POJ 3087 Shuffle'm Up （模拟）

Shuffle‘m Up

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 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 S₁and 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

Greater New York 2006

一道模拟题，

解题思路：这个直接模拟（或者退化的BFS或DFS吧，就一种情况，汗。~）题目所要求的Shuffle的过程就行了，只是要注意搜索失败的条件。当S1和S2又回到初始化状态时就是搜索失败的条件，为什么会这样呢？把所有的情况放在一个集合里面，Shuffle会模拟到所有情况，直到回到初始状态，无限循环下去。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define M 110
char mat[220],S1[M],S2[M];
char map[1000][220];
int main()
{
  int n,m,i,j,x,k;
  while(scanf("%d",&m)!=EOF&&m)
  {
      for(k=1;k<=m;k++)
      {
          memset(map,‘0‘,sizeof(map));
          memset(mat,‘0‘,sizeof(mat));
          scanf("%d%s%s%s",&n,S1,S2,mat);
          x=1;
          while(1)
          {
             for(i=0;i<n;i++)
             {
                 map[x][2*i]=S2[i];  //S2放下面。
                 map[x][2*i+1]=S1[i]; //S1放上面。
             }
            map[x][2*n]=NULL;          //要截断，不然根本匹配不上。
            if(strcmp(map[x],mat)==0)
            break;
            for(i=0;i<x;i++)
            if(strcmp(map[i],map[x])==0)
            {x=-1;break;}     //这个break没跳出while只是跳出for循环。
            for(i=0;i<=n-1;i++)
            S1[i]=map[x][i];
            for(i=n;i<2*n;i++)
            S2[i-n]=map[x][i];
            if(x==-1) break;
            x++;
          }
          printf("%d %d\n",k,x);
   }
  }
  return 0;
}

POJ 3087 Shuffle'm Up （模拟）,布布扣,bubuko.com

POJ 3087 Shuffle'm Up （模拟）