POJ - 1979 （dfs）

POJ - 1979 （dfs）

题目正文如下：

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 //可走的

‘#’ - a red tile //不可走的

‘@’ - a man on a black tile(appears exactly once in a data set) //人的起始位置

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).
//对于每个数据，您的程序应该输出一行，其中包含他可以从初始瓷砖(包括它本身)到达的瓷砖数量。

Sample Input

//用截图形状看得比较清楚

Sample Output

45
59
6
13

题目分析：
其实题目就是要求对每个输入的地图，从’@‘开始进行深度搜索，把所有能走的’.'都走一遍，每走一格答案就加一， 刚开始站的位置算一格。

AC代码如下：

#include<stdio.h>
#include<string.h>
int W,H,sum;	//行数，列数，答案
char room[20][20];  //地图
void dfs(int x,int y)
{	
	int dx,dy,nx,ny;
	sum++;					//计算砖块个数
	room[x][y]='#';				//搜索过的将该位置置换为#，防止再次搜索
	for(dx=-1;dx<=1;dx++)
		for(dy=-1;dy<=1;dy++)
	{	nx=x+dx;
		ny=y+dy;
		if(dx==0||dy==0)		//控制上下左右四个方向
		{if(0<=nx&&nx<H&&0<=ny&&ny<W&&room[nx][ny]=='.') dfs(nx,ny);}	//如果没有出界且符合条件，则dfs
	}
}
int main()
{	int i,j;
	void dfs(int x,int y);
	while(~scanf("%d %d",&W,&H))
	{	sum=0;				//重置砖块个数
		memset(room,0,sizeof(room));	//重置数组
		if(!W&&!H) break;		//都为0，则退出
		for(i=0;i<H;i++)
		{	for(j=0;j<W;j++)
			scanf(" %c",&room[i][j]);	//输入
		}
		for(i=0;i<H;i++)
			for(j=0;j<W;j++)
			if(room[i][j]=='@') dfs(i,j);	//寻找@，并从@开始深度优先搜索
		printf("%d\n",sum);
	}
	return 0;
}