Robot Motion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10072    Accepted Submission(s): 4697

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
DFS简单题 
 #include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<cstdlib>
#include<string>
#define eps 0.000000001
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int N=;
char mp[N][N];
int visited[N][N];
int m,n;
int flag;
int step,loopstep;
void DFS(int x,int y){
if(x<||x>=n||y<||y>=m){
return ;
}
if(visited[x][y]!=){
flag=;
loopstep=step-visited[x][y]+;
step=visited[x][y]-;
return ;
}
step++;
visited[x][y]=step;
if(mp[x][y]=='N')DFS(x-,y);
else if(mp[x][y]=='S')DFS(x+,y);
else if(mp[x][y]=='E')DFS(x,y+);
else if(mp[x][y]=='W')DFS(x,y-);
}
int main(){
while(scanf("%d%d",&n,&m)!=EOF){
if(n==&&m==)break;
int k;
scanf("%d",&k);
flag=;
for(int i=;i<n;i++)scanf("%s",mp[i]);
loopstep=step=;
memset(visited,,sizeof(visited));
DFS(,k-);
if(flag==) printf("%d step(s) to exit\n", step);
else
printf("%d step(s) before a loop of %d step(s)\n", step,loopstep); }
}
 
Sample Output
10 step(s) to exit
3 step(s) before a loop of 8 step(s)

hdu 1035(DFS)的更多相关文章

  1. hdu 1342(DFS)

    Lotto Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submi ...

  2. F - JDG HDU - 2112 (最短路)&& E - IGNB HDU - 1242 (dfs)

    经过锦囊相助,海东集团终于度过了危机,从此,HDU的发展就一直顺风顺水,到了2050年,集团已经相当规模了,据说进入了钱江肉丝经济开发区500强.这时候,XHD夫妇也退居了二线,并在风景秀美的诸暨市浬 ...

  3. F - Auxiliary Set HDU - 5927 (dfs判断lca)

    题目链接: F - Auxiliary Set HDU - 5927 学习网址:https://blog.csdn.net/yiqzq/article/details/81952369题目大意一棵节点 ...

  4. hdu - 1072(dfs剪枝或bfs)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1072 思路:深搜每一个节点,并且进行剪枝,记录每一步上一次的s1,s2:如果之前走过的时间小于这一次, ...

  5. HDU——2647Reward(DFS或差分约束)

    Reward Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Sub ...

  6. P - Sudoku Killer HDU - 1426(dfs + map统计数据)

    P - Sudoku Killer HDU - 1426 自从2006年3月10日至11日的首届数独世界锦标赛以后,数独这项游戏越来越受到人们的喜爱和重视. 据说,在2008北京奥运会上,会将数独列为 ...

  7. hdu 1142(DFS+dijkstra)

    #include<iostream> #include<cstdio> #include<cmath> #include<map> #include&l ...

  8. hdu 1015(DFS)

    Safecracker Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total ...

  9. hdu 1181(DFS)变 形 课

    变形课 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submis ...

随机推荐

  1. 开发日记(项目中SQL查询的优化)

    今天发现自己之前写的一些SQL查询在执行效率方面非常不理想,于是尝试做了些改进. 需求为查询国地税表和税源表中,国税有而税源没有的条目数,之前的查询如下: SELECT COUNT(NAME)     ...

  2. jQuery实现页面锚点滚动效果

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  3. PB连接SQLITE

    sqlite3数据库,简单而功能强大,比起ini文件保存用户设置,更简单安全,为什么使用数据库存用户设置,由开发者自己去想吧进入话题:pb中可以用ole DB方式在不注册odbc的情况下直接连接数据库 ...

  4. CSS——宠物demo

    注意:ul中自带padding值,需要清除. <!DOCTYPE html> <html lang="en"> <head> <meta ...

  5. day11-函数对象、名称空间和作用域

    目录 函数对象 函数的嵌套 名称空间和作用域 内置名称空间 全局名称空间 局部名称空间 作用域 全局作用域 局部作用域 global和nonlocal 函数对象 在Python中,一切皆对象,函数也是 ...

  6. SQL查询性能优化

    使用高效的查询 使用 EXISTS 代替 IN -- 查询A表中同时存在B表的数据 -- 慢 SELECT * FROM Class_A WHERE id IN (SELECT id FROM Cla ...

  7. animation与transition区别

    transition: 过渡属性 过渡所需要时间 过渡动画函数 过渡延迟时间:默认值分别为:all 0 ease 0 1.局限性: 1)只能设置一个属性 2)需要伪类/事件触发才执行 3)只能设置动画 ...

  8. jquery动态生成二维码添加自定义logo

    动态生成二维码中间带logo. jquery.qrcode.js 动态生成二维码api很简单. 引入jquer(版本任意),引入jquery.qrcode.js 不需要中间带logo这样就可以了.带l ...

  9. HDU1465 不容易系列之一&&HDU4535吉哥系列故事——礼尚往来

    HDU1465不容易系列之一 Problem Description 大家常常感慨,要做好一件事情真的不容易,确实,失败比成功容易多了!做好“一件”事情尚且不易,若想永远成功而总从不失败,那更是难上加 ...

  10. hdu2016 数据的交换输出【C++】

    数据的交换输出 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Sub ...