Description

A diamond puzzle is played on a tessellated hexagon like the one shown in Figure 1 below. And in this problem the faces produced by the tessellation are identified as they are numbered in the same figure. If two faces share a side, they are called neighboring faces. Thus, even-numbered faces have three neighboring faces, while odd-numbered faces have only two. At any point during the play of the puzzle, six of the seven faces hold a unique digit ranging from 1 to 6, and the other one is empty. A move in the puzzle is to move a digit from one face to a neighboring empty one.

Starting from any configuration, some series of moves can always make the puzzle look identical to either one shown in Figures 2 and 3. Your task is to calculate the minimum number of moves to make it become the one inFigure 2.

Input

Output

Sample Input

3
1324506
2410653
0123456

Sample Output

10
-1
0

渣渣水平看了题解，果断bfs走起。ps：博客写的太简单了，因为时间比较紧。所以以后时间充足的话还是详细点好。

题意：由当前状态得到最终状态需要最小步数。只允许在空格处移动=-=，从磨状态开始搜

所以就记录空格所在位子以及由此空格可以移动到的位置。比如：空格在位置2，则3，0，1号位置都可以移动到此位置。

另外的对于我来说就是STL里map的使用了，映射关系map<string,int>visit记录有木有被访问过，map<string,int>ans记录到谋一状态所需的最小步数。下面贴挫码。思路来源：点击打开链接

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<queue>
#include<map>
using namespace std;
struct node{
    char s[10];
    int step;
};
map<string,int>visit;
map<string,int>ans;
char s1[10],s2[10]="0123456";
int dr[8][5]={{2,4,6,-1},{2,6,-1},{1,3,0,-1},{2,4,-1},{3,5,0,-1},{4,6,-1},{1,5,0,-1}};//记录从0开始的每个位置如果为0时其他可以移动到此位置的点的位置
void bfs()
{
    queue<node>q;
    node st,ed;
    strcpy(st.s,s2);
    st.step=0;
    visit[s2]=1;
    q.push(st);
    while(!q.empty())
    {
        st=q.front();
        q.pop();

        ans[st.s]=st.step;
        int temp;
        for(int i=0;i<7;i++)//找到0的位置
        {
            if(st.s[i]=='0')
            {
               temp=i;
               break;
            }
        }
        for(int i=0;dr[temp][i]!=-1;i++)//对于可以到0的位置依次搜索
        {
            char ss[10];
            strcpy(ss,st.s);
            ss[temp]=st.s[dr[temp][i]];//交换位置
            ss[dr[temp][i]]='0';
            if(!visit[ss])//没有被访问过
            {
                visit[ss]=1;
                strcpy(ed.s,ss);
                ed.step=st.step+1;
                q.push(ed);
            }
        }
    }
}
int main()
{
    bfs();
    int n;
    scanf("%d",&n);
    while(n--)
    {
        getchar();
        scanf("%s",s1);
        if(!strcmp(s1,s2))
            printf("%d\n",0);
        else
            printf("%d\n",ans[s1]==0?-1:ans[s1]);
    }
    return 0;
}