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

依照要求模拟一遍。注意分配的顺序。

模拟退火呀

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<map>
using namespace std;
char s1[110],s2[110];
char s[110*2];
int n,c; int main()
{
cin>>n;
for(int t=1;t<=n;t++)
{
cin>>c>>s1>>s2>>s;
map<string,bool>visit;//标志映射
visit[s]=true;
int step=0;
while(true)
{
char str[110*2];
int len=0;
for(int i=0;i<c;i++)
{
str[len++]=s2[i];
str[len++]=s1[i];
}
str[len]='\0';
step++;
if(!strcmp(str,s))
{
cout<<t<<" "<<step<<endl;
break;
}
else if(visit[str]&&strcmp(str,s))//出现过但又不是目标状态退出
{
cout<<t<<" "<<-1<<endl;
break;
}
visit[str]=true;
for(int i=0;i<c;i++)
s1[i]=str[i];//发s1
s1[c]='\0';
for(int i=c,k=0;i<2*c;i++)
s2[k++]=str[i];//发s2
s2[c]='\0';
}
}
return 0;
}

POJ 3087 Shuffle&#39;m Up(模拟退火)的更多相关文章

  1. POJ 3087 Shuffle&#39;m Up(模拟)

    Shuffle'm Up Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7404   Accepted: 3421 Desc ...

  2. 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 ...

  3. 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 ...

  4. 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. */ ...

  5. POJ 3087 Shuffle'm Up

    Shuffle'm Up Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit ...

  6. POJ 3087 Shuffle'm Up 线性同余,暴力 难度:2

    http://poj.org/problem?id=3087 设:s1={A1,A2,A3,...Ac} s2={Ac+1,Ac+2,Ac+3,....A2c} 则 合在一起成为 Ac+1,A1,Ac ...

  7. 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 ...

  8. 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 ...

  9. POJ 3087 Shuffle'm Up DFS

    link:http://poj.org/problem?id=3087 题意:给你两串字串(必定偶数长),按照扑克牌那样的洗法(每次从S2堆底中拿第一张,再从S1堆底拿一张放在上面),洗好后的一堆可以 ...

随机推荐

  1. SSL 认证之后,request.getScheme()获取不到https的问题记录

    通过浏览器输入https://www.xxx.com,request.getScheme()获取到的确实http而不是https通过request.getRequestURL()拿到的也是http:/ ...

  2. UITabBarController 详解

    // UITabBarController 标签视图控制 // 主要管理没有层级关系的视图控制器 // 1. ViewControllers 所有被管理的视图控制器, 都在这个数组中 // 2. se ...

  3. 【iOS开发】canOpenURL: failed for URL

    控制台输出 如图是在我启动一个 Xcode 7 + iOS 9 的 App 之后,控制台的输出. 这在 Xcode 6.4 + iOS 8 时,是不会有的情况,原因是[为了强制增强数据访问安全, iO ...

  4. jquery ajax 不执行赋值,return没有返回值的解决方法

    大家先看一段简单的jquery ajax 返回值的js 复制代码 代码如下: function getReturnAjax{ $.ajax({ type:"POST", url:& ...

  5. centos7 安装python3和pip3

    centos7默认是安装的python2.7以及对于的pip 如果要使用python3并且保留python2请看以下步骤 sudo yum -y install epel-release sudo y ...

  6. 《Linux操作系统编译构建指南》

    在线阅读地址:http://www.doc88.com/p-5126905896771.html Linux编译构建定制qq群: 521902245 文件夹...0 前言...3 第零章 绪论...5 ...

  7. wampserver 下载链接没反应的解决办法

    可能有很多小伙伴和我一样使用wampserver时,下载链接点击就是没有反应,当时我以为是因为网络原因,链接没有加载出来,或者是链接的请求不能得到响应,结果百度了一下才发现被“习惯”坑了一把,wamp ...

  8. CocoSourcesCS 2

    CocoSourcesCS 2 /*------------------------------------------------------------------------- DFA.cs - ...

  9. [React] Detect user activity with a custom useIdle React Hook

    If the user hasn't used your application for a few minutes, you may want to log them out of the appl ...

  10. 爪哇国新游记之七----使用ArrayList统计水果出现次数

    之前学习制作了DArray,了解ArrayList就容易了. /** * 用于存储水果名及数量 * */ public class Fruit{ private String name; public ...