            }
        }
            n=dfs(a,b);
            printf("%d\n",n);
    }
    return 0;
}