HDOJ Red and Black 1312(入门级DFS)

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10410    Accepted Submission(s): 6505

Problem 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
   
   
   
   

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

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#include<stdio.h>
#include<stdlib.h>
int zb[4][2]={-1,0,1,0,0,-1,0,1};
char s[25][25];
int w,h,step;//这三个值一定要用全局变量。 
int pd(int x, int y)
{
	int xx,yy;
	if(x>=w||x<0||y>=h||y<0)return 0;
	for(int i=0;i<4;i++)//检查该点上下左右的点是否符合题目要求。 
	{
		xx=x+zb[i][0];
		yy=y+zb[i][1];
		if(xx>=w||xx<0||yy>=h||yy<0||s[xx][yy]=='#')continue;//检查坐标为xx，yy的点。 
		step++;
		s[xx][yy]='#';//如果该点已经检查过，就把它变成'#',防止再次被检查。 
		pd(xx,yy);
	}
}
int main()
{
	int pd(int x,int y);
	while(scanf("%d%d",&w,&h),w!=0&&h!=0)
	{
	   getchar(); 
	   int x1,y1;
		for(int i=0;i<h;i++)
		  {
		  for(int j=0;j<w;j++)
	      {
	      	scanf("%c",&s[j][i]);
		    if(s[j][i]=='@')
		    {
		    	x1=j;
		    	y1=i;
			}
		  }
		  getchar();//记得吸收换行符。 
		  }
		s[x1][y1]='#';
		step=1;
		pd(x1,y1);
		printf("%d\n",step);
	}
	return 0;
}