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【HDU】1312--Red and Black(DFS)Red and Black

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Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 21053    Accepted Submission(s): 12830

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  
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int w,h;
int v[25][25];
char str[25][25];
int ans=0;
int fx[4]={0,0,1,-1};
int fy[4]={1,-1,0,0};
void dfs(int x,int y)
{
	v[x][y]=1;
	for(int i=0;i<4;i++)
	{
		int xx=x+fx[i],yy=y+fy[i];
		if(xx>=0&&xx<h&&yy>=0&&yy<w&&!v[xx][yy]&&str[xx][yy]=='.')
		{
			ans++;dfs(xx,yy);
		}
	}
}
int main()
{
	while(~scanf("%d%d%*c",&w,&h),w,h)
	{
		ans=0;
		for(int i=0;i<h;i++)
			scanf("%s",str[i]);
		memset(v,0,sizeof(v));
		for(int i=0;i<h;i++)
			for(int j=0;j<w;j++)
			{
				if(!v[i][j]&&str[i][j]=='@')
				{
					ans++;dfs(i,j);
				}
			} 
		printf("%d\n",ans);	
		
	}
return 0;
}
           

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