hdu 1312 Red and Black

Problem 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) 

 

 

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

 

 

#include <iostream>

using namespace std;
int x,y,k,t;
char data[25][25];

int f(int i,int j)
{
    if(i<1||i>y||j<1||j>x)//处理边界问题
    return 0;
    if(data[i][j]!='#')
    {
        data[i][j]='#';
        return 1+f(i-1,j)+f(i+1,j)+f(i,j-1)+f(i,j+1);
    }
    else//上/下/左/右/不能走了
    return 0;
}

int main()
{
    while(cin>>x>>y)
    {
        if(x==0&&y==0) break;
        k=-1;t=-1;
        for(int i=1;i<=y;i++)
        {
            for(int j=1;j<=x;j++)
           {
               cin>>data[i][j];
               if(data[i][j]=='@')
               {
                 k=i;t=j;
               }
           }
        }
        cout<<f(k,t)<<endl;
    }
    return 0;
}