Red and Black（DFS深搜实现）

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

 

 

 

 

深搜实现方案：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
char map[110][110];
int  vis[110][100];
int  dir[8][2]={{0,1},{0,-1},{1,0},{-1,0}};
int n,m,num;
void DFS(int x,int y)
{
    int a,b,i;
    vis[x][y]=1;
    num++;
    for(i=0;i<8;i++)
    {
        a=x+dir[i][0];
        b=y+dir[i][1];
        if(a>=0&&a<m&&b>=0&&b<n&&vis[a][b]==0&&map[a][b]=='.')
        {
            DFS(a,b);
        }
    }
}
int main()
{
    int i,j,x,y;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        getchar();
        if(m==0&&n==0)
            break;
        memset(map,0,sizeof(map));
        memset(vis,0,sizeof(vis));
        for(i=0; i<m; i++)
        {
            scanf("%s",map[i]);
        }
        for(i=0; i<m; i++)
        {
            for(j=0; j<n; j++)
            {
                if(map[i][j]=='@')///只有一个人
                {
                   x=i;
                   y=j;
                }
            }
        }
        num=0;
        DFS(x,y);
        printf("%d
",num);
    }
    return 0;
}