POJ 3087 Shuffle'm Up

素颜马尾好姑娘i 2022-06-18 10:17 118阅读 0赞

Shuffle'm Up

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;10637</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;4918</td> 
  </tr> 
 </tbody> 
</table>

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:

![3087_1.png][]

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"。

代码：

#include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdiO>
    #include <map>
    #include <string>
    #include <queue>
    
    using namespace std;
    
    typedef long long ll;
    map<string,int>m;
    string s;
    int n;
    
    int bfs(string s1,string s2)
    {
        queue<string>q;
        string s12 = "";
        for(int i=0;i<n;i++)
        {
            s12 += s2[i];
            s12 += s1[i];
        }
        m[s12] = 1;
        q.push(s12);
        while(!q.empty())
        {
            string s3 = q.front();
            q.pop();
            if(s3 == s)
                return m[s3];
            s1 = s3.substr(0,n);
            s2 = s3.substr(n,n);
            string s12 = "";
            for(int i=0;i<n;i++)
            {
                s12 += s2[i];
                s12 += s1[i];
            }
            if(m[s12] > 0)//判断是否无尽循环
                return -1;
            m[s12] = m[s3] + 1;//统计次数
            q.push(s12);//入队
        }
        return -1;
    }
    int main()
    {
        int t;
        cin>>t;
        for(int i=1;i<=t;i++)
        {
            cin>>n;
            string s1,s2;
            cin>>s1>>s2>>s;
            int ans = bfs(s1,s2);
            cout<<i<<" "<<ans<<endl;
        }
        return 0;
    }

[3087_1.png]: /images/20220618/155d03de6d154e6bae313320146379df.png