Red and Black

Red and Black

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

45
59
6
13

深搜

#include
#include
#include

using namespace std;

char a[25][25];
int flag[25][25];
int sum;

void dfs(int x,int y);

int main()
{

    int n,m,i,j;
    while(cin>>m>>n&&(m||n))
    {
        getchar();
        sum=0;
        memset(flag,0,sizeof(flag));
        memset(a,'#',sizeof(a));
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=m; j++)
            {
                cin>>a[i][j];
            }
            getchar();
        }
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=m; j++)
            {
                if(a[i][j]=='@')
                {
                    sum++;
                    flag[i][j]=1;
                    dfs(i,j);
                    i=n;
                    j=m;
                }
            }
        }
        cout<return 0;
}

void dfs(int x,int y)
{
    if(a[x-1][y]=='.' && flag[x-1][y]==0)
    {
        sum++;
        flag[x-1][y]=1;
        dfs(x-1,y);
    }
    if(a[x][y-1]=='.' && flag[x][y-1]==0)
    {
        sum++;
        flag[x][y-1]=1;
        dfs(x,y-1);
    }
    if(a[x][y+1]=='.' && flag[x][y+1]==0)
    {
        sum++;
        flag[x][y+1]=1;
        dfs(x,y+1);
    }
    if(a[x+1][y]=='.' && flag[x+1][y]==0)
    {
        sum++;
        flag[x+1][y]=1;
        dfs(x+1,y);
    }
}