codeforces 60B Serial Time!

Serial Time!

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

The Cereal Guy's friend Serial Guy likes to watch soap operas. An episode is about to start, and he hasn't washed his plate yet. But he decided to at least put in under the tap to
be filled with water. The plate can be represented by a parallelepiped k × n × m,
that is, it hask layers (the first layer is the upper one), each of which is a rectangle n × m with
empty squares ('.') and obstacles ('#'). The water can only be present in the empty squares. The tap is positioned above the square (x, y) of
the first layer, it is guaranteed that this square is empty. Every minute a cubical unit of water falls into the plate. Find out in how many minutes the Serial Guy should unglue himself from the soap opera and turn the water off for it not to overfill the
plate. That is, you should find the moment of time when the plate is absolutely full and is going to be overfilled in the next moment.
Note: the water fills all the area within reach (see sample 4). Water flows in each of the
6 directions, through faces of 1 × 1 × 1 cubes.

Input
The first line contains three numbers k, n, m (1 ≤ k, n, m ≤ 10)
which are the sizes of the plate. Then follow k rectangles consisting of nlines
each containing m characters '.' or '#', which represents the "layers" of the plate in the order from
the top to the bottom. The rectangles are separated by empty lines (see the samples). The last line contains x and y (1 ≤ x ≤ n, 1 ≤ y ≤ m)
which are the tap's coordinates. x is the number of the line and y is
the number of the column. Lines of each layer are numbered from left to right by the integers from 1 to n,
columns of each layer are numbered from top to bottom by the integers from 1 to m.

Output
The answer should contain a single number, showing in how many minutes the plate will be filled.

Sample test(s)

input

1 1 1

.

1 1

output

1

input

2 1 1.

#

1 1

output

1

input

2 2 2

.#
##

..
..

1 1

output

5

input

3 2 2

#.
##

#.
.#

..
..

1 2

output

7

input

3 3 3

.#.
###
##.

.##
###
##.

...
...
...

1 1

output

13

#include<cstdio>
int K,N,M,ans;
char a[12][12][12];
void dfs(int z,int x,int y)
{
if(z<0||z>=K||x<0||x>=N||y<0||y>=M||a[z][x][y]=='#') return;
ans++;
a[z][x][y]='#';
dfs(z,x,y+1);
dfs(z,x,y-1);
dfs(z,x+1,y);
dfs(z,x-1,y);
dfs(z+1,x,y);
dfs(z-1,x,y);
}

int main()
{
int i,j;
scanf("%d%d%d",&K,&N,&M);
for(i=0;i<K;i++)
for(j=0;j<N;j++)
scanf("%s",a[i][j]);
scanf("%d%d",&i,&j);
dfs(0,i-1,j-1);
printf("%d\n",ans);
return 0;
}