A - Red and Black(3.2.1)（小递归）

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
 
思路：使用递归，对@所在的位置四个方向查找，一直到不满足条件h>=n||l>=m||h<0||l<0||str[h][l]=='#'||boo[h][l]==1为止。。。
 
#include <iostream>
#include <string>
#include<algorithm>    //头文件
using namespace std;
bool boo[25][25];
string str[21];
int jishu=0;
int m,n,h=0,l=0;//h为行，l为列
void search(int h,int l)
{
 if(h>=n||l>=m||h<0||l<0||str[h][l]=='#'||boo[h][l]==1)  //此处如果没有||boo[h][l]==1就会死循环，但是却不显示一直进行，而是程序莫名其妙的结束，以后注意！！！！！！！！！！！！！！
  return ;
 boo[h][l]=1;
 jishu++;
 
search(h,l-1);
 search(h,l+1);
 search(h-1,l);
 search(h+1,l);
}
int main()
{
 int i,j;
 cin>>m>>n;//m是一个字符串的长度 n是字符串数目
 while(m!=0&&n!=0)
 { jishu=0;
       h=0,l=0;
  for(i=0;i<n;i++)
  {
   cin>>str[i];
   for(j=0;j<m;j++)
    if(str[i][j]=='@')
    { 
     h=i;
     l=j;
    }
  }
  memset(boo,0,25*25);
  search(h,l);
  cout<<jishu<<endl;
  cin>>m>>n;
 }
 return 0;
}