Robot Motion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7974    Accepted Submission(s):
3685

Problem 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
 
Sample Output
10 step(s) to exit
3 step(s) before a loop of 8 step(s)
做题时两个地方被坑到:
1、刚开始的时候横纵坐标搞反了  
2、输出的结果错误后来发现输入的时候数组是从0开始的而在搜索判断中却是从数组中1 开始判断
#include<stdio.h>
#include<string.h>
#define MAX 110
char map[MAX][MAX];//所要走的地图
int vis[MAX][MAX];//用来记录当前走了多少步
int n,m,t;
int x2,y2;//用于记录当前步的上一步
void dfs(int x1,int y1)
{
if(x1>n||x1<1||y1<1||y1>m)//如果超出地图范围则证明已经走出去了
{
printf("%d step(s) to exit\n",vis[x2][y2]);
return ;
}
else if(vis[x1][y1])//如果不为0则证明此处已经走过形成环
{
printf("%d step(s) before a loop of %d step(s)\n",vis[x1][y1]-1,vis[x2][y2]-vis[x1][y1]+1);
return ;
}
vis[x1][y1]=vis[x2][y2]+1;//当前走的步数是上一步加1
x2=x1;y2=y1;
if(map[x1][y1]=='W')//向左走
y1-=1;
else if(map[x1][y1]=='S')//向下走
x1+=1;
else if(map[x1][y1]=='E')//向右走
y1+=1;
else if(map[x1][y1]=='N')//向上走
x1-=1;
dfs(x1,y1);
}
int main()
{
int j,i,s,k;
int x1,x2,y1,y2;
while(scanf("%d%d",&n,&m)&&n!=0&&m!=0)
{
scanf("%d",&t);
for(i=1;i<=n;i++)
{
getchar();
for(j=1;j<=m;j++)
{
scanf("%c",&map[i][j]);
}
}
x1=1;y1=t;//起点
x2=x1;y2=y1;
memset(vis,0,sizeof(vis));//数组清零
dfs(x1,y1);
}
return 0;
}

  

hdoj 1035 Robot Motion的更多相关文章

  1. HDOJ(HDU).1035 Robot Motion (DFS)

    HDOJ(HDU).1035 Robot Motion [从零开始DFS(4)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DF ...

  2. [ACM] hdu 1035 Robot Motion (模拟或DFS)

    Robot Motion Problem Description A robot has been programmed to follow the instructions in its path. ...

  3. hdu 1035 Robot Motion(模拟)

    Problem Description A robot has been programmed to follow the instructions in its path. Instructions ...

  4. hdu 1035 Robot Motion(dfs)

    虽然做出来了,还是很失望的!!! 加油!!!还是慢慢来吧!!! >>>>>>>>>>>>>>>>> ...

  5. 题解报告:hdu 1035 Robot Motion(简单搜索一遍)

    Problem Description A robot has been programmed to follow the instructions in its path. Instructions ...

  6. HDU 1035 Robot Motion(dfs + 模拟)

    嗯... 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1035 这道题比较简单,但自己一直被卡,原因就是在读入mp这张字符图的时候用了scanf被卡. ...

  7. (step 4.3.5)hdu 1035(Robot Motion——DFS)

    题目大意:输入三个整数n,m,k,分别表示在接下来有一个n行m列的地图.一个机器人从第一行的第k列进入.问机器人经过多少步才能出来.如果出现了循环 则输出循环的步数 解题思路:DFS 代码如下(有详细 ...

  8. HDU-1035 Robot Motion

    http://acm.hdu.edu.cn/showproblem.php?pid=1035 Robot Motion Time Limit: 2000/1000 MS (Java/Others)   ...

  9. hdu1035 Robot Motion (DFS)

    Robot Motion Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Tot ...

随机推荐

  1. Expat Parser解析xml文件

    Expat 解析器是基于事件的解析器. 基于事件的解析器集中在 XML 文档的内容,而不是它们的结构.正因为如此,基于事件的解析器能够比基于树的解析器更快地访问数据. 请看下面的 XML 片段: &l ...

  2. 快速设置IP的脚本

    @echo off cls ::set NAME="本地连接" set NAME="无线网络连接" set IP=192.168.1.55 set MASK=2 ...

  3. Django生产环境的部署-Apache-mod_wsgi

    httpd.conf配置 ServerSignature On ServerTokens Full Define APACHE24 Apache2.4 Define SERVER_BASE_DIR & ...

  4. [swift] NSClassFromString 无法获得该类

    在写OC的时候需要用 NSClassFromString(classStringName)获得一个类,如果存在就用这个类型来声明一个对象, 但是在swift的时候却往往得不到这个类,为什么呢? 从截图 ...

  5. 定位 - CoreLocation - 打印位置信息

    1. 导入框架 <CoreLocation.framework>, 引入头文件 import <CoreLocation/CoreLocation.h>; 2. 创建管理者对象 ...

  6. 个人学习笔记--MyBatis-的搭建及第一个程序

    1.导入Jar包 2.设置全局配置文件 <?xml version="1.0" encoding="UTF-8" ?> <!DOCTYPE c ...

  7. 两种方法,获取磁盘剩余空间--PYTHON

    import ctypes import os import platform import sys def get_free_space_mb(folder): """ ...

  8. EasyUI 调用getSelections方法只能获取到一行的原因

    $('#tt').datagrid({ url: 'GetDataJosn', title: 'DataGrid', width: 800, height: 300, pageSize: 10, id ...

  9. 使用@ResponseBody 出现错误Could not find acceptable representation

    org.springframework.web.HttpMediaTypeNotAcceptableException: Could not find acceptable representatio ...

  10. 结论: blocking_query 是当前堵塞其他会话正在运行的SQL.而不是原始堵塞SQL

    查看当前session线程号 mysql> select connection_id(); +-----------------+ | connection_id() | +---------- ...