POJ-3087 Shuffle'm Up (模拟)
Description
A common pastime for poker players at a poker table is to shuffle stacks of chips. Shuffling chips is performed by starting with two stacks of poker chips, S1 and S2, each stack containing C chips. Each stack may contain chips of several different colors.
The actual shuffle operation is performed by interleaving a chip from S1 with a chip from S2 as shown below for C = 5:

The single resultant stack, S12, contains 2 * C chips. The bottommost chip of S12 is the bottommost chip from S2. On top of that chip, is the bottommost chip from S1. The interleaving process continues taking the 2nd chip from the bottom of S2 and placing that on S12, followed by the 2nd chip from the bottom of S1 and so on until the topmost chip from S1 is placed on top of S12.
After the shuffle operation, S12 is split into 2 new stacks by taking the bottommost C chips from S12 to form a new S1 and the topmost C chips from S12 to form a new S2. The shuffle operation may then be repeated to form a new S12.
For this problem, you will write a program to determine if a particular resultant stack S12 can be formed by shuffling two stacks some number of times.
Input
The first line of input contains a single integer N, (1 ≤ N ≤ 1000) which is the number of datasets that follow.
Each dataset consists of four lines of input. The first line of a dataset specifies an integer C, (1 ≤ C ≤ 100) which is the number of chips in each initial stack (S1 and S2). The second line of each dataset specifies the colors of each of the C chips in stack S1, starting with the bottommost chip. The third line of each dataset specifies the colors of each of the C chips in stack S2 starting with the bottommost chip. Colors are expressed as a single uppercase letter (A through H). There are no blanks or separators between the chip colors. The fourth line of each dataset contains 2 * C uppercase letters (A through H), representing the colors of the desired result of the shuffling of S1 and S2 zero or more times. The bottommost chip’s color is specified first.
Output
Output for each dataset consists of a single line that displays the dataset number (1 though N), a space, and an integer value which is the minimum number of shuffle operations required to get the desired resultant stack. If the desired result can not be reached using the input for the dataset, display the value negative 1 (−1) for the number of shuffle operations.
Sample Input
2
4
AHAH
HAHA
HHAAAAHH
3
CDE
CDE
EEDDCC
Sample Output
1 2
2 -1 题目分析:按题意模拟即可。 代码如下:
# include<iostream>
# include<cstdio>
# include<map>
# include<string>
# include<cstring>
# include<algorithm>
using namespace std;
string unit(string s1,string s2)
{
string res="";
int i=,j=,l=s1.size();
for(int k=;k<*l;++k){
if(k&)
res+=s1[i++];
else
res+=s2[j++];
}
return res;
}
void solve(string s1,string s2,string S,int l)
{
map<string,int>mp;
int cnt=;
string s=unit(s1,s2);
while(){
++mp[s];
++cnt;
if(s==S){
printf("%d\n",cnt);
return ;
}
s1=s.substr(,l);
s2=s.substr(l,l);
s=unit(s1,s2);
if(mp[s]>=)
break;
}
printf("-1\n");
}
int main()
{
int T,l,cas=;
string s1,s2,s;
scanf("%d\n",&T);
while(T--)
{
cin>>l;
cin>>s1>>s2>>s;
printf("%d ",++cas);
solve(s1,s2,s,l);
}
return ;
}
POJ-3087 Shuffle'm Up (模拟)的更多相关文章
- POJ 3087 Shuffle'm Up (模拟+map)
题目链接:http://poj.org/problem?id=3087 题目大意:已知两堆牌s1和s2的初始状态, 其牌数均为c,按给定规则能将他们相互交叉组合成一堆牌s12,再将s12的最底下的c块 ...
- poj 3087 Shuffle'm Up (模拟过程)
Description A common pastime for poker players at a poker table is to shuffle stacks of chips. Shuff ...
- POJ 3087 Shuffle'm Up 模拟,看着不像搜索啊
题意:给定s1,s1两副扑克,顺序从下到上.依次将s2,s1的扑克一张一张混合.例如s1,ABC; s2,DEF. 则第一次混合后为DAEBFC. 然后令前半段为s1, 后半段为s2. 如果可以变换成 ...
- POJ.3087 Shuffle'm Up (模拟)
POJ.3087 Shuffle'm Up (模拟) 题意分析 给定两个长度为len的字符串s1和s2, 接着给出一个长度为len*2的字符串s12. 将字符串s1和s2通过一定的变换变成s12,找到 ...
- POJ 3087 Shuffle'm Up(洗牌)
POJ 3087 Shuffle'm Up(洗牌) Time Limit: 1000MS Memory Limit: 65536K Description - 题目描述 A common pas ...
- DFS POJ 3087 Shuffle'm Up
题目传送门 /* 题意:两块扑克牌按照顺序叠起来后,把下半部分给第一块,上半部给第二块,一直持续下去,直到叠成指定的样子 DFS:直接模拟搜索,用map记录该字符串是否被搜过.读懂题目是关键. */ ...
- POJ 3087 Shuffle'm Up
Shuffle'm Up Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit ...
- poj 3087 Shuffle'm Up ( map 模拟 )
题目:http://poj.org/problem?id=3087 题意:已知两堆牌s1和s2的初始状态, 其牌数均为c,按给定规则能将他们相互交叉组合成一堆牌s12,再将s12的最底下的c块牌归为s ...
- POJ 3087 Shuffle'm Up(模拟)
Shuffle'm Up Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7404 Accepted: 3421 Desc ...
- POJ 3087 Shuffle'm Up【模拟/map/string】
Shuffle'm Up Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 14471 Accepted: 6633 Descrip ...
随机推荐
- 【基于EF Core的Code First模式的DotNetCore快速开发框架】完成对DB First代码生成的支持
前言 距离上一篇文章<基于EF Core的Code First模式的DotNetCore快速开发框架>已过去大半个年头,时光荏苒,岁月如梭...比较尴尬的是,在这大半个年头里,除了日常带娃 ...
- Django框架----数据库表的单表查询
一.添加表记录 对于单表有两种方式 # 添加数据的两种方式 # 方式一:实例化对象就是一条表记录 Frank_obj = models.Student(name ="海东",cou ...
- ab命令压力测试
网站性能压力测试是服务器网站性能调优过程中必不可缺少的一环.只有让服务器处在高压情况下,才能真正体现出软件.硬件等各种设置不当所暴露出的问题. 性能测试工具目前最常见的有以下几种:ab.http_lo ...
- 关于sqlite使用场景
对于sqlite,实际中从来没有用过,也几乎没有考虑过其使用场景,更不要说专门去研究它了,今天看最新的数据库流行度排行榜的时候,发现sqlite的长期趋势好像一直在第十位左右徘徊,特地搜索了下其使用场 ...
- jenkins 安装 + maven + git部署
1. 安装JDK 2. 安装maven 3. 安装git 4. 安装tomcat tar zxvf apache-tomcat-8.5.14.tar.gz 找到tomcat-->config-- ...
- CentOS 7下搭建配置SVN服务器
CentOS 7下搭建配置SVN服务器 1. 安装 CentOS通过yum安装subversion. $ sudo yum install subversion subversion安装在/bin目录 ...
- Visual Status各个版本官网下载
网址:https://www.visualstudio.com/zh-hans/vs/older-downloads/
- [VS 2015] VS2015 完整ISO镜像包
区别 :https://www.visualstudio.com/zh-cn/products/compare-visual-studio-2015-products-vs 完整ISO镜像:下载 VS ...
- Linux内存管理--虚拟地址、逻辑地址、线性地址和物理地址的区别(二)【转】
本文转载自:http://blog.csdn.net/yusiguyuan/article/details/9668363 这篇文章中介绍了四个名词的概念,下面针对四个地址的转换进行分析 CPU将一个 ...
- 《js高级程序设计》--第三章数据类型
一.关键字 二.保留字 三.数据类型 (数据类型具有动态性) 1.Undefined 声明变量却未对其加以初始化(赋值) 2.Null null值表示一个空对象指针,而这也正是使用typeof操作 ...