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题目链接:
Time Limit: 2 Seconds Memory Limit: 65536 KBProblem 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)
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
Problem solving report:
Description: 一个男人站在黑色的瓷砖上。从瓷砖中，他可以移动到四个相邻瓷砖中的一个。但他不能在红瓦上移动，他只能在黑色瓷砖上移动。求他可以达到的黑色瓷砖的数量。
Problem solving: 简单的DFS搜索。。。#include#define MAXN 20int n, m, step;char map[MAXN][MAXN];int dir[4][2] = { {0, -1}, {1, 0}, {0, 1}, {-1 ,0}};int DFS(int x, int y) { step++; map[x][y] = '#'; for (int k = 0; k < 4; k++) { int tx = x + dir[k][1]; int ty = y + dir[k][0]; if (tx >= 0 && tx < n && ty >= 0 && ty < m && map[tx][ty] == '.') DFS(tx, ty); } return step;}int main() { int ex, ey; while (scanf("%d%d", &m, &n), n + m) { step = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { scanf(" %c", &map[i][j]); if (map[i][j] == '@') ex = i, ey = j; } } printf("%d\n", DFS(ex, ey)); } return 0;}
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