hdu 1312 Red and Black

Red and Black

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 258   Accepted Submission(s) : 139

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

 

题意：一个人从@点出发，求他所能到达的'.'的数目，'#'不可走，@本身算1个点。

思路：搜索入门题。

#include<stdio.h>
#include<string.h>
char as[22][22];
int hash[22][22];
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
int w,h,ans;
void dfs(int x,int y)
{
	int i,xx,yy;
	hash[x][y]=1;
	for(i=0;i<4;i++)
	{
		xx=x+dx[i];
		yy=y+dy[i];
		if(xx>=0&&xx<h&&yy>=0&&yy<w&&as[xx][yy]=='.'&&!hash[xx][yy])
		{
			ans++;
			dfs(xx,yy);
		}
	}
}
int main()
{
	int i,j,x,y;
	while(scanf("%d%d",&w,&h)!=EOF)
	{
		if(w==0&&h==0)
			break;
		for(i=0;i<h;i++)
			scanf("%s",as[i]);
		for(i=0;i<h;i++)
		{
			for(j=0;j<w;j++)
			{
				hash[i][j]=0;
				if(as[i][j]=='@')
				{
					x=i;
					y=j;
				}
			}
		}
		ans=1;
		dfs(x,y);
		printf("%d\n",ans);
	}
	return 0;
}