DFS深搜poj1979

深搜，从一点向各处搜找到所有能走的地方。

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12004		Accepted: 6097

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
13

#include<iostream>
using namespace std;
char map[22][22];
int sum,l,h;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
bool border(int x,int y)
{
    if(x<0||x>=h||y<0||y>=l) return 0;
    return 1;
}
void search(int x,int y)
{
    int i;
    int xx,yy;
    sum++;
    map[x][y]='#';
    for(i=0;i<4;i++)
    {
        xx=x+dir[i][0];
        yy=y+dir[i][1];
        if(border(xx,yy)&&map[xx][yy]=='.')
        search(xx,yy);
    }
}
int main()
{
    int i,j;
    int x0,y0;
    while(cin>>l>>h)
    {
        sum=0;
        if(l==0&&h==0)break;
        for(i=0;i<h;i++)
        {
            for(j=0;j<l;j++)
            {
                cin>>map[i][j];
                if(map[i][j]=='@')
                {
                    x0=i;
                    y0=j;
                }
            }
        }
        search(x0,y0);
        cout<<sum<<endl;
    }
    return 0;
}