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
题意：
房顶有红色黑色瓦片，@为起始位置站在黑瓦片上，‘.’代表黑瓦片‘#’代表红瓦片，只能上下左右四个方向移动只能走黑色瓦片，计算能到达的黑色瓦片的个数。
思路：
还是把走过的点标记为‘#’，利用深度优先搜索计算，和计算油田的个数类似。
一次就过了，开心啦啦啦。
代码：
#include<stdio.h>
char s[25][25];
int m,n,sum;
void dfs(int x,int y)
{
int nx,ny,k;
int next[4][2]={0,1, 1,0, 0,-1, -1,0};
for(k=0;k<4;k++)
{
nx=x+next[k][0];
ny=y+next[k][1];
if(nx<0||nx>=n||ny<0||ny>=m)
continue;
if(s[nx][ny]=='.')
{
s[nx][ny]='#';
sum++;
dfs(nx,ny);
}
}
}
int main()
{
int i,j;
while(~scanf("%d%d",&m,&n))
{
if(m==0&n==0)
break;
sum=0;
for(i=0;i<n;i++)
scanf("%s",s[i]);
for(i=0;i<n;i++)
for(j=0;j<m;j++)
{
if(s[i][j]=='@')
{
s[i][j]='#';
dfs(i,j);
sum++;
}
}
printf("%d\n",sum);
}
return 0;
}