POJ 1979 Red and Black (DFS)

Red and Black

http://poj.org/problem?id=1979

Time Limit:  1000MS

Memory Limit: 30000K

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

45 59 6 13

Source

Japan 2004 Domestic

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完整代码：

/*16ms,388KB*/

#include<cstdio>
#include<cstring>

char grid[25][25];
bool vis[25][25];
int cnt, x, y;

void dfs(int i, int j)
{
	if (grid[i][j] == '#' || vis[i][j]) return;
	vis[i][j] = true;
	++cnt;
	dfs(i - 1, j); dfs(i + 1, j); dfs(i, j - 1); dfs(i, j + 1);
}

int main()
{
	int i, j, xx, yy;
	while (scanf("%d%d\n", &x, &y), x)
	{
		memset(grid, '#', sizeof(grid));
		for (i = 1; i <= y; ++i)
		{
			for (j = 1; j <= x; ++j)
			{
				grid[i][j] = getchar();
				if (grid[i][j] == '@')
					xx = i, yy = j;
			}
			getchar();
		}
		memset(vis, 0, sizeof(vis));
		cnt = 0;
		dfs(xx, yy);
		printf("%d\n", cnt);
	}
	return 0;
}