Dice - HDU 5012 搜索

最新推荐文章于 2021-01-19 20:52:36 发布

提比-我有特殊的AC技巧

最新推荐文章于 2021-01-19 20:52:36 发布

阅读量611

点赞数

CC 4.0 BY-SA版权

分类专栏：搜索 HDU

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/39459143

HDU 同时被 2 个专栏收录

133 篇文章

订阅专栏

搜索

12 篇文章

订阅专栏

探讨了通过四种特定操作将两个特殊骰子从不同初始状态变为完全相同状态所需的最少步骤数。采用广度优先搜索算法解决问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Dice

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 676 Accepted Submission(s): 385

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 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 5 6 4 3
1 2 3 4 5 6
1 4 2 5 3 6

Sample Output

0
3
-1

题意：一次翻转有4种方式，问你将骰子的一种方式变成另一种方式至少需要多少步。

思路：广搜就可以了。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
int vis[150000],vc[810][1000][7];
int t,pow7[10],num[7],ans,op;
bool flag;
int suan(int pos)
{
    int i,k=0;
    for(i=1;i<=6;i++)
       k+=num[i]*pow7[i];
    if(k==op)
    {
        flag=true;
        ans=pos+1;
    }
    if(vis[k]==t)
      return 0;
    else
    {
        vis[k]=t;
        return 1;
    }
}
void solve(int pos,int num2)
{
    if(flag)
      return;
    int i,j,k;
    //11111111111
    num[1]=vc[pos][num2][3];
    num[2]=vc[pos][num2][4];
    num[3]=vc[pos][num2][2];
    num[4]=vc[pos][num2][1];
    num[5]=vc[pos][num2][5];
    num[6]=vc[pos][num2][6];

    k=suan(pos);
    if(k==1)
    {
        vc[pos+1][0][0]++;
        k=vc[pos+1][0][0];
        for(i=1;i<=6;i++)
           vc[pos+1][k][i]=num[i];
    }
    num[1]=vc[pos][num2][5];
    num[2]=vc[pos][num2][6];
    num[3]=vc[pos][num2][3];
    num[4]=vc[pos][num2][4];
    num[5]=vc[pos][num2][2];
    num[6]=vc[pos][num2][1];
    k=suan(pos);
    if(k==1)
    {
        vc[pos+1][0][0]++;
        k=vc[pos+1][0][0];
        for(i=1;i<=6;i++)
           vc[pos+1][k][i]=num[i];
    }
    num[1]=vc[pos][num2][6];
    num[2]=vc[pos][num2][5];
    num[3]=vc[pos][num2][3];
    num[4]=vc[pos][num2][4];
    num[5]=vc[pos][num2][1];
    num[6]=vc[pos][num2][2];

    k=suan(pos);
    if(k==1)
    {
        vc[pos+1][0][0]++;
        k=vc[pos+1][0][0];
        for(i=1;i<=6;i++)
           vc[pos+1][k][i]=num[i];
    }
    num[1]=vc[pos][num2][4];
    num[2]=vc[pos][num2][3];
    num[3]=vc[pos][num2][1];
    num[4]=vc[pos][num2][2];
    num[5]=vc[pos][num2][5];
    num[6]=vc[pos][num2][6];

    k=suan(pos);
    if(k==1)
    {
        vc[pos+1][0][0]++;
        k=vc[pos+1][0][0];
        for(i=1;i<=6;i++)
           vc[pos+1][k][i]=num[i];
    }
}
int main()
{
    int n,m,i,j,k;
    pow7[1]=1;
    for(i=2;i<=6;i++)
       pow7[i]=pow7[i-1]*7;
    while(~scanf("%d",&vc[0][1][1]))
    {
        t++;
        for(i=2;i<=6;i++)
           scanf("%d",&vc[0][1][i]);
        vc[0][0][0]=1;

        k=0;
        for(i=1;i<=6;i++)
           scanf("%d",&num[i]);
        for(i=1;i<=6;i++)
           k+=num[i]*pow7[i];
        op=k;


        k=0;
        for(i=1;i<=6;i++)
           k+=vc[0][1][i]*pow7[i];
        if(k==op)
        {
            printf("0\n");
            continue;
        }

        for(i=1;i<=800;i++)
           vc[i][0][0]=0;
        flag=false;
        for(i=0;i<=800;i++)
        {
            //printf("solve %d\n",i);
            if(flag ||vc[i][0][0]==0)
              break;
            for(j=1;j<=vc[i][0][0];j++)
            {
                if(flag)
                  break;
                solve(i,j);
            }
        }
        if(flag)
          printf("%d\n",ans);
        else
          printf("-1\n");
    }
}