题目地址： http://poj.org/problem?id=1979 或者 https://vjudge.net/problem/OpenJ_Bailian-2816
Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 46793 | Accepted: 25201 |
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
4559613
一开始我的代码没有在每次输入前将地板置零，导致在输入11 6这组数据时出错，因为在之前输入11 9 这组数据时已经将11 6 外面的地板置为了.或者#， 之后
再测试11 6 时，由于没有置空地板，所以11 9 时在11 6 外面的地板仍然保留了下来，导致11 6 这组测试数据的结果产生错误。
正确代码：
#includeusing 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;}