HDU-1312题解

题目描述

据说翻译是个好东西（这句与题目无关?）

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

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

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

Sample Output

1 2 3 4

45 59 6 13

解题思路

一道深搜水题 嗯我就是来水题解的

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#include <iostream> #include <string> using namespace std; char map[22][22]; int n, m; int result = 1; int x, y; void dfs(int a, int b) { map[a][b] = '#'; result++; if (a - 1 > 0 && map[a - 1][b] == '.') { dfs(a - 1, b); } if (a + 1 <= n && map[a + 1][b] == '.') { dfs(a + 1, b); } if (b - 1 > 0 && map[a][b - 1] == '.') { dfs(a, b - 1); } if (b + 1 <= m && map[a][b + 1] == '.') { dfs(a, b + 1); } } int main() { while (cin >> m >> n) { if (n == 0 && m == 0) break; for (int i = 1; i <= n; ++i) { for (int j = 1; j <= m; ++j) { char s; cin >> s; if (s == '@') { x = i; y = j; } map[i][j] = s; } } map[x][y] = '#'; if (x-1 > 0 && map[x - 1][y] == '.') { dfs(x - 1, y); } if (x + 1 <= n && map[x + 1][y] == '.') { dfs(x + 1, y); } if (y - 1 > 0 && map[x][y - 1] == '.') { dfs(x, y - 1); } if (y + 1 <= m && map[x][y + 1] == '.') { dfs(x, y + 1); } cout << result << endl; result = 1; } return 0; }