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            poj3087

            Shuffle'm Up

            Time Limit: 1000MS Memory Limit: 65536K
            Total Submissions: 3434 Accepted: 1602

            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
            題意就是兩坨牌,洗牌,看能不能到達所求的牌的狀態,如果能到達,最少幾步
            但是這題有個規定是,洗牌的規則是唯一的,所以完全可以用模擬來做
            我們可以把牌插起來,然后再平分成兩半,然后再插,直到能到達所求狀態或者出現了重復狀態
            或者從目標狀態出發找原來的狀態,同上
            在判重的時候不必記錄已經走過的狀態,只要記錄走第一步后的狀態,如果狀態和第一步后的狀態一樣的話,說明不能得到解
             1#include<stdio.h>
             2#include<string.h>
             3#include<math.h>
             4char s1[105],s2[150],s3[300],tmp[300],case1[150],case2[150];
             5char str1[150],str2[150];
             6int c;
             7int getans()
             8{
             9    int num,len,i;
            10    len=c*2;
            11    num=0;
            12    strcpy(tmp,s3);
            13    while(1)
            14    {
            15        for(i=0; i<len; i++)
            16            if(i%2==0)
            17            {
            18                str2[i/2]=tmp[i];
            19            }

            20            else if(i%2==1)
            21            {
            22                str1[i/2]=tmp[i];
            23            }

            24        str1[c]='\0';
            25        str2[c]='\0';
            26        if(num==0)
            27        {
            28            strcpy(case1,str1);
            29            strcpy(case2,str2);
            30        }

            31        num++;
            32        /*printf("%s\n",str1);
            33        printf("%s\n",str2);
            34        system("pause");*/

            35        if((strcmp(str1,s1)==0&&strcmp(str2,s2)==0))
            36            return num;
            37        else if(num!=1&&(strcmp(case1,str1)==0&&strcmp(case2,str2)==0))
            38        {
            39            return -1;
            40        }

            41        for(i=0; i<c; i++)
            42        {
            43            tmp[i]=str1[i];
            44        }

            45        for(i=c; i<len; i++)
            46        {
            47            tmp[i]=str2[i-c];
            48        }

            49        tmp[len]='\0';
            50    }

            51    /*if((strcmp(str1,s1)==0&&strcmp(str2,s2)==0))
            52        return num;
            53    else return -1;*/

            54}

            55int main()
            56{
            57    int t,i;
            58    scanf("%d",&t);
            59    for(i=1; i<=t; i++)
            60    {
            61        scanf("%d",&c);
            62        scanf("%s",&s1);
            63        scanf("%s",&s2);
            64        scanf("%s",&s3);
            65        printf("%d %d\n",i,getans());
            66    }

            67    return 0;
            68}

            69
            70/*
            71!(strcmp(str1,s1)==0&&strcmp(str2,s2)==0)
            72*/

            73

            posted on 2012-03-15 20:58 jh818012 閱讀(169) 評論(0)  編輯 收藏 引用

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            • 1.?re: poj1426
            • 我嚓,,輝哥,,居然搜到你的題解了
            • --season
            • 2.?re: poj3083
            • @王私江
              (8+i)&3 相當于是 取余3的意思 因為 3 的 二進制是 000011 和(8+i)
            • --游客
            • 3.?re: poj3414[未登錄]
            • @王私江
              0ms
            • --jh818012
            • 4.?re: poj3414
            • 200+行,跑了多少ms呢?我的130+行哦,你菜啦,哈哈。
            • --王私江
            • 5.?re: poj1426
            • 評論內容較長,點擊標題查看
            • --王私江
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