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[][];
 int  vis[][];
 int  dir[][]={{,},{,-},{,},{-,}};
 int n,m,num;
 void DFS(int x,int y)
 {
     int a,b,i;
     vis[x][y]=;
     num++;
     for(i=;i<;i++)
     {
         a=x+dir[i][];
         b=y+dir[i][];
         if(a>=&&a<m&&b>=&&b<n&&vis[a][b]==&&map[a][b]=='.')
         {
             DFS(a,b);
         }
     }
 }
 int main()
 {
     int i,j,x,y;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         getchar();
         if(m==&&n==)
             break;
         memset(map,,sizeof(map));
         memset(vis,,sizeof(vis));
         for(i=; i<m; i++)
         {
             scanf("%s",map[i]);
         }
         for(i=; i<m; i++)
         {
             for(j=; j<n; j++)
             {
                 if(map[i][j]=='@')///只有一个人
                 {
                    x=i;
                    y=j;
                 }
             }
         }
         num=;
         DFS(x,y);
         printf("%d\n",num);
     }
     return ;
 }