DFS训练-Red and Black~解题报告

Red and Black

题目描述：

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:

Sample Output:

题目大意：

给出一张图，图上面有黑砖"."与红砖“#”与初始点“@”，红砖是不能踩的，只有黑砖和初始点才能踩，然后问你在这张图怎么走，才能走的最多黑砖，初始点也是黑砖，所以这也要记录。

思路分析：

这道题直接从初始点“@”开始DFS，然后每次所走的路线记录最多黑砖数量，然后在输出就行了。

AC代码：

用时：15ms

#include<cstdio>
#include<string.h>
int x,y;//地图的行数与列数 
char c1[22][22];//记录地图 
int c2[22][22];//标记所走的位置是否走过 
int x1,y1,max;//x1与y1为初始点@的位置，max为黑砖最大数量。 
int div[4][4]={{1,0},{0,1},{-1,0},{0,-1}};//方向分别为上下左右 
int n;
void LemonScanf()
{
	int d,e;
	for(d=0;d<x;d++)
	{
		scanf("%s",c1[d]);
	}
	for(d=0;d<x;d++)
	{
		for(e=0;e<y;e++)
		{
			if(c1[d][e]=='@')
			{
				x1=d;
				y1=e;
			}
		}
	}
}
void LemonDFS(int x1,int y1)
{
	int i;
	if(n>max)//比较数量 
	{
		max=n;
	}
	for(i=0;i<4;i++)
	{
		int x2=x1+div[i][0];
		int y2=y1+div[i][1];
		if(x2<0 || y2<0 || x2>=x || y2>=y || c1[x2][y2]=='#' || c2[x2][y2])
		//判断是否越界与是否是红砖与是否走过该位置 
		continue;
		if(c1[x2][y2]=='.')
		{
			n++;//记录所踩黑砖数量 
			c2[x2][y2]=1;//标记该位置 
			LemonDFS(x2,y2);//继续深搜 
		}
	}
	return;
}
int main()
{
	while(scanf("%d %d",&y,&x)!=EOF)
	{
		max=0;
		n=1;
		if(x==0&&y==0)break;
		memset(c2,0,sizeof(c2));//这里是为了刷新c2的数据，防止对后面测试组造成错乱 
		LemonScanf();
		LemonDFS(x1,y1);
		printf("%d\n",max);//输出最大黑砖数量 
	}
}