第四届 山东省ACM Rubik’s cube(爆搜)

Rubik’s cube

Time Limit: 2000MS Memory Limit: 65536KB

Submit Statistic

Problem Description

    Flabby is addicted to Rubik’s cube. He has many kinds of Rubik’s cube and plays them well. He can reorder a Rubik’s cube in a few seconds. He is famous for playing Rubik’s cube and attracts many young girls. One of his friends, Ant, is jealous of him and decides to let him be embarrassed in public. Ant makes a small Rubik’s cube by himself. This small magic is 2*2*2 and is drawn in blue and red, as shown in Figure 1. 
                                                              
    This Rubik’s cube can be rotated as well as other kinds of magic cube. Each of six faces can rotated clockwise and counterclockwise. Each 90 degree rotation on any face is counted as one step. The Rubik’s cube which has been reordered has only one color on each face.
     Ant is a clever boy. Sometimes, he can make a Rubik’s cube which can be reordered in a few steps. He can also make a Rubik’s cube which can’t be reordered in any way.
                                                            
                                                                 
Flabby knows what Ant thinks in his mind. He knows that you are a good programmer and asks you for help. Tell him whether this special Rubik’s cube can be reordered in a few steps.

Input

    In the input file, the first line is an integer T which is the number of test case. For each test case, there is 6 lines of integers describe the Rubik’s cube. For each line, there are four integers. Pij which is shown in Figure 3 corresponds to the jth integer of the ith line. If Pij is 0, it means the corresponding place is red. If Pij is 1, it means the corresponding place is blue. 

Output

    If the magic cube can be reordered in n steps, output the minimum n in a line. If the magic cube cannot be reordered, output “IMPOSSIBLE!” in a line. 

Example Input

3
0 0 0 0
0 1 0 1
1 1 0 0
1 0 1 0
1 1 0 0
1 1 1 1
0 0 1 1
1 1 0 0
0 1 0 1
1 0 0 0
1 1 1 1
0 1 0 0
1 0 1 1
0 1 0 0
0 0 0 0
1 1 1 1
1 0 0 0
0 1 1 1

Example Output

1
IMPOSSIBLE!
8

Hint

 

Author

 2013年山东省第四届ACM大学生程序设计竞赛

借鉴网址：http://blog.csdn.net/binwin20/article/details/9073941

思路:魔方的旋转方法总共有12种,

只考虑前面的四个正方形,上面向左旋转,跟下面向右旋转是一个效果,方法数可以除2..

向右旋转一次等于向左旋转两次,可以一起处理,方法数再减半.三种情况.代码很短.

只有两种颜色,很多重复状态.用map当hash页不会超时.100ms水过.比标程代码短时间少..

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <string>
#define LL long long
#define DB double
#define SF scanf
#define PF printf
#define N 1<<24
#define bug cout<<"bug"<<endl;
using namespace std;
map<int,int> mp;
struct nod{
    int a[25];
    int dis;
};
int cc[20][15]={
    {0,1,4,5,20,21,12,13},
    {8,11,10,9},
    {18,16,2,0,9,10,21,23},
    {13,12,14,15},
    {4,6,17,16,15,13,9,8},
    {0,1,3,2},
};
void turn(nod &t,int c)
{
    c<<=1;
    int tmp = t.a[cc[c][7]];
    for(int i=7;i>0;i--)
    {
        t.a[cc[c][i]]=t.a[cc[c][i-1]];
    }
    t.a[cc[c][0]] = tmp;
    tmp = t.a[cc[c][7]];
    for(int i=7;i>0;i--)
    {
        t.a[cc[c][i]]=t.a[cc[c][i-1]];
    }
    t.a[cc[c][0]] = tmp;

    tmp = t.a[cc[c|1][3]];
    for(int i=3;i>0;i--)
    {
        t.a[cc[c|1][i]]=t.a[cc[c|1][i-1]];
    }
    t.a[cc[c|1][0]] = tmp;
}
nod in;
queue<nod> que;
int gethas(nod &t)
{
    int ret = 0;
    for(int i=0;i<24;i++)
    {
        ret = t.a[i]+(ret<<1);
    }return ret;
}
bool ok(nod &t)
{
    for(int i=0;i<24;i+=4)
    {
        for(int j=1;j<4;j++)
        if(t.a[i]!=t.a[i+j]) return false;
    }return true;
}
int solve()
{
    int con = 0;
    for(int i=0;i<24;i++)
    if(in.a[i]) con++;
    if(con%4) return -1;
    in.dis = 0;
    while(!que.empty()) que.pop();
    que.push(in);
    mp.clear();
    mp[gethas(in)] = 1;
    nod e,t;
    if(ok(in)) return t.dis;
    while(!que.empty())
    {
        e = que.front();que.pop();
        for(int i=0;i<3;i++)
        {
            t = e;t.dis++;
            turn(t,i);
            if(mp.find(gethas(t))==mp.end())
            {
                mp[gethas(t)] = 1;//bug
                if(ok(t)) return t.dis;
                que.push(t);
            }
            turn(t,i);
            turn(t,i);
            if(mp.find(gethas(t))==mp.end())
            {
                mp[gethas(t)] = 1;//bug
                if(ok(t)) return t.dis;
                que.push(t);
            }
        }
    }return -1;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    int cas;SF("%d",&cas);
    while(cas--)
    {
        for(int i=0;i<24;i++)
        SF("%d",&in.a[i]);
        int k = solve();
        if(k==-1) printf("IMPOSSIBLE!
");
        else printf("%d
",k);
    }
    return 0;
}

官方标程

/**
   23
   10
   23
   10
 010101
 323232
   01
   32
    6
    3
   412
    5
 */
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;
char md[6][4];
#define MAX 20000000
bool flag[16][16][16][16][16][16];
struct M
{
    short int num[6];
    short int lv;
    void print()
    {
        int i;
        for(i = 0; i < 6; i++)
            printf("%d ",num[i]);
        printf("
");
        return;
    }
}queue[MAX];
M (*funp[6])(M m);
bool check(M m)
{
    int i;
    for(i = 0; i < 6; i++)
    {
        if(m.num[i] != 15 && m.num[i] != 0)
            return false;
    }
    return true;
}
M turnY_L(M tm)
{
    M m;
    m.num[0] = (tm.num[0] >> 1) | ((tm.num[0] & 1) << 3);
    m.num[1] = (tm.num[1] & 6) | ((tm.num[4] & 1) << 3) | ((tm.num[4] & 2) >> 1);
    m.num[2] = (tm.num[2] & 12) | ((tm.num[1] & 1) << 1) | ((tm.num[1] & 8) >> 3);
    m.num[3] = (tm.num[3] & 9) | ((tm.num[2] & 2) << 1) | ((tm.num[2] & 1) << 1);
    m.num[4] = (tm.num[4] & 12) | ((tm.num[3] & 4) >> 1) | ((tm.num[3] & 2) >> 1);
    m.num[5] = tm.num[5];
    return m;
}
M turnY_R(M tm)
{
    M m;
    m.num[0] = ((tm.num[0] & 7) << 1) | ((tm.num[0] & 8) >> 3);
    m.num[1] = (tm.num[1] & 6) | ((tm.num[2] & 2) >> 1) | ((tm.num[2] & 1) << 3);
    m.num[2] = (tm.num[2] & 12) | ((tm.num[3] & 4) >> 1) | ((tm.num[3] & 2) >> 1);
    m.num[3] = (tm.num[3] & 9) | ((tm.num[4] & 1) << 1) | ((tm.num[4] & 2) << 1);
    m.num[4] = (tm.num[4] & 12) | ((tm.num[1] & 1) << 1) | ((tm.num[1] & 8) >> 3);
    m.num[5] = tm.num[5];
    return m;
}

/*------------------------------------------------------------------------------------*/

M turnX_L(M tm)
{
    M m;
    m.num[1] = (tm.num[1] >> 1) | ((tm.num[1] & 1) << 3);
    m.num[0] = (tm.num[0] & 9) | ((tm.num[2] & 1) << 2) | ((tm.num[2] & 8) >> 2);
    m.num[2] = (tm.num[2] & 6) | ((tm.num[5] & 9));
    m.num[3] = tm.num[3];
    m.num[4] = (tm.num[4] & 9) | ((tm.num[0] & 6));
    m.num[5] = (tm.num[5] & 6) | ((tm.num[4] & 4) >> 2) | ((tm.num[4] & 2) << 2);
    return m;
}
M turnX_R(M tm)
{
    M m;
    m.num[1] = ((tm.num[1] & 7) << 1) | ((tm.num[1] & 8) >> 3);
    m.num[0] = (tm.num[0] & 9) | ((tm.num[4] & 6));
    m.num[2] = (tm.num[2] & 6) | ((tm.num[0] & 4) >> 2) | ((tm.num[0] & 2) << 2);
    m.num[3] = tm.num[3];
    m.num[4] = (tm.num[4] & 9) | ((tm.num[5] & 8) >> 2) | ((tm.num[5] & 1) << 2);
    m.num[5] = (tm.num[5] & 6) | ((tm.num[2] & 9));
    return m;
}

M turnZ_L(M tm)
{
    M m;
    m.num[4] = (tm.num[4] >> 1) | ((tm.num[4] & 1) << 3);
    m.num[0] = (tm.num[0] & 3) | (tm.num[1] & 12);
    m.num[1] = (tm.num[1] & 3) | (tm.num[5] & 12);
    m.num[2] = tm.num[2];
    m.num[3] = (tm.num[3] & 3) | (tm.num[0] & 12);
    m.num[5] = (tm.num[5] & 3) | (tm.num[3] & 12);
    return m;
}
M turnZ_R(M tm)
{
    M m;
    m.num[4] = ((tm.num[4] & 7)<< 1) | ((tm.num[4] & 8) >> 3);
    m.num[0] = (tm.num[0] & 3) | (tm.num[3] & 12);
    m.num[1] = (tm.num[1] & 3) | (tm.num[0] & 12);
    m.num[2] = tm.num[2];
    m.num[3] = (tm.num[3] & 3) | (tm.num[5] & 12);
    m.num[5] = (tm.num[5] & 3) | (tm.num[1] & 12);
    return m;
}


void record_flag(int num1,int num2,int num3,int num4,int num5,int num6)
{
    //printf("%d %d %d %d %d %d
",num1,num2,num3,num4,num5,num6);
    flag[num1][num2][num3][num4][num5][num6] = 1;
    flag[num4][num1][(num3>>1)|((num3&1)<<3)][num6][((num5&7)<<1)+((num3&8)>>3)][num2] = 1;
    flag[num6][num4][((num3&3)<<2)+((num3&12)>>2)][num2][((num5&3)<<2)+((num5&12)>>2)][num1] = 1;
    flag[num2][num6][((num3&7)<<1)|((num3&8)>>3)][num1][(num5>>1)+((num5&1)<<3)][num4] = 1;
    return;
}
int Search(M m)
{
    funp[0] = turnX_L;
    funp[1] = turnX_R;
    funp[2] = turnY_L;
    funp[3] = turnY_R;
    funp[4] = turnZ_L;
    funp[5] = turnZ_R;
    M tmp,tm;
    int front,rear,i;
    front = rear = 0;
    memset(flag,0,sizeof(flag));
    m.lv = 0;
    queue[rear++] = m;
    record_flag(m.num[0],m.num[1],m.num[2],m.num[3],m.num[4],m.num[5]);
    while(front < rear)
    {
        tmp = queue[front++];
        if(check(tmp))
            return tmp.lv;
        for(i = 0; i < 6; i++)
        {
            tm = funp[i](tmp);
            tm.lv = tmp.lv + 1;
            if(flag[tm.num[0]][tm.num[1]][tm.num[2]][tm.num[3]][tm.num[4]][tm.num[5]] == 0)
            {
                queue[rear++] = tm;
                record_flag(tm.num[0],tm.num[1],tm.num[2],tm.num[3],tm.num[4],tm.num[5]);
            }
        }
    }
    return -1;
}

int main()
{
    int T,i,n1,n2,n3,n4,ans;
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    scanf("%d",&T);
    M m;
    while(T--)
    {
        for(i = 0; i < 6; i++)
        {
            scanf("%d%d%d%d",&n1,&n2,&n3,&n4);
            m.num[i] = n1 + (n2 << 1) + (n3 << 3) + (n4 << 2);
          //  printf("%d
",m.num[i]);
        }
        ans = Search(m);
        if(ans != -1)
            printf("%d
", ans);
        else
            printf("IMPOSSIBLE!
");
    }
    return 0;
}

官方数据生成代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <ctime>
#include <cstdlib>
using namespace std;
int a[24] = {0,0,1,1,1,1,0,0,0,1,0,1,1,0,1,0,1,1,0,1,0,1,0,0};
int main()
{
    int t1,t2,i;
    freopen("data.in","w",stdout);
    int T = 10;
    printf("%d
",T);
    srand(0);
    while(T--)
    {

        t1 = rand()%24;
        do
        {
            t2 = ((rand()%24)*(rand()%24))%24;
        }while(a[t1]==a[t2]);
        a[t1] = a[t1] ^ a[t2];
        a[t2] = a[t1] ^ a[t2];
        a[t1] = a[t1] ^ a[t2];
        for(i = 0; i < 24; i++)
        {
            printf("%d",a[i]);
            if(i % 4 == 3)
                printf("
");
            else
                printf(" ");
        }
        printf("
");
    }
    return 0;
}