zoj1004 Anagrams by Stack

本文详细解析了Anagrams算法问题,通过提供输入样例和输出样例,阐述了解决此类问题的方法和步骤。包括如何使用堆栈操作将一个单词转换为另一个单词的安格拉姆,并且给出了有效序列的操作指令,以及对这些序列进行排序输出的过程。
Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

 Status

Description

How can anagrams result from sequences of stack operations? There are two sequences of stack operators which can convert TROT to TORT:

[
i i i i o o o o
i o i i o o i o
]

where i stands for Push and o stands for Pop. Your program should, given pairs of words produce sequences of stack operations which convert the first word to the second.

Input

The input will consist of several lines of input. The first line of each pair of input lines is to be considered as a source word (which does not include the end-of-line character). The second line (again, not including the end-of-line character) of each pair is a target word. The end of input is marked by end of file.

Output

For each input pair, your program should produce a sorted list of valid sequences of i and o which produce the target word from the source word. Each list should be delimited by

[
]
and the sequences should be printed in "dictionary order". Within each sequence, each   i  and   o  is followed by a single space and each sequence is terminated by a new line.
Process

A stack is a data storage and retrieval structure permitting two operations:

Push - to insert an item and  
Pop - to retrieve the most recently pushed item

We will use the symbol i (in) for push and o (out) for pop operations for an initially empty stack of characters. Given an input word, some sequences of push and pop operations are valid in that every character of the word is both pushed and popped, and furthermore, no attempt is ever made to pop the empty stack. For example, if the word FOO is input, then the sequence:

i i o i o ois valid, but
i i ois not (it's too short), neither is
i i o o o i(there's an illegal pop of an empty stack)

Valid sequences yield rearrangements of the letters in an input word. For example, the input word FOO and the sequence i i o i o oproduce the anagram OOF. So also would the sequence i i i o o o. You are to write a program to input pairs of words and output all the valid sequences of i and o which will produce the second member of each pair from the first.

Sample Input

madam
adamm
bahama
bahama
long
short
eric
rice

Sample Output

[
i i i i o o o i o o 
i i i i o o o o i o 
i i o i o i o i o o 
i i o i o i o o i o 
]
[
i o i i i o o i i o o o 
i o i i i o o o i o i o 
i o i o i o i i i o o o 
i o i o i o i o i o i o 
]
[
]
[
i i o i o i o o 
]

Source

Zhejiang University Local Contest 2001
 
 
 
分析:
内容概要:本文介绍了一个基于多传感器融合的定位系统设计方案,采用GPS、里程计和电子罗盘作为定位传感器,利用扩展卡尔曼滤波(EKF)算法对多源传感器数据进行融合处理,最终输出目标的滤波后位置信息,并提供了完整的Matlab代码实现。该方法有效提升了定位精度与稳定性,尤其适用于存在单一传感器误差或信号丢失的复杂环境,如自动驾驶、移动采用GPS、里程计和电子罗盘作为定位传感器,EKF作为多传感器的融合算法,最终输出目标的滤波位置(Matlab代码实现)机器人导航等领域。文中详细阐述了各传感器的数据建模方式、状态转移与观测方程构建,以及EKF算法的具体实现步骤,具有较强的工程实践价值。; 适合人群:具备一定Matlab编程基础,熟悉传感器原理和滤波算法的高校研究生、科研人员及从事自动驾驶、机器人导航等相关领域的工程技术人员。; 使用场景及目标:①学习和掌握多传感器融合的基本理论与实现方法;②应用于移动机器人、无人车、无人机等系统的高精度定位与导航开发;③作为EKF算法在实际工程中应用的教学案例或项目参考; 阅读建议:建议读者结合Matlab代码逐行理解算法实现过程,重点关注状态预测与观测更新模块的设计逻辑,可尝试引入真实传感器数据或仿真噪声环境以验证算法鲁棒性,并进一步拓展至UKF、PF等更高级滤波算法的研究与对比。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值