Red and Black
Time Limit: 1000MS |
|
Memory Limit: 30000K |
Total Submissions: 22633 |
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Accepted: 12217 |
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
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1005;
char map[maxn][maxn];
int n, m;
int sum;
int dx[] = { -1,1,0,0 };
int dy[] = { 0,0,1,-1 };
int dfs(int x, int y)
{
map[x][y] = '#';
sum++;
for (int i = 0; i < 4; i++)
{
int fx = x + dx[i];
int fy = y + dy[i];
if (fx >= 0 && fx < m&&fy >= 0 && fy < n&&map[fx][fy] == '.')
{
dfs(fx, fy);
}
}
return sum;
}
int main()
{
cin >> n >> m;
int pi, pj;
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
cin >> map[i][j];
if (map[i][j] == '@')
{
pi = i;
pj = j;
}
}
}
dfs(pi, pj);
printf("%d\n", sum);
}