前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >HDUOJ-----Robot Motion

HDUOJ-----Robot Motion

作者头像
Gxjun
发布2018-03-21 11:30:18
6870
发布2018-03-21 11:30:18
举报
文章被收录于专栏:mlml

Robot Motion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5011    Accepted Submission(s): 2321

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)

Source

Mid-Central USA 1999

 模拟题..

代码语言:javascript
复制
 1 #include<cstdio>
 2 #include<cstring>
 3 const int maxn=1000;
 4  char maze[11][maxn];
 5  int record[11][maxn];
 6  int main()
 7  {
 8      int r,c,pos,i,j;
 9     while(scanf("%d%d",&r,&c),r+c)
10     {
11         scanf("%d",&pos);
12         memset(maze,'\0',sizeof maze);
13         memset(record,0,sizeof record);
14        for( i=0;i<r;i++)
15        {
16         scanf("%s",maze[i]);
17        }
18        int newr=0,newc=pos-1;
19        bool judge=true;
20        while(newc>=0&&newr>=0&&maze[newr][newc]!=0)
21        {  
22                record[newr][newc]++;
23              if(record[newr][newc]==3)
24              {  
25                  judge=false;
26                  break;
27              }
28             
29            switch(maze[newr][newc])
30            {
31              case 'N': newr--; break;  //up
32              case 'S': newr++; break;  //down
33              case 'E': newc++; break;  //right
34              case 'W': newc--; break;  //left
35            
36            }
37        }
38        int step=0,circle=0;
39        for( i=0;i<r;i++)
40        {
41            for( j=0;j<c;j++)
42            {
43               if(record[i][j]==1) 
44                   step++;
45               else 
46                   if(record[i][j]!=0)
47                       circle++;
48                 
49            }
50        }
51        
52       if(judge)
53            printf("%d step(s) to exit\n",step);
54        else
55            printf("%d step(s) before a loop of %d step(s)\n",step,circle);
56     }
57     return 0;
58  }
本文参与 腾讯云自媒体分享计划,分享自作者个人站点/博客。
原始发表:2013-07-19 ,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体分享计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
目录
  • Robot Motion
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档