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
解题思路：
如果说，你写搜索的题写的挺多了话，那么这道题可能你看到样例，就能猜到这是一道要用DFS的题，题目的大致意思就是说给你一个图，‘@’代表这个人初始的位置；‘#’代表红色瓷砖，表示不能走；‘.’代表黑色瓷砖，可以走。最后要计算的是遍历走完整幅图，一共可以经过多少块黑色瓷砖。
程序代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int w,h,ans;
char map[25][25];
int num1[4]={-1,1,0,0};
int num2[4]={0,0,-1,1};
void dfs(int x,int y)
{
map[x][y]='#';
ans++;
for(int i=0;i<4;i++)
{
int tx=x+num1[i];
int ty=y+num2[i];
if(tx>=0&&tx<h&&ty>=0&&ty<w&&map[tx][ty]=='.')
dfs(tx,ty);
}
return ;
}
int main()
{
while(cin>>w>>h)
{
if(w==0&&h==0)
break;
ans=0;
int x,y;
for(int i=0;i<h;i++)
for(int j=0;j<w;j++)
cin>>map[i][j];
for(int i=0;i<h;i++)
{
for(int j=0;j<w;j++)
{
if(map[i][j]=='@')
{
x=i;
y=j;
}
}
}
dfs(x,y);
cout<<ans<<endl;
}
return 0;
}