Organize Your Train part II
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 6478   Accepted: 1871

Description

RJ Freight, a Japanese railroad company for freight operations has recently constructed exchange lines at Hazawa, Yokohama. The layout of the lines is shown in Figure 1.


Figure 1: Layout of the exchange lines

A freight train consists of 2 to 72 freight cars. There are 26 types of freight cars, which are denoted by 26 lowercase letters from "a" to "z". The cars of the same type are indistinguishable from each other, and each car's direction doesn't matter either. Thus, a string of lowercase letters of length 2 to 72 is sufficient to completely express the configuration of a train.

Upon arrival at the exchange lines, a train is divided into two sub-trains at an arbitrary position (prior to entering the storage lines). Each of the sub-trains may have its direction reversed (using the reversal line). Finally, the two sub-trains are connected in either order to form the final configuration. Note that the reversal operation is optional for each of the sub-trains.

For example, if the arrival configuration is "abcd", the train is split into two sub-trains of either 3:1, 2:2 or 1:3 cars. For each of the splitting, possible final configurations are as follows ("+" indicates final concatenation position):

  [3:1]
abc+d cba+d d+abc d+cba
[2:2]
ab+cd ab+dc ba+cd ba+dc cd+ab cd+ba dc+ab dc+ba
[1:3]
a+bcd a+dcb bcd+a dcb+a

Excluding duplicates, 12 distinct configurations are possible.

Given an arrival configuration, answer the number of distinct configurations which can be constructed using the exchange lines described above.

Input

The entire input looks like the following.

the number of datasets = m
1st dataset 
2nd dataset 
... 
m-th dataset

Each dataset represents an arriving train, and is a string of 2 to 72 lowercase letters in an input line.

Output

For each dataset, output the number of possible train configurations in a line. No other characters should appear in the output.

Sample Input

4
aa
abba
abcd
abcde

Sample Output

1
6
12
18

Source

 
题目大意:就是一个字符串拆成两部分,然后任何一个可以旋转,然后两个的位置可以变动,所以总共有8种组合,问总共能拼成多少个不同的字符串。
#include<iostream>
#include<cstdio>
#include<cstring> using namespace std; struct Trie{
int cnt;
int next[];
}root[]; char str[];
int len,top,ans; void init(int k){
for(int i=;i<;i++)
root[k].next[i]=-;
root[k].cnt=;
} void InsertTrie(int p){
for(int i=;i<len;i++){
int id=str[i]-'a';
if(root[p].next[id]==-){
root[p].next[id]=top;
init(top);
top++;
}
p=root[p].next[id];
}
if(root[p].cnt==){
ans++;
root[p].cnt++;
}
} int main(){ //freopen("input.txt","r",stdin); char s1[],s2[],s3[],s4[],s5[];
int head,t;
scanf("%d",&t);
while(t--){
scanf("%s",s1);
len=strlen(s1);
ans=;
head=;
top=;
init(head);
int i,j;
for(i=;i<len;i++){
for(j=;j<i;j++){
s2[j]=s1[j];
s4[i--j]=s2[j];
}
s2[i]='\0'; s4[i]='\0';
for(j=i;j<len;j++){
s3[j-i]=s1[j];
s5[len--j]=s3[j-i];
}
s3[j-i]='\0'; s5[j-i]='\0';
strcpy(str,s2); strcat(str,s3);
InsertTrie(head); strcpy(str,s2); strcat(str,s5);
InsertTrie(head); strcpy(str,s3); strcat(str,s2);
InsertTrie(head); strcpy(str,s3); strcat(str,s4);
InsertTrie(head); strcpy(str,s4); strcat(str,s3);
InsertTrie(head); strcpy(str,s4); strcat(str,s5);
InsertTrie(head); strcpy(str,s5); strcat(str,s2);
InsertTrie(head); strcpy(str,s5); strcat(str,s4);
InsertTrie(head);
}
printf("%d\n",ans);
}
return ;
}

POJ 3007 Organize Your Train part II (字典树 静态)的更多相关文章

  1. POJ 3007 Organize Your Train part II

    题意: 如上图所示,将一个字符串进行分割,反转等操作后不同字符串的个数: 例如字符串abba:可以按三种比例分割:1:3:2:2:3:1 部分反转可以得到如下所有的字符串: 去掉重复可以得到六个不同的 ...

  2. Organize Your Train part II 字典树(此题专卡STL)

    Organize Your Train part II Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8787   Acce ...

  3. poj 3007 Organize Your Train part II(静态字典树哈希)

    Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6700 Accepted: 1922 Description RJ Freigh ...

  4. poj 3007 Organize Your Train part II(二叉排序树)

    题目:http://poj.org/problem?id=3007 题意:按照图示的改变字符串,问有多少种..字符串.. 思路:分几种排序的方法,,刚开始用map 超时(map效率不高啊..),后来搜 ...

  5. POJ 3007 Organize Your Train part II(哈希链地址法)

    http://poj.org/problem?id=3007 题意 :给你一个字符串,让你无论从什么地方分割,把这个字符串分成两部分s1和s2,然后再求出s3和s4,让你进行组合,看能出来多少种不同的 ...

  6. POJ 3007:Organize Your Train part II

    Organize Your Train part II Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7561   Acce ...

  7. poj 2503 Babelfish(Map、Hash、字典树)

    题目链接:http://poj.org/bbs?problem_id=2503 思路分析: 题目数据数据量为10^5, 为查找问题,使用Hash或Map等查找树可以解决,也可以使用字典树查找. 代码( ...

  8. ACM学习历程—POJ 3764 The xor-longest Path(xor && 字典树 && 贪心)

    题目链接:http://poj.org/problem?id=3764 题目大意是在树上求一条路径,使得xor和最大. 由于是在树上,所以两个结点之间应有唯一路径. 而xor(u, v) = xor( ...

  9. nyoj 230/poj 2513 彩色棒 并查集+字典树+欧拉回路

    题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=230 题意:给你许许多多的木棍,没条木棍两端有两种颜色,问你在将木棍相连时,接触的端点颜色 ...

随机推荐

  1. 深度学习材料:从感知机到深度网络A Deep Learning Tutorial: From Perceptrons to Deep Networks

    In recent years, there’s been a resurgence in the field of Artificial Intelligence. It’s spread beyo ...

  2. Coursera课程《大家的编程》(Python入门)中课程目录

    Getting Started with Python Getting Started with Python is the first course in the specialization Py ...

  3. poj 1007 Quoit Design(分治)

    Quoit Design Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) To ...

  4. 充满未来和科幻的界面设计FUI在国内还没有起步在国外早起相当成熟

    所谓FUI可以是幻想界面(Fantasy User Interfaces).科幻界面(Fictional User Interfaces).假界面(Fake User Interfaces).未来主义 ...

  5. C/C++ 语言获取文件大小

    在C语言中测试文件的大小,主要使用二个标准函数. 1.fseek 函数原型:int fseek ( FILE * stream, long int offset, int origin ); 参数说明 ...

  6. 通过小实例谈谈javascript的间隔调用和延时调用

    用 setInterval方法可以以指定的间隔实现循环调用函数,直到clearInterval方法取消循环 用clearInterval方法取消循环时,必须将setInterval方法的调用赋值给一个 ...

  7. Netdata Linux下性能实时监测工具

    导读 本文将介绍一款非常好用的工具——Netdata,这是一款Linux性能实时监测工具,为一款开源工具,我对其英文文档进行了翻译,水平有限,有翻译错误的地方欢迎大家指出,希望本文对大家有所帮助,谢谢 ...

  8. FM遇到错误RQP-DEF-0354和QE-DEF-0144

    版本:Cognos 10.2.1 系统:Win10 操作过程:在FM调用了一个存储过程,其中引用了前端page页面的参数如下图所示,在验证和保存查询主题的时候一直提示参数没有替换值,错误 信息如下图所 ...

  9. berkelydb学习

    http://www.oracle.com/technetwork/cn/java/seltzer-berkeleydb-sql-085418-zhs.html 官网中文学习网址

  10. [模式识别].(希腊)西奥多里蒂斯&lt;第四版&gt;笔记5之__特征选取

    1,引言 有关模式识别的一个主要问题是维数灾难.我们将在第7章看到维数非常easy变得非常大. 减少维数的必要性有几方面的原因.计算复杂度是一个方面.还有一个有关分类器的泛化性能. 因此,本章的主要任 ...