POJ1979 Red and Black(DFS)

题意：

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

 

分析:

也没啥说的，入门DFS，加一个全局变量res记录一下步数就行，写过很多次了。

 

AC代码：

#include
#include
using namespace std;
#define MAX 30
int W,H;
int result;
char Graph[MAX][MAX];
int direction[4][2]={{0,1},{0,-1},{-1,0},{1,0}};

void DFS(int p,int q)
{
	if(p>=0 && p<=W-1 && q>=0 &&q<=H-1 && Graph[p][q]=='.')
	{
		result++;
		Graph[p][q]='#';
	}
	else return;
	for(int i = 0;i<4;i++)
	{
		int x=p+direction[i][0];
		int y=q+direction[i][1];
		DFS(x,y);
	}
}

int main()
{
	int i,j,p,q;
	while(cin>>H>>W)
	{
		result=0;
		memset(Graph,0,sizeof(Graph));
		if(H==0 && W==0)
			break;
		for(i=0;i>Graph[i][j];
				if(Graph[i][j]=='@')
				{
					p=i;
					q=j;
					Graph[i][j]='.';
				}
			}
		}
		DFS(p,q);
		cout<