HDU - 1312 ： Red and Black

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.
InputThe 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)
OutputFor 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

@可以通过上下左右通到的.有几个。可以DFS也可以BFS，只需把图遍历即可。

#include<iostream>
#include<string.h>

using namespace std;

char a[22][22];
int s;

void DFS(int x1,int y1)
{
    if(a[x1][y1]=='.'||a[x1][y1]=='@')
    {
        a[x1][y1]='#';
        s++;
    DFS(x1+1,y1);
    DFS(x1-1,y1);
    DFS(x1,y1+1);
    DFS(x1,y1-1);
    return;
    }
}

int main()
{
    int x,y;
    while(scanf("%d %d",&y,&x),x||y)
    {
        int x1,y1;
        s=0;
        memset(a,0,sizeof(a));
        for(int i=1;i<=x;i++)
        {
            getchar();
            for(int j=1;j<=y;j++)
            {
                scanf("%c",&a[i][j]);
                //cout<<i<<" "<<j<<endl;
                if(a[i][j]=='@')
                {
                    x1=i;
                    y1=j;
                }
            }
        }
        DFS(x1,y1);
        cout<<s<<endl;
    }
    
    return 0;
}