poj3087(模拟)Shuffle'm Up（简单学习一下map）

本文介绍了一个关于洗牌算法的比赛题目，玩家需要通过一系列操作将两堆扑克芯片混合并重新分配，以达到特定的目标状态。文章详细解释了洗牌过程，并提出了一种算法来判断是否能通过多次洗牌达到期望的结果。

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Shuffle'm Up

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9277		Accepted: 4256

Description

A common pastime(娱乐) for poker(拨火棍) players at a poker table is to shuffle(洗牌) stacks(堆) of chips. Shuffling chips is performed by starting with two stacks of poker chips, S₁ and S₂, each stack(堆) containing C chips. Each stack may contain chips of several different colors.

The actual shuffle(洗牌) operation is performed by interleaving(交错) a chip from S₁ with a chip from S₂ as shown below for C = 5:

The single resultant(结果的) stack, S₁₂, contains 2 * C chips. The bottommost(最低的) chip of S₁₂ is the bottommost(最低的) chip from S₂. On top of that chip, is the bottommost chip from S₁. The interleaving(交错) 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(洗牌) operation, S₁₂ is split into 2 new stacks(堆) by taking the bottommost(最低的) C chips from S₁₂ to form a new S₁ and the topmost C chips from S₁₂ to form a new S₂. The shuffle(洗牌) operation may then be repeated to form a new S₁₂.

For this problem, you will write a program to determine if a particular resultant(结果的) stack(堆) S₁₂ can be formed by shuffling two stacks some number of times.

Input

The first line of input(投入) contains a single integer(整数) N, (1 ≤ N ≤ 1000) which is the number of datasets that follow.

Each dataset consists of four lines of input(投入). The first line of a dataset specifies(指定) an integer(整数) C, (1 ≤ C ≤ 100) which is the number of chips in each initial stack ((堆)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(最低的) chip. The third line of each dataset specifies the colors of each of the C chips in stack S₂ starting with the bottommost(最低的) chip. Colors are expressed as a single uppercase(以大写字母印刷) letter (A through H). There are no blanks or separators(分离器) 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(支吾的) of S₁ and S₂ zero or more times. The bottommost chip’s color is specified f(指定)irst.

Output

Output(输出) for each dataset consists of a single line that displays the dataset number (1 though N), a space, and an integer(整数) value which is the minimum(最小的) number of shuffle(洗牌) operations required to get the desired resultant(结果的) stack(堆). If the desired result can not be reached using the input(投入) for the dataset, display the value negative(负的) 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

#include<stdio.h>
#include<map>
#include<string>
#include<string.h>
using namespace std;
int main()
{
    int t;
    int n;
    char s1[1000],s2[1000],s12[1000],ss[1000];
    scanf("%d",&t);
    int cou=0;
    while(t--)
    {
        cou++;
        int flag=0;
        scanf("%d",&n);
        map<string,bool>vis;
        scanf("%s%s%s",s1,s2,ss);

        vis[ss]=1;
        int step=0;
        while(1)
        {
               step++;
               int c=0;
               for(int i=0;i<n;i++)
              {
                s12[c++]=s2[i];
                s12[c++]=s1[i];
              }
               s12[c]='\0';
              if(strcmp(s12,ss)==0)
              {
                  flag=1;
                  //printf("%d\n",step);
                  break;
              }

              if(vis[s12]==1)
              {

                  //printf("-1\n");
                  break;
              }

              vis[s12]=1;

              for(int i=0;i<n;i++)
              {
                  s1[i]=s12[i];

              }

               c=0;
              for(int i=n;i<2*n;i++)
              {
                  s2[c++]=s12[i];
              }
              s2[c]='\0';
        }
        if(flag==1)
        printf("%d %d\n",cou,step);
        else printf("%d -1\n",cou);

    }
    return 0;
}