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：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 #include<iostream>
 5 using namespace std;
 6 char map[110][110];
 7 int  vis[110][100];
 8 int  dir[8][2]={{0,1},{0,-1},{1,0},{-1,0}};
 9 int n,m,num;
10 void DFS(int x,int y)
11 {
12     int a,b,i;
13     vis[x][y]=1;
14     num++;
15     for(i=0;i<8;i++)
16     {
17         a=x+dir[i][0];
18         b=y+dir[i][1];
19         if(a>=0&&a<m&&b>=0&&b<n&&vis[a][b]==0&&map[a][b]=='.')
20         {
21             DFS(a,b);
22         }
23     }
24 }
25 int main()
26 {
27     int i,j,x,y;
28     while(scanf("%d%d",&n,&m)!=EOF)
29     {
30         getchar();
31         if(m==0&&n==0)
32             break;
33         memset(map,0,sizeof(map));
34         memset(vis,0,sizeof(vis));
35         for(i=0; i<m; i++)
36         {
37             scanf("%s",map[i]);
38         }
39         for(i=0; i<m; i++)
40         {
41             for(j=0; j<n; j++)
42             {
43                 if(map[i][j]=='@')///只有一个人
44                 {
45                    x=i;
46                    y=j;
47                 }
48             }
49         }
50         num=0;
51         DFS(x,y);
52         printf("%d\n",num);
53     }
54     return 0;
55 }