Codeforces #617 (Div. 3) C. Yet Another Walking Robot
There is a robot on a coordinate plane. Initially, the robot is located at the point (0,0)(0,0) . Its path is described as a string ss of length nn consisting of characters 'L', 'R', 'U', 'D'.
Each of these characters corresponds to some move:
- 'L' (left): means that the robot moves from the point (x,y)(x,y) to the point (x−1,y)(x−1,y) ;
- 'R' (right): means that the robot moves from the point (x,y)(x,y) to the point (x+1,y)(x+1,y) ;
- 'U' (up): means that the robot moves from the point (x,y)(x,y) to the point (x,y+1)(x,y+1) ;
- 'D' (down): means that the robot moves from the point (x,y)(x,y) to the point (x,y−1)(x,y−1) .
The company that created this robot asked you to optimize the path of the robot somehow. To do this, you can remove any non-empty substring of the path. But this company doesn't want their customers to notice the change in the robot behavior. It means that if before the optimization the robot ended its path at the point (xe,ye)(xe,ye) , then after optimization (i.e. removing some single substring from ss ) the robot also ends its path at the point (xe,ye)(xe,ye) .
This optimization is a low-budget project so you need to remove the shortest possible non-empty substring to optimize the robot's path such that the endpoint of his path doesn't change. It is possible that you can't optimize the path. Also, it is possible that after the optimization the target path is an empty string (i.e. deleted substring is the whole string ss ).
Recall that the substring of ss is such string that can be obtained from ss by removing some amount of characters (possibly, zero) from the prefix and some amount of characters (possibly, zero) from the suffix. For example, the substrings of "LURLLR" are "LU", "LR", "LURLLR", "URL", but not "RR" and "UL".
You have to answer tt independent test cases.
The first line of the input contains one integer tt (1≤t≤10001≤t≤1000 ) — the number of test cases.
The next 2t2t lines describe test cases. Each test case is given on two lines. The first line of the test case contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105 ) — the length of the robot's path. The second line of the test case contains one string ss consisting of nn characters 'L', 'R', 'U', 'D' — the robot's path.
It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105 (∑n≤2⋅105∑n≤2⋅105 ).
For each test case, print the answer on it. If you cannot remove such non-empty substring that the endpoint of the robot's path doesn't change, print -1. Otherwise, print two integers ll and rr such that 1≤l≤r≤n1≤l≤r≤n — endpoints of the substring you remove. The value r−l+1r−l+1 should be minimum possible. If there are several answers, print any of them.
4
4
LRUD
4
LURD
5
RRUDU
5
LLDDR
1 2
1 4
3 4
-1
数据结构题,考stl的map怎么用。用map<pair<int,int>,int>存储走过的地方,是坐标点到字符串下标的映射。只要发现下一个走到的点已经出现在字典里了,说明这一段走的路就是在兜圈子,可以直接删掉,这时候更新答案的值同时把这个点的坐标对应的值更新成最新的(因为以后要是再一次经过的话到当前位置的字符串长度肯定小于到更新前位置的字符串的长度)。最后输出答案即可。
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
map<pair<int,int>,int>m;
while(t--)
{
int n;
scanf("%d",&n);
int i;
m.clear();//不要忘记清空
m[{,}]=;
int x=,y=;
int ans=;
int beg=-,end=-;
char s[];
scanf("%s",s);
for(i=;i<=n;i++)
{
char c=s[i-];
if(c=='L')
{
x-=;
}
else if(c=='R')
{
x+=;
}
else if(c=='U')
{
y+=;
}
else if(c=='D')
{
y-=;
}
if(m.find({x,y})!=m.end())
{
if(ans>i-m[{x,y}]+)
{
ans=i-m[{x,y}]+;//更新答案
beg=m[{x,y}]+;
end=i;
m[{x,y}]=i;
}
else
{
m[{x,y}]=i;//ans不必更新的话只更新这个点对应的值即可
}
}
else
{
m[{x,y}]=i;//之前没发现的话直接添加进字典 ,经过第i次操作到达(x,y)
}
}
if(ans!=&&beg!=-)cout<<beg<<' '<<end<<endl;
else cout<<-<<endl;
}
return ;
}
Codeforces #617 (Div. 3) C. Yet Another Walking Robot的更多相关文章
- Codeforces #617 (Div. 3) D. Fight with Monsters(贪心,排序)
There are nn monsters standing in a row numbered from 11 to nn . The ii -th monster has hihi health ...
- Codeforces #617 (Div. 3)B. Food Buying
Mishka wants to buy some food in the nearby shop. Initially, he has ss burles on his card. Mishka ca ...
- Codeforces Round #605 (Div. 3) B. Snow Walking Robot(构造)
链接: https://codeforces.com/contest/1272/problem/B 题意: Recently you have bought a snow walking robot ...
- Codeforces Round #617 (Div. 3) 补题记录
1296A - Array with Odd Sum 题意:可以改变数组中的一个数的值成另外一个数组中的数,问能不能使数组的和是个奇数 思路:签到,如果本来数组的和就是个奇数,那就OK 如果不是,就需 ...
- Codeforces Round #617 (Div. 3) 题解
又是隔了一年才来补题的我 A.B水题就不用说了 C - Yet Another Walking Robot C题我居然卡了一会,最后决定用map水,结果出来看了看题解,居然真的是map...没想到会出 ...
- [CF百场计划]Codeforces Round #617 (Div. 3)
A. Array with Odd Sum Description You are given an array \(a\) consisting of \(n\) integers. In one ...
- C. Yet Another Walking Robot Round #617 (Div. 3)()(map + 前后相同状态的存储)
C. Yet Another Walking Robot time limit per test 1 second memory limit per test 256 megabytes input ...
- Codeforces #344 Div.2
Codeforces #344 Div.2 Interview 题目描述:求两个序列的子序列或操作的和的最大值 solution 签到题 时间复杂度:\(O(n^2)\) Print Check 题目 ...
- Codeforces #345 Div.1
Codeforces #345 Div.1 打CF有助于提高做题的正确率. Watchmen 题目描述:求欧拉距离等于曼哈顿距离的点对个数. solution 签到题,其实就是求有多少对点在同一行或同 ...
随机推荐
- Linux - CentOS7 命令行快捷键简介
1. 概述 CentOS7 下输入命令的一些快捷键 2. 快捷键 1. 移动 单个字符 ctrl + b/f 单个单词 alt + b/f 行首行尾 ctrl + a/e 2. 编辑 删除单个字符 c ...
- 【音乐欣赏】《Wrong》 - Far Out / Emilia Ali
曲名:Wrong 作者:Far Out / Emilia Ali [00:16.03]Ride body on mine [00:18.07]Griping your waist as I was o ...
- find 报错 find: paths must precede expression:
编写shell脚本,报错,如下面 [root@localhost backup]#find ./ -name mysqldump* -mtime +3 -delete [root@localhost ...
- MVC简要介绍
(转自:http://www.cnbeta.com/articles/107924.htm) MVC不是一种设计模式(design pattern),它是一种架构模式(Architectural pa ...
- 程序员过年必备 -- Auto.js微信自动抢红包
打开微信就不用管了: - 自动打开未读消息 - 自动滑动屏幕检测红包 - 自动跳过无效红包 基于Auto JS,apk版本4.01: - 大多数动作均基于控件 - 极个别点击基于动态抓取的坐标 - 这 ...
- linux日常运维工作
Linux的使用环境也日趋成熟,各种开源产品络绎不绝,大有百花齐放的盛景,那么当Linux落地企业,回归工作时,我们还要面对这Linux运维方面的诸多问题,今天我们特意组织一场有关Linux 在企业运 ...
- stopWatch 用法
package com.example.stopwatch; import org.springframework.util.StopWatch; public class TestStopWatch ...
- 201771010135 杨蓉庆《面对对象程序设计(java)》第十五周学习总结
1.实验目的与要求 (1) 掌握Java应用程序的打包操作: (2) 了解应用程序存储配置信息的两种方法: (3) 掌握基于JNLP协议的java Web Start应用程序的发布方法: (5) 掌握 ...
- 如何切换虚拟机(centos6)和windows
通过设置热键,选择Ctrl+Alt+Fx即可.重启linux之后按Ctrl+Alt+Fx切换不同的终端的就可以了 图一. 图二.
- mcast_set_ttl函数
#include <errno.h> #include <net/if.h> #include <sys/socket.h> #include <netine ...