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
几天没A题了，先拿道水题热热手。
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;
}
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