hdu 5012 Dice BFS 2014 ACM/ICPC Asia Regional Xi'an Online

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5012

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

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

比赛的时候想复杂了，看了题解，发现真的是个很easy的BFS....T_T

方向数组：

e: 0 1 2 3 4 5

左：3 2 0 1 4 5
右：2 3 1 0 4 5
前：5 4 2 3 0 1
后：4 5 2 3 1 0

代码：

#include <algorithm>#include <cstdlib>#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <cctype>#include <cmath>#include <stack>#include <queue>#include <list>#include <map>#include <set>using namespace std;#define min2(x, y)     min(x, y)#define max2(x, y)     max(x, y)#define min3(x, y, z)  min(x, min(y, z))#define max3(x, y, z)  max3(x, max(y, z))#define clr(x, y)      memset(x, y, sizeof(x))#define fr(i,n)        for(int i = 0; i < n; i++)#define fr1(i,n)       for(int i = 1; i < n; i++)#define upfr(i,j,n)    for(int i = j; i <= n; i++)#define dowfr(i,j,n)   for(int i = n; i >= j; i--)#define scf(n)         scanf("%d",&n)#define ptf(n)         printf("%d",n)#define ptfs(s)        printf("%s",s)#define ptln()         printf("\n")#define srt(a,n)       sort(a,n)#define LL long long#define pi acos(-1.0)#define inf 1 << 31-1#define eps 0.00001#define maxn 6#define maxm 7struct node{    int arr[maxn];    int step;}s,e;int is_ok(node a, node b){    fr(i,maxn)    if(a.arr[i] != b.arr[i])        return 0;    return 1;}int vis[maxm][maxm][maxm][maxm][maxm][maxm];node turn(node a,int dir){    node b;    if(dir == 0)    {        b.arr[0] = a.arr[3];        b.arr[1] = a.arr[2];        b.arr[2] = a.arr[0];        b.arr[3] = a.arr[1];        b.arr[4] = a.arr[4];        b.arr[5] = a.arr[5];    }    if(dir == 1)    {        b.arr[0] = a.arr[2];        b.arr[1] = a.arr[3];        b.arr[2] = a.arr[1];        b.arr[3] = a.arr[0];        b.arr[4] = a.arr[4];        b.arr[5] = a.arr[5];    }    if(dir == 2)    {        b.arr[0] = a.arr[5];        b.arr[1] = a.arr[4];        b.arr[2] = a.arr[2];        b.arr[3] = a.arr[3];        b.arr[4] = a.arr[0];        b.arr[5] = a.arr[1];    }    if(dir == 3)    {        b.arr[0] = a.arr[4];        b.arr[1] = a.arr[5];        b.arr[2] = a.arr[2];        b.arr[3] = a.arr[3];        b.arr[4] = a.arr[1];        b.arr[5] = a.arr[0];    }    return b;}int  bfs(){    clr(vis,0);    queue <node> q;    q.push(s);    vis[s.arr[0]][s.arr[1]][s.arr[2]][s.arr[3]][s.arr[4]][s.arr[5]] = 1;    node now, next;    while(!q.empty())    {        now = q.front();        q.pop();        if(is_ok(e, now))            return now.step;        fr(i,4)        {            next = turn(now, i);            if(!vis[next.arr[0]][next.arr[1]][next.arr[2]][next.arr[3]][next.arr[4]][next.arr[5]])                {                    next.step = now.step + 1;                    vis[next.arr[0]][next.arr[1]][next.arr[2]][next.arr[3]][next.arr[4]][next.arr[5]] = 1;                    q.push(next);                }        }    }    return -1;}int main(){    //freopen("in.txt", "r", stdin);    clr(s.arr, 0);    clr(e.arr, 0);    while(scf(s.arr[0])==1)    {        fr1(i,6)        scf(s.arr[i]);        s.step = 0;        fr(i,6)        scf(e.arr[i]);        ptf(bfs());        ptln();    }    return 0;}