hdu 5012 Dice

Dice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 154    Accepted Submission(s): 89

Problem Description

There are 2 special dices on the table. On each face of the dice, a distinct number was written. Consider a₁.a₂,a₃,a₄,a₅,a₆ to be numbers written on top face, bottom face, left face, right face, front face and back face of dice A. Similarly, consider b₁.b₂,b₃,b₄,b₅,b₆ to be numbers on specific faces of dice B. It’s guaranteed that all numbers written on dices are integers no smaller than 1 and no more than 6 while a_i ≠ a_j and b_i ≠ b_j for all i ≠ j. Specially, sum of numbers on opposite faces may not be 7.

At the beginning, the two dices may face different(which means there exist some i, a_i ≠ b_i). Ddy wants to make the two dices look the same from all directions(which means for all i, a_i = b_i) only by the following four rotation operations.(Please read the picture for more information)


Now Ddy wants to calculate the minimal steps that he has to take to achieve his goal.

 

Input

There are multiple test cases. Please process till EOF. 

For each case, the first line consists of six integers a₁,a₂,a₃,a₄,a₅,a₆, representing the numbers on dice A. 

The second line consists of six integers b₁,b₂,b₃,b₄,b₅,b₆, representing the numbers on dice B.

 

Output

For each test case, print a line with a number representing the answer. If there’s no way to make two dices exactly the same, output -1.

 

Sample Input

1 2 3 4 5 61 2 3 4 5 61 2 3 4 5 61 2 5 6 4 31 2 3 4 5 61 4 2 5 3 6

 

Sample Output

03-1

 

题意：给你两个筛子的序列（上下左右），通过左右上下的旋转方式使得两个筛子完全一样，求最小的步数，如不能达到输出-1.

广搜，对每个状态的筛子，进行上下左右的搜索，并标记。

#include <iostream>#include <cmath>#include <queue>#include <stdio.h>using namespace std;int minstep;char a[8]="\0";struct node{    int Count;    char b[8];};bool visit[654322];void BFS(node num){    queue<node>Q;    Q.push(num);    int x=atoi(num.b);    visit[x]=1;    while (!Q.empty())    {        node h=Q.front();        Q.pop();        if(strcmp(h.b,a)==0)        {            if(minstep>h.Count)            {                minstep=h.Count;            }        }        else        {            //上            char temp[8]="\0";            temp[0]=h.b[4];            temp[1]=h.b[5];            temp[2]=h.b[2];            temp[3]=h.b[3];            temp[4]=h.b[1];            temp[5]=h.b[0];            int x=atoi(temp);            if(!visit[x])            {                visit[x]=1;                node hd=h;                strcpy(hd.b, temp);                hd.Count++;                Q.push(hd);                            }            //下            temp[0]=h.b[5];            temp[1]=h.b[4];            temp[2]=h.b[2];            temp[3]=h.b[3];            temp[4]=h.b[0];            temp[5]=h.b[1];            x=atoi(temp);            if(!visit[x])            {                visit[x]=1;                node hd=h;                strcpy(hd.b, temp);                hd.Count++;                Q.push(hd);            }                        //左            temp[0]=h.b[3];            temp[1]=h.b[2];            temp[2]=h.b[0];            temp[3]=h.b[1];            temp[4]=h.b[4];            temp[5]=h.b[5];            x=atoi(temp);            if(!visit[x])            {                visit[x]=1;                node hd=h;                strcpy(hd.b, temp);                hd.Count++;                Q.push(hd);            }                        //右            temp[0]=h.b[2];            temp[1]=h.b[3];            temp[2]=h.b[1];            temp[3]=h.b[0];            temp[4]=h.b[4];            temp[5]=h.b[5];            x=atoi(temp);            if(!visit[x])            {                visit[x]=1;                node hd=h;                strcpy(hd.b, temp);                hd.Count++;                Q.push(hd);            }        }    }}int main(){    while (cin>>a[0])    {        node num;        minstep=100000;        memset(visit, 0, sizeof(visit));        for (int i=1; i<=5; i++)        {            cin>>a[i];        }        num.b[6]='\0';        for (int i=0; i<=5; i++)        {            cin>>num.b[i];        }                num.Count=0;                BFS(num);       if(minstep==100000)       {           cout<<"-1"<<endl;       }       else       {           cout<<minstep<<endl;       }    }    return 0;}