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 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
Output
Sample Input
Sample Output
#include<bits/stdc++.h>
using namespace std;
const int maxn=;
int row,col,st,step,cnt[maxn][maxn];char s[maxn][maxn];
void dfs(int x,int y){
while(true){//遍历一次即可
if(x<||x>=row||y<||y>=col){cout<<step<<" step(s) to exit"<<endl;return;}
if(cnt[x][y]>-){cout<<cnt[x][y]<<" step(s) before a loop of "<<(step-cnt[x][y])<<" step(s)"<<endl;return;}
//只要cnt[x][y]>-1,说明该点已经遍历过了,直接退出
cnt[x][y]=step++;
if(s[x][y]=='E')y++;
else if(s[x][y]=='W')y--;
else if(s[x][y]=='N')x--;
else x++;
}
}
int main(){
while(~scanf("%d%d%d",&row,&col,&st)&&(row+col+st)){
memset(cnt,-,sizeof(cnt));//全部初始化为-1
for(int i=;i<row;++i){
getchar();
for(int j=;j<col;++j)
scanf("%c",&s[i][j]);
}
step=;//初始值为0
dfs(,st-);
}
return ;
}
题解报告:hdu 1035 Robot Motion(简单搜索一遍)的更多相关文章
- 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 ...
- (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 ...
- [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. ...
- 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!!! >>>>>>>>>>>>>>>>> ...
- 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. ...
- hdu 1035 Robot Motion(模拟)
Problem Description A robot has been programmed to follow the instructions in its path. Instructions ...
- hdoj 1035 Robot Motion
Robot Motion Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Tota ...
- DFS与BFS题解:[kaungbin]带你飞 简单搜索 解题报告
DFS and BFS 在解题前我们还是大致讲一下dfs与bfs的.(我感觉我不会bfs) 1.DFS dfs(深度优先算法) 正如其名,dfs是相当的深度,不走到最深处绝不回头的那种. 深度优先搜 ...
- POJ1573(Robot Motion)--简单模拟+简单dfs
题目在这里 题意 : 问你按照图中所给的提示走,多少步能走出来??? 其实只要根据这个提示走下去就行了.模拟每一步就OK,因为下一步的操作和上一步一样,所以简单dfs.如果出现loop状态,只要记忆每 ...
随机推荐
- Linux下汇编语言学习笔记41 ---
这是17年暑假学习Linux汇编语言的笔记记录,参考书目为清华大学出版社 Jeff Duntemann著 梁晓辉译<汇编语言基于Linux环境>的书,喜欢看原版书的同学可以看<Ass ...
- [bzoj1692][Usaco2007 Dec]队列变换_后缀数组_贪心
队列变换 bzoj-1692 Usaco-2007 Dec 题目大意:给定一个长度为$n$的字符串.每次从头或尾取出一个字符加到另一个字符串里.要求变换后生成的字符串字典序最小,求字典序最小的字符串. ...
- Evaluate Reverse Polish Notation(逆波兰式)
Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, ...
- <项目><day11>查看用户浏览过的商品
<项目>查看用户浏览过的商品 1.创建一个entity包储存实体对象 1.1创建一个Product的类存储实体对象 对象具有以下属性,并添加set和get方法,含参和不含参的构造方法,to ...
- json数组显示格式
{“colorAndImg”:[{"颜色":“红色”,"地址":“www.sohu.com”}, {“颜色”:“绿色,“地址”:“www.sohu.com”}] ...
- How to force immediate stop of threads in Jmeter servers如何在jmeter执行完,立即停止jmeter
https://stackoverflow.com/questions/38900315/how-to-force-immediate-stop-of-threads-in-jmeter-server ...
- Jenkins + SVN搭建php持续集成
目标需求 开发提交代码到SVN,jenkins在分发服务器上执行'svn update',分发服务器在把代码同步到webserver,实现持续集成 流程 配置jenkins 一.jenkins所需插件 ...
- 【LeetCode-面试算法经典-Java实现】【067-Add Binary(二进制加法)】
[067-Add Binary(二进制加法)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 Given two binary strings, return thei ...
- [VueJS + Typescript] Decouple Dependencies Using IoC Containers in Vue with TypeScript and InversifyJS
Using Object Oriented Programming, OOP, style allows us to apply Inversion of Control, IoC, and more ...
- 【Unix编程】进程间通信(IPC)
进程间通信(IPC,InterProcess Communication)是指在不同进程之间传播或交换信息.IPC的方式通常有管道(包括无名管道和命名管道).消息队列.信号量.共享存储.Socket. ...