poj 1979 dfs

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 35582		Accepted: 19262

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#[email protected]#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
[email protected]
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
struct node
{
    int x;
    int y;
    node(){}
    node(int xx,int yy):x(xx),y(yy){}
};
int n,m;
char g[20][20];
int visit[20][20];
int ans;
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};

void dfs(node start)
{
    visit[start.x][start.y]=1;
    ++ans;
    for(int i=0;i<4;i++)
    {
        node np(start.x+dir[i][0],start.y+dir[i][1]);
        if(np.x>=0&&np.x<n&&np.y>=0&&np.y<m&&!visit[np.x][np.y]&&g[np.x][np.y]!='#')
            dfs(np);
    }
}

int main()
{
    while(cin>>m>>n&&(m+n))
    {
        memset(visit,0,sizeof(visit));
        node start;
        int cur=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
            {
                cin>>g[i][j];
                if(g[i][j]=='@')
                {
                    start.x=i;start.y=j;
                }
            }
        ans=0;
        dfs(start);
        cout<<ans<<endl;
    }
    return 0;
}