POJ 1979 深度优先搜索

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Language:
Red and Black
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****

  #include<stdio.h>
    #include<string.h>
    const int MAXN=21;
    int step=0;
    int row,line;
    char picture[MAXN][MAXN];
    int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
    //操作指南0：向下 1：右 2：上 3：左 
    void dfs(int x,int y);
    int main()
    {int x,y;
        scanf("%d%d",&line,&row);
        while(row!=0&&line!=0) 
        {memset(picture,0,MAXN*MAXN*sizeof(char));
         getchar();
    	for(int i=0;i<row;i++)
                {for(int j=0;j<line;j++)
                 {scanf("%c",&picture[i][j]);
                   //get the position of @
    			   if(picture[i][j]=='@') {x=j;y=i;}
    				 }
                 	getchar();
    	               }
    	        step=0;
    	             dfs(x,y);
    				 printf("%d\n",step+1);  
    	    scanf("%d%d",&line,&row);}
    return 0;
    }
    
    void dfs(int x,int y){
         picture[y][x]='#';
    	if(y+1<row&&picture[y+dy[0]][x+dx[0]]!='#')  
    	{	step++;
    	     dfs(x+dx[0],y+dy[0]);
    	}
    	 
    	if(x+1<line&&picture[y+dy[1]][x+dx[1]]!='#')
    	{step++;
    	     dfs(x+dx[1],y+dy[1]);
    	}
    	    
    	if(y-1>=0&&picture[y+dy[2]][x+dx[2]]!='#')
    	{
    	step++;
    	 dfs(x+dx[2],y+dy[2]);
    	}
        
    	if(x-1>=0&&picture[y+dy[3]][x+dx[3]]!='#')
    	{
    	step++;
    	 dfs(x+dx[3],y+dy[3]);}
    	}//递归可以回退