POJ 3087 Shuffle'm Up 【简单搜索】【bfs】【模拟】

Shuffle'm Up

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15347		Accepted: 7010

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 2nd chip from the bottom of S2 and placing that on S12, followed by the 2nd 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：有两个扑克牌堆：s1和s2

两个牌堆均有c张牌，每次洗牌将s2牌堆最下面那张牌放到最下面，然后再放上s1最下面那张牌。这样可以形成一个新的牌堆：s12。 然后再把s12最下面n张牌抽出组成新的s1堆，上面的n张牌组成s2堆。然后再次进行同样的操作，问操作多少次后能行程所给的序列。

如果不能则输出-1。

 

这题相当于一个只有一条通道的bfs(因为洗牌方式是确定的)，所以我们只需要注意循环跳出条件即可。也就是当s1,s2同时分成了到以前出现过的序列，即出现了循环节。注意，必须是两者同时到达循环节。这里我们用map和pair进行记录。

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;
#define LL long long
#define M(a,b) memset(a,b,sizeof(a))
const int MAXN = 305;
const int INF =  0x3f3f3f3f;
int t,n,flag,num;
int Case = 1;
char str1[MAXN],str2[MAXN],ans[MAXN],ans2[MAXN];
map< pair<string,string>,int>M;
int main()
{
    cin>>t;
    while(t--)
    {
        M.clear();///注意清空
        num = flag = 0;
        scanf("%d",&n);
        scanf("%s",str1);
        scanf("%s",str2);
        scanf("%s",ans);
        M[make_pair(str1,str2)] = 1;///初始情况标记为1

        while(strcmp(ans2,ans)!=0)
        {
            num++;
            ///模拟洗牌

            for(int i = 0; i<n; i++)
            {
                ans2[i*2] = str2[i];
            }
            for(int i = 0; i<n; i++)
            {
                ans2[i*2+1] = str1[i];
            }
            ans2[2*n] = '\0';
            ///模拟拆牌
            for(int i=0; i<n; i++)
            {
                str1[i] = ans2[i];
            }
            str1[n] = '\0';
            for(int i=n; i<2*n; i++)
            {
                str2[i-n] = ans2[i];
            }
            str2[n] = '\0';

            if(M[make_pair(str1,str2)]==1)///判断是否有循环节
            {
                flag = 1;
                break;
            }
            else
            {
                M[make_pair(str1,str2)] = 1;
            }
        }
        if(flag)
        {
            printf("%d -1\n",Case++);
        }
        else
        {
            printf("%d %d\n",Case++,num);
        }
    }
    return 0;
}
//S1 :AHAH
//S2 :HAHA
//S12 : HAAHHAAH

//S1: HAAH
//S2: HAAH
//S12:HHAAAAHH