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.
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
Source
题目类型：简单DFS/BFS
算法分析：统计调用DFS或者是BFS时走过的节点即可(也就是扩展的节点数)
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#include <iostream> #include <fstream> #include <cstring> using namespace std; const int maxn = 36 + 6; const int dx[] = {0, 0, -1, 1}; const int dy[] = {-1, 1, 0, 0}; char ans[maxn][maxn]; bool visited[maxn][maxn]; int row, col, s_x, s_y, sum; inline bool OkMove (int x, int y) { if (x >= 0 && x <= row - 1 && y >= 0 && y <= col - 1 && !visited[x][y] && ans[x][y] == '.') return true; return false; } void DFS (int x, int y) { int i; for (i = 0; i < 4; i++) { if (OkMove (x + dx[i], y + dy[i])) { int val_x = x + dx[i], val_y = y + dy[i]; visited[val_x][val_y] = true; sum++; DFS (val_x, val_y); } } } int main() { // ifstream cin ("aaa.txt"); while (cin >> col >> row) { if (row == 0 && col == 0) break; int i; for (i = 0; i < row; i++) { int j; for (j = 0; j < col; j++) { cin >> ans[i][j]; if (ans[i][j] == '@') { s_x = i; s_y = j; } } } sum = 1; memset (visited, false, sizeof (visited)); visited[s_x][s_y] = true; DFS (s_x, s_y); cout << sum << endl; } return 0; } |
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