Robot Motion
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 12856   Accepted: 6240

Description


A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in a grid. The possible instructions are

N north (up the page) 
S south (down the page) 
E east (to the right on the page) 
W west (to the left on the page)

For example, suppose the robot starts on the north (top) side of Grid 1 and starts south (down). The path the robot follows is shown. The robot goes through 10 instructions in the grid before leaving the grid.

Compare what happens in Grid 2: the robot goes through 3 instructions only once, and then starts a loop through 8 instructions, and never exits.

You are to write a program that determines how long it takes a robot to get out of the grid or how the robot loops around. 

Input

There will be one or more grids for robots to navigate. The data for each is in the following form. On the first line are three integers separated by blanks: the number of rows in the grid, the number of columns in the grid, and the number of the column in
which the robot enters from the north. The possible entry columns are numbered starting with one at the left. Then come the rows of the direction instructions. Each grid will have at least one and at most 10 rows and columns of instructions. The lines of instructions
contain only the characters N, S, E, or W with no blanks. The end of input is indicated by a row containing 0 0 0.

Output

For each grid in the input there is one line of output. Either the robot follows a certain number of instructions and exits the grid on any one the four sides or else the robot follows the instructions on a certain number of locations once, and then the instructions
on some number of locations repeatedly. The sample input below corresponds to the two grids above and illustrates the two forms of output. The word "step" is always immediately followed by "(s)" whether or not the number before it is 1.

Sample Input

3 6 5

NEESWE

WWWESS

SNWWWW

4 5 1

SESWE

EESNW

NWEEN

EWSEN

0 0 0

Sample Output

10 step(s) to exit

3 step(s) before a loop of 8 step(s)

Source

题目链接：POJ 1573

简单BFS，过不了的是试一下这组样例……

Try this test case:

2 2 1

SW

EN

The right answer should be:

0 step(s) before a loop of 4 step(s)

由于刚开始的步数为0，加了一个vis数组，用一个数组记录每一次到的地方的步数即可，若访问过就说明有回路，若可以走出界则说明有出口……

代码：

#include<iostream>

#include<algorithm>

#include<cstdlib>

#include<sstream>

#include<cstring>

#include<bitset>

#include<cstdio>

#include<string>

#include<deque>

#include<stack>

#include<cmath>

#include<queue>

#include<set>

#include<map>

using namespace std;

#define INF 0x3f3f3f3f

#define CLR(x,y) memset(x,y,sizeof(x))

#define LC(x) (x<<1)

#define RC(x) ((x<<1)+1)

#define MID(x,y) ((x+y)>>1)

typedef pair<int,int> pii;

typedef long long LL;

const double PI=acos(-1.0);

const int N=15;

char pos[N][N];

int his_step[N][N];

int vis[N][N];

struct info

{

	int x;

	int y;

	int step;

	info operator+(info b)

	{

		b.x+=x;

		b.y+=y;

		b.step+=step;

		return b;

	}

};

info S,direct[4]={{0,1,1},{0,-1,1},{1,0,1},{-1,0,1}};

int n,m,c;

void init()

{

	CLR(pos,0);

	CLR(his_step,0);

	CLR(vis,0);

}

int main(void)

{

	int i,j,k,ans1,ans2;

	while (~scanf("%d%d%d",&n,&m,&c)&&(n||m||c))

	{

		init();

		ans1=-1;

		ans2=-1;

		for (i=0; i<n; ++i)

			scanf("%s",pos[i]);

		S.x=0;

		S.y=c-1;

		S.step=0;

		queue<info>Q;

		Q.push(S);

		while (!Q.empty())

		{

			info now=Q.front();

			Q.pop();

			if(vis[now.x][now.y])

			{

				ans1=his_step[now.x][now.y];

				ans2=now.step-his_step[now.x][now.y];

				break;

			}

			else

			{

				his_step[now.x][now.y]=now.step;

				vis[now.x][now.y]=1;

			}

			info v;

			switch(pos[now.x][now.y])

			{

				case 'N':v=now+direct[3];break;

				case 'S':v=now+direct[2];break;

				case 'W':v=now+direct[1];break;

				case 'E':v=now+direct[0];break;

			}

			if(v.x<0||v.x>=n||v.y<0||v.y>=m)

			{

				ans1=v.step;

				break;

			}

			Q.push(v);

		}

		if(ans2==-1)

			printf("%d step(s) to exit\n",ans1);

		else

			printf("%d step(s) before a loop of %d step(s)\n",ans1,ans2);

	}

	return 0;

}

POJ 1573 Robot Motion（BFS）的更多相关文章

  1. POJ 1573 Robot Motion（模拟）

    题目代号:POJ 1573 题目链接:http://poj.org/problem?id=1573 Language: Default Robot Motion Time Limit: 1000MS ...

  2. 模拟 POJ 1573 Robot Motion

    题目地址:http://poj.org/problem?id=1573 /* 题意:给定地图和起始位置,robot(上下左右)一步一步去走,问走出地图的步数 如果是死循环,输出走进死循环之前的步数和死 ...

  3. POJ 1573 Robot Motion

    Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12978   Accepted: 6290 Des ...

  4. poj 1573 Robot Motion【模拟题 写个while循环一直到机器人跳出来】

                                                                                                         ...

  5. poj1573 Robot Motion（DFS）

    题目链接 http://poj.org/problem?id=1573 题意 一个机器人在给定的迷宫中行走,在迷宫中的特定位置只能按照特定的方向行走,有两种情况:①机器人按照方向序列走出迷宫,这时输出 ...

  6. POJ 3984 迷宫问题（BFS）

    迷宫问题 Description 定义一个二维数组: int maze[5][5] = { 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, ...

  7. POJ 2435Navigating the City（bfs）

    题意:给你一个地图,’+’代表十字路口,‘-’‘|’表示街道,‘.’表示建筑物,‘s’,’E’ 起点和终点.输出从起点到终点的的 最短路径(包括方向和沿该方向的经过的十字路口数) 分析:ans[i][ ...

  8. poj 3669 Meteor Shower（bfs）

    Description Bessie hears that an extraordinary meteor shower is coming; reports say that these meteo ...

  9. poj1573&amp;&amp;hdu1035 Robot Motion（模拟）

    转载请注明出处:http://blog.csdn.net/u012860063? viewmode=contents 题目链接: HDU:pid=1035">http://acm.hd ...

随机推荐

  1. 【mysql】执行mysql脚本

    来源:http://zhidao.baidu.com/link?url=p78PlTJZlheB4uSKCkmZApg-3qrTIBCS3yI5LbGJLEAnUuO3-4GE6dLqq1LWC_kn ...

  2. 基于gitosis的Git云端服务器配置

    (本文需要自己实践,由于时间关系,我仅仅是做了整理和快速的练习,至于笔记中的账号和ip域名都是我参考文章中的.如果读者有任何问题欢迎留言和发邮件到luoquantao@126.com) 硬件:云端阿里 ...

  3. ScrollView与TableView实现选择效果

    在cocos2dx中,ScrollView与TableView都可以实现选择效果,其中ScrollView较为原始,TableView的格子大小可以不与winSize一样大. ScrollView实现 ...

  4. kvm虚似机监控

    [root@ok Desktop]# yum install virt-top [root@ok Desktop]# virt-top virt-top 22:53:44 - x86_64 4/4CP ...

  5. PHP数组的使用方法小结

    数组就是一组数据的集合,把一系列数据组织起来,形成一个可操作的整体.数组的每个实体都包含两项:键和值. 一.什么是数组数组就是一组数据的集合,把一系列数据组织起来,形成一个可操作的整体.数组的每个实体 ...

  6. (转)云存储：阿里云OSS 、又拍云和 七牛 的比较

    阿里OSS:好处就是,那是一套完整的体系,存储,数据库,CDN,服务器,阿里都可以给你全包.缺点,费用对于没有盈利的网站来说太高了,好像定位就是给那些高端客户使用的,而且CDN,OSS的流量是分开收费 ...

  7. ThinkPHP实现移动端访问自动切换主题模板

    ThinkPHP的模板主题机制,如果只是在PC,只要需修改 DEFAULT_THEME (新版模板主题默认是空,表示不启用模板主题功能)配置项就可以方便的实现多模板主题切换. 但对于移动端和PC端,也 ...

  8. hdu 1728 bfs **

    简单bfs,记录好状态即可 #include<cstdio> #include<iostream> #include<algorithm> #include< ...

  9. PHP扩展开发

    安装好php,进入安装目录. zbseoag@ubuntu:/usr/local/php-5.6.28/ext$ ./ext_skel --extname=mytest zbseoag@ubuntu: ...

  10. Touch 事件的分发和消费机制

    Android 中与 Touch 事件相关的方法包括:dispatchTouchEvent(MotionEvent ev).onInterceptTouchEvent(MotionEvent ev). ...