poj 3087

http://poj.org/problem?id=3087

Shuffle'm Up

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13803		Accepted: 6356

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 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

Greater New York 2006

 

题意：有两堆牌s1,s2。每次两堆牌交替，s2的第一张放在最下面，然后s1的第一张放在上面，如题目中的图。

要求s1与s2 的操作完成后，与输入第三行相同。假如第一次完成操作后，与第三个不相同，则，将底部C张牌作为s1，上面C张作为s2.

若无法找到输出-1。

思路：一遍一遍模拟，将每种状态存进map中，若map中出现过，即输出-1.

#include<cstdio>
#include<cstring>
#include<map>
#include<string>
using namespace std;
const int maxn=1006;
char s1[maxn],s2[maxn],s12[maxn*2],d1[maxn*2];///

void INIT()
{
    memset( s1, 0, sizeof s1);
    memset( s2, 0, sizeof s2);
    memset( s12, 0, sizeof s12);
    memset( d1, 0, sizeof d1);
}

int main()
{
    int n;
    while( ~scanf("%d",&n)){
        for(int jkl=1;jkl<=n;jkl++){
            INIT();
            map< string,int> fin;
            fin.clear();
            int c,ans=1,fol=0;
            scanf("%d%s%s%s",&c,s1,s2,s12);///输入c，s1，s2，s12
            printf("%d ",jkl);
            for(;;){
                fol=0;
                for(int i=0;i<c;i++){
                    d1[fol++]=s2[i];
                    d1[fol++]=s1[i];
                }
                fol=0;
                for(int i=0;i<c;i++){
                    s1[i]=d1[fol++];
                }
                for(int i=0;i<c;i++){
                    s2[i]=d1[fol++];
                }
                if(strcmp(s12,d1)==0){
                    printf("%d
",ans);
                    break;
                }
                if(fin[d1]==1){
                    printf("-1
");
                    break;
                }
                fin[d1]=1;
                ans++;
            }
        }
    }
    return 0;
}