Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 46793		Accepted: 25201

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.

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.

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).

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

45
59
6
13

#include <iostream>using namespace std;char Floor[30][30];    //地板
int visited[30][30];    //访问标记,0表示未访问，1表示已访问
int num = 0;   //瓷砖数void dfs(int i, int j)
{visited[i][j] = 1;    //标记为已访问++num;if (Floor[i - 1][j] == '.' && !visited[i - 1][j])dfs(i - 1, j);    //往上走if (Floor[i][j - 1] == '.' && !visited[i][j - 1])dfs(i, j - 1);    //往左走if (Floor[i][j + 1] == '.' && !visited[i][j + 1])dfs(i, j + 1);    //往右走if (Floor[i + 1][j] == '.' && !visited[i + 1][j])dfs(i + 1, j);    //往下走

}int main()
{int W, H;while (cin >> W >> H && (W != 0 || H != 0))    //W是列数，H是行数
    {num = 0;    //将访问的黑瓷砖数置零for (int i = 0; i < 30; ++i)for (int j = 0; j < 30; ++j)Floor[i][j] = '#';        //将地板置零for (int i = 0; i < 30; ++i)for (int j = 0; j < 30; ++j)visited[i][j] = 0;        //将地板的访问状态置零int start_i, start_j;    //起点坐标//创建地板,二维数组的第1行和第1列不用，并且地板初始为30×30，足够大，//从而避免初始点落在边界上调用dfs时产生的数组越界问题for (int i = 1; i <= H; ++i)for (int j = 1; j <= W; ++j){cin >> Floor[i][j];if (Floor[i][j] == '@')    //记录下起点坐标
                {start_i = i;start_j = j;}}dfs(start_i, start_j);cout << num << endl;}return 0;}