POJ3087:Shuffle'm Up(模拟)
http://poj.org/problem?id=3087
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块牌归为s1,最顶的c块牌归为s2,依此循环下去。现在输入s1和s2的初始状态 以及 预想的最终状态
s12问s1 s2经过多少次洗牌之后,最终能达到状态s12,若永远不可能相同,则输出"-1"。
题目解析:
水题一道却放在bfs训练计划中,害我不敢做。最主要的部分就是状态记录,然后判重。
若s1和s2在洗牌后的状态,是前面洗牌时已经出现过的一个状态,且这个状态不是预想的状态S12,就说明无论怎样再洗牌都不可能达到S12了,因为这个洗牌操作已经陷入了一个“环”。
如果状态没有重复过,则一直模拟洗牌,直至s12出现。
记录状态可以用map<string,int>q。
Map的缺省值为0
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <map>
using namespace std;
int n;
int main()
{
int T,sum,K=;
scanf("%d",&T);
char s1[],s2[],s12[];
char d[];
while(T--)
{
++K;
sum=;
scanf("%d",&n);
scanf("%*c%s%s%s",s1,s2,s12);
map<string,int>q;
q[s12]=;
while()
{
int z=;
for(int i=; i<n; i++)
{
d[z++]=s2[i];
d[z++]=s1[i];
}
d[z]='\0';
sum++;
if(strcmp(d,s12)==)
{
printf("%d %d\n",K,sum);
break;
}
else if(q[d]!=)
{
printf("%d -1\n",K);
break;
}
q[d]=;
for(int i=; i<n; i++)
{
s1[i]=d[i];
}
s1[n]='\0';
for(int i=,j=n; j<z; j++,i++)
s2[i]=d[j];
s2[n]='\0';
}
}
return ;
}
POJ3087:Shuffle'm Up(模拟)的更多相关文章
- POJ3087 Shuffle'm Up(模拟)
题目链接. AC代码如下; #include <iostream> #include <cstdio> #include <cstring> #include &l ...
- poj3087 Shuffle'm Up(模拟)
Shuffle'm Up Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10766 Accepted: 4976 Des ...
- POJ-3087 Shuffle'm Up (模拟)
Description A common pastime for poker players at a poker table is to shuffle stacks of chips. Shuff ...
- POJ3087 Shuffle'm Up 简单模拟
题意:就是给你两副扑克,然后一张盖一张洗牌,不断重复这个过程,看能不能达到目标的扑克顺序 分析:然后就模拟下,-1的情况就是有循环节 #include<cstdio> #include&l ...
- 【POJ - 3087】Shuffle'm Up(模拟)
Shuffle'm Up 直接写中文了 Descriptions: 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, ...
- 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 ...
- POJ3087 Shuffle'm Up —— 打表找规律 / map判重
题目链接:http://poj.org/problem?id=3087 Shuffle'm Up Time Limit: 1000MS Memory Limit: 65536K Total Sub ...
- poj3087 Shuffle'm Up
Description A common pastime for poker players at a poker table is to shuffle stacks of chips. Shuff ...
- POJ 3078 - Shuffle'm Up - [模拟题]
题目链接:http://poj.org/problem?id=3087 Description A common pastime for poker players at a poker table ...
随机推荐
- [转]java中判断字符串是否为数字的三种方法
1用JAVA自带的函数public static boolean isNumeric(String str){ for (int i = str.length();--i>=0;){ ...
- WP8.1学习系列(第九章)——透视Pivot开发指南
Windows Phone 8 的 Pivot 控件 2014/6/18 适用于:Windows Phone 8 和 Windows Phone Silverlight 8.1 | Windows P ...
- Material Design系列第四篇——Defining Shadows and Clipping Views
Defining Shadows and Clipping Views This lesson teaches you to Assign Elevation to Your Views Custom ...
- Email standards
https://www.fastmail.com/help/technical/standards.html Email structure These RFCs define the way ema ...
- jumpserver的安装
原文地址:http://docs.jumpserver.org/zh/docs/step_by_step.html 为了保证服务器安全,加个堡垒机,所有ssh连接都通过堡垒机来完成,堡垒机也需要有身份 ...
- SVN服务端安装
1 首先安装SVN和Subversion. 安装文件可自行百度. 2 在服务端创建版本库. 我的安装目录是c:\Program Files(x86)\Subversion. 安装完成后在安装目录下sh ...
- 云存储命令行工具---libs3
ceph 的客户端有很多,有s3cmd.cloudberryExplorer等,今天介绍另一个libs3 一. 安装 Libs3是RGW s3接口的命令行工具,与s3cmd类似,使用C++生成. 1. ...
- 为自定义的View添加长按事件
以前开发画板组件时,要添加一个长按监听事件,这个画板实际上就是继承自View的一个自定义组件. 首先,设置好长按事件发生时要触发的操作: private class LongPressRunnable ...
- HDU-1394 Minimum Inversion Number(线段树求逆序数)
Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Ot ...
- c++从文件中读取一行数据并保存在数组中
从txt文本中读取数据存入数组中 #include <iostream> #include <fstream> #include <string> #include ...