POJ 3087 Shuffle'm Up

Posted 2020-07-24 tech-chen

POJ 3087 Shuffle‘m Up

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8843		Accepted: 4078

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 2^nd chip from the bottom of S₂ and placing that on S₁₂, followed by the 2^nd 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

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"。

 

 1 /*这个模拟即可，注意要判断重复状态，因为洗牌的规则是一定的，如果洗牌的状态出现重复，则说明一定洗不到了。
 2 还有要注意调用系统自带的字符串函数的时候，字符串函数的末尾必须有‘\0‘结束符才可以有效的。自己for循环复制的字符数组结尾时没有‘\0‘的，要自己补上
 3 */
 4 #include<iostream>
 5 using namespace std;
 6 #include<cstdio>
 7 #define N 201
 8 #include<map>
 9 #include<cstring>
10 int test;
11 int main()
12 {
13     scanf("%d",&test);
14     for(int i=1;i<=test;++i)
15     {
16         char s1[N],s2[N],goals[N];
17         int len,ans=0;
18         bool flag=true;
19         map<string,bool>ma;
20         scanf("%d%s%s%s",&len,s1,s2,goals);
21         while(1)
22         {
23             char temp[N];
24             for(int i=0;i<len;++i)
25             {
26                 temp[2*i+1]=s1[i];
27                 temp[2*i]=s2[i];
28             }
29             temp[2*len]=‘\0‘;//
30             ans++;
31             if(!strcmp(temp,goals))
32             {
33                 break;
34             }
35             if(ma[temp])
36             {
37                 flag=false;
38                 break;
39             }
40             ma[temp]=true;
41             for(int i=0;i<len;++i)
42             {
43                 s2[i]=temp[i+len];
44                 s1[i]=temp[i];
45             }
46             s1[len]=s2[len]=‘\0‘;
47         }
48         if(flag)
49           printf("%d %d\n",i,ans);
50         else printf("%d %d\n",i,-1);
51     }
52     return 0;
53 }