POJ 3087 Shuffle'm Up

Shuffle'm Up

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3087

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 * Cuppercase 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

 
注：本人英语很渣，题目大意大多来自百度~=0=

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"。

Input:

2//有两个测试样例
4//每堆牌有几张
AHAH//s1
HAHA//s2
HHAAAAHH//需要达到的状态
3
CDE
CDE
EEDDCC
Output

1 2 //需要输出第几个样例 第几次洗牌达到要求
2 -1

本来是被分类到广搜的， 在网上看到大神用模拟 就试着他们的方法写了下 很愉快地A掉了

下面是代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cstdlib>
#include <cmath>
#include <cctype>
#define N 210

using namespace std;

struct node
{
    char s[N];
    struct node *next;
}*head, *last, *q;

int main()
{
    int c, sample, ans, T, i;
    char s1[N], s2[N], str[N], temp[N];//s1, s2第几堆牌， str需要达到的状态， temp用来保存每次洗牌后的状态
    scanf("%d", &T);

    for(sample = 1; sample <= T; sample++)//输出时需要输出第几个样例，sample来保存
    {
        scanf("%d %s %s %s", &c, s1, s2, str);

        head = (node *)malloc(sizeof(node));
        head->next = NULL;
        last = head;
        ans = 0;

        while(1)
        {
            ++ans;
            for(i = 0; i < c; i++)
            {
                temp[2 * i] = s2[i];
                temp[2 * i + 1] = s1[i];
            }
            temp[2 * i] = '';

            if(!strcmp(temp, str))//已经得到想要的状态
            {
                printf("%d %d
", sample, ans);
                break;
            }

            bool flag = false;
            q = head;
            while(q->next != NULL) {//查找此状态有没有出现过
                if(!strcmp(q->s, temp)){
                    flag = true;
                    break;
                }
                q = q->next;
            }

            if(flag) {//状态重复出现，达不到想要状态
                printf("%d -1
", sample);
                break;
            }

            q = (node *)malloc(sizeof (node));
            strcpy(q->s, temp);
            last->next = q;
            last=q;
            q->next=NULL;

            for(i = 0; i < c; i++) {
                s1[i] = temp[i];
                s2[i] = temp[i + c];
            }
        }
    }
}