Poj3087
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8842 | Accepted: 4077 |
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 newS12.
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 theC 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 throughH). 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
Source
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
#define N 1010
char s1[N],s2[N],str[N],shuffle[N],goal[N];
int x,n;
inline int dfs(){
int step=,x,y,i;
for(;;){
step++;
x=y=;
for(int i=;i<*n;i++){
if(i%==)
shuffle[i]=s2[x++];
else
shuffle[i]=s1[y++];
}
shuffle[*n]='\0';
if(strcmp(shuffle,goal)==) return step;
if(strcmp(shuffle,str)==) return -;
strncpy(s1,shuffle,n);
s1[n]='\0';
strncpy(s2,shuffle+n,n);
s2[n]='\0';
}
}
int main(){
cin>>x;
for(int t=;t<=x;t++){
cin>>n;
cin>>s1>>s2>>goal;
strncpy(str,s1,n);
strncpy(str+n,s2,n);
str[*n]='\0';
cout<<t<<" "<<dfs()<<endl;
}
return ;
}
Poj3087的更多相关文章
- 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 ...
- POJ3087 Shuffle'm Up(模拟)
题目链接. AC代码如下; #include <iostream> #include <cstdio> #include <cstring> #include &l ...
- 搜索:POJ2251&POJ1426&POJ3087&POJ2488
图的遍历也称为搜索,就是从图中某个顶点出发,沿着一些边遍历图中所有的顶点,且每个顶点仅被访问一次,遍历可采取两种不同的方式:深度优先搜索(DFS)和广度优先搜索(BFS). 1.DFS算法思想` 从顶 ...
- poj3087 Shuffle'm Up(bfs)
http://poj.org/problem?id=3087 注意复制字符串的时候,要在末尾加上'\0',否则导致strcmp出错. 还有就是开数组大小的时候看清楚一点,别开错了debug了好久. # ...
- poj3087 Shuffle'm Up(模拟)
Shuffle'm Up Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10766 Accepted: 4976 Des ...
- 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 ...
- 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(KB1-G 简单搜索)
Shuffle'm Up Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10366 Accepted: 4800 Des ...
- POJ3087(模拟)
#include"iostream" #include"string" #include"map" using namespace std; ...
随机推荐
- FAT AP v200R005 配置二层透明模式(web&命令行,开局)
背景: vlan123:用户业务vlan,192.168.1.0/24 Vlan2001:管理vlan,172.168.129.0/24 vlan1:默认vlan,不建议使用. 注意事项: 配置服务集 ...
- maven 动态版本 aliyun阿里云Maven仓库地址——加速你的maven构建
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/20 ...
- Laravel之路(事务)mysql事务
其实关于mysql的事务(原声mysql语句),我在我的博客里面有提到(mysql的文章分类下) 今天看下基于laravel框架ORM的处理 准备: 表必须是InnoDB引擎 DB::beginTra ...
- UINavigationController 返回按钮去掉文字
[[UIBarButtonItem appearance] setBackButtonTitlePositionAdjustment:UIOffsetMake(0, -60) forBarMetric ...
- Android架构须知
1.了解不同版本号的特性包含IDE的. 如:AsyncTask3.0之后和之前的差别.Android 5.0的新的API.Android 6.0 不能用HttpClient .AS2.0的新特性 等等 ...
- atitit.jquery tmpl模板总结 .doc
atitit.jquery tmpl模板总结 .doc 1. atitit.动态模版解析1 1.1. Jquery.tmpl.js1 1.2. 比起anrular js方便啊.1 2. 动态模板引擎解 ...
- 使用SQLite
SQLite是一种嵌入式数据库,它的数据库就是一个文件.由于SQLite本身是C写的,而且体积很小,所以,经常被集成到各种应用程序中,甚至在iOS和Android的App中都可以集成. Python就 ...
- Greatest Number 山东省第一届省赛
Greatest Number Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描述 Saya likes math, because ...
- c# 多线程里面创建byte数组发生内存溢出异常求解
在多线程里面读取一个400多M的Xml文件,首先将其读入FileStream里面,然后,在执行 byte [] bts = new byte[fs.Length]; 这句代码时,出现内存溢出的异常,求 ...
- CSS学习笔记(3)--表格边框
http://www.alixixi.com/web/a/2009082657736.shtml 对于很多初学HTML的人来说,表格<table>是最常用的标签了,但对于表格边框的控制,很 ...