DFS解决POJ 1979

 

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;
int used[25][25];
char map[25][25];
int d[4][2] = {{-1,0}, {1,0}, {0,-1}, {0,1}};//4个方向的走法下，上，左， 右
int n, m;
int count;//记录走过的总数
void dfs(int i, int j)
{
    count++;
    used[i][j]=1;
    int ii;
    for(ii=0; ii<4; ii++)
    {
        if(i+d[ii][0]>=0 && i+d[ii][0]<n && j+d[ii][1]>=0 && j+d[ii][1]<m && map[i+d[ii][0]][j+d[ii][1]]=='.' && !used[i+d[ii][0]][j+d[ii][1]])
        {
            dfs(i+d[ii][0],j+d[ii][1]);
        }
    }
}
int main()
{
    int start_i, start_j, i, j;
    while(cin>>m>>n && (n+m))
    {
        memset(used, 0, sizeof(used));
        for(i=0; i<n; i++)
        {
            for(j=0; j<m; j++)
            {
                cin>>map[i][j];
                if(map[i][j]=='@')
                {
                    start_i = i;
                    start_j = j;
                }
            }
        }
        count = 0;
        dfs(start_i, start_j);
        cout<<count<<endl;
    }
    return 0;
}