BFS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：

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：

就是给出一个地图，图上有红砖和黑砖以及一个初始点，分别为“#”与“.“与”@“,并且红砖不能踩，问你从初始点开始走，问最多能踩多少黑砖。

思路分析：

这道题直接用BFS套模版进去，用整数数组标记位置是否走过即可，BFS和DFS，没想到BFS的时间居然只要0ms真的很块，但是空间郁闷了。

AC代码：

用时：0MS

#include
#include
#include
#include
#include
using namespace std;
struct num1//创建数据结构 
{
	int x;
	int y;
	int num;
};
int LemonJudge(int x2,int y2);
char c1[22][22];//构造地图 
int vis[22][22];//标记位置是否走过 
int b,c;//地图的行与列 
int x1,y1;//初始的位置 
int sum;//黑砖数量 
int die[5][2]={{1,0},{0,1},{-1,0},{0,-1}};//方向变量 
void LemonScanf()//输入地图 
{
	int d,e;
	for(d=0;d<c;d++)
	{
		scanf("%s",c1[d]);
	}
	for(d=0;d<c;d++)
	{
		for(e=0;e<b;e++)
		{
			if(c1[d][e]=='@')//找初始点 
			{
				x1=d;
				y1=e;
				break;
			}
		}
	}
}
int LemonJudge(int x2,int y2)
{
	if(x2<0 || y2<0 || x2>=c || y2>=b || c1[x2][y2]=='#' || vis[x2][y2])
	{
		return 1;
	}
	return 0;
}
void LemonBFS()
{
	
	num1 next,Lemon;
	int i;
	Lemon.x=x1;//记录第一个点 
	Lemon.y=y1;//记录第一个点的纵坐标 
	queue<num1>q;//创造队列 
	q.push(Lemon);//将Lemon的数据结构放入队列中 
	vis[Lemon.x][Lemon.y]=1;//标记初始位置走过 
	while(!q.empty())//如果队列无元素，就结束循环 
	{
		Lemon=q.front();//访问对头元素
		q.pop();//删除顶元素 
		for(i=0;i<4;i++)
		{
			next.x=Lemon.x+die[i][0];//记录下一次的坐标 
			next.y=Lemon.y+die[i][1];
			if(LemonJudge(next.x,next.y))//判断条件 
			{
				continue;
			}
			vis[next.x][next.y]=1;//标记 
			sum++;
			q.push(next);//将next数据结构放入新的队列中 
		}
	}
}
int main()
{
	while(scanf("%d %d",&b,&c)!=EOF)
	{
		if(b==0 && c==0)break;
		memset(vis,0,sizeof(vis));
		//刷新数据 
		sum=1;
		LemonScanf();
		LemonBFS();
		printf("%d\n",sum);
	}
}