POJ1979深搜问题

Red and Black
Time Limit: 1000MS  Memory Limit: 30000K 
Total Submissions: 17584  Accepted: 9279 

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

Source

Japan 2004 Domestic



#include<iostream>
using namespace std;
#define maxvex 21
char c[maxvex][maxvex];
struct {int x,y;}d[4]={{1,0},{-1,0},{0,1},{0,-1}};
int ans,line,row,newx,newy;
void find(int i,int j)
{
    ans++;
    c[i][j]='#';
    for(int k=0;k<4;k++)
    {
        newx=i+d[k].x;
        newy=j+d[k].y;
        if(newx>=0&&newx<line&&newy>=0 && newy<row&&c[newx][newy]=='.')
            find(newx,newy);
    }
}
int main()
{
    int i,j,tag;
    while(scanf("%d%d",&row,&line)&&line&&row)
    {
        tag=1;
        ans=0;
        for(i=0;i<line;i++)
            cin>>c[i];
        for(i=0;i<line;i++)
        {
            for(j=0;j<row;j++)
                if(c[i][j]=='@')
                {
                    find(i,j);
                    break;
                    tag=0;
                }
                if(tag==0)
                    break;
        }
    printf("%d\n",ans);
    }
    return 0;
}