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清北学堂模拟赛d5t6 cube

时间:2017-10-11 19:21:45      阅读:181      评论:0      收藏:0      [点我收藏+]

标签:col   names   printf   code   scanf   font   mes   emc   memcpy   

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题面有误!10,11,12操作类别为A,13,14,15类别为B,16,17,18类别为C.

分析:一道大暴力,每次记录一下走了多少步,上一步操作类别是啥就可以了.最后只需要写6种操作,每一次操作进行4次就还原了,所以不用memcpy再来转.

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int n, ans[10], cnt, c[30];
bool stop = false;

void turn1()
{
    int tmp, tmp2;
    tmp = c[1];
    c[1] = c[3];
    c[3] = c[4];
    c[4] = c[2];
    c[2] = tmp;
    tmp = c[23];
    tmp2 = c[24];
    c[23] = c[10];
    c[24] = c[9];
    c[9] = c[5];
    c[10] = c[6];
    c[5] = c[13];
    c[6] = c[14];
    c[13] = tmp2;
    c[14] = tmp;
}

void turn2()
{
    int tmp,tmp2;
    tmp = c[9];
    c[9] = c[11];
    c[11] = c[12];
    c[12] = c[10];
    c[10] = tmp;
    tmp = c[1];
    tmp2 = c[3];
    c[1] = c[21];
    c[3] = c[23];
    c[23] = c[19];
    c[21] = c[17];
    c[19] = c[7];
    c[17] = c[5];
    c[7] = tmp2;
    c[5] = tmp;
}

void turn3()
{
    int tmp = c[5], tmp2;
    c[5] = c[7];
    c[7] = c[8];
    c[8] = c[6];
    c[6] = tmp;
    tmp = c[17];
    tmp2 = c[18];
    c[17] = c[15];
    c[18] = c[13];
    c[15] = c[4];
    c[13] = c[3];
    c[3] = c[12];
    c[4] = c[10];
    c[10] = tmp;
    c[12] = tmp2;
}

void turn4()
{
    int tmp = c[13], tmp2;
    c[13] = c[15];
    c[15] = c[16];
    c[16] = c[14];
    c[14] = tmp;
    tmp = c[6];
    tmp2 = c[8];
    c[6] = c[18];
    c[8] = c[20];
    c[18] = c[22];
    c[20] = c[24];
    c[22] = c[2];
    c[24] = c[4];
    c[2] = tmp;
    c[4] = tmp2;
}

void turn5()
{
    int tmp = c[21], tmp2;
    c[21] = c[23];
    c[23] = c[24];
    c[24] = c[22];
    c[22] = tmp;
    tmp = c[19];
    tmp2 = c[20];
    c[19] = c[9];
    c[20] = c[11];
    c[9] = c[2];
    c[11] = c[1];
    c[1] = c[14];
    c[2] = c[16];
    c[14] = tmp2;
    c[16] = tmp;
}

void turn6()
{
    int tmp = c[17], tmp2;
    c[17] = c[19];
    c[19] = c[20];
    c[20] = c[18];
    c[18] = tmp;
    tmp = c[21];
    tmp2 = c[22];
    c[21] = c[16];
    c[22] = c[15];
    c[16] = c[8];
    c[15] = c[7];
    c[8] = c[12];
    c[7] = c[11];
    c[12] = tmp;
    c[11] = tmp2;
}

bool check()
{
    return c[1] == c[2] && c[2] == c[3] && c[3] == c[4] &&
        c[5] == c[6] && c[6] == c[7] && c[7] == c[8] &&
        c[9] == c[10] && c[10] == c[11] && c[11] == c[12] &&
        c[13] == c[14] && c[14] == c[15] && c[15] == c[16] &&
        c[17] == c[18] && c[18] == c[19] && c[19] == c[20] &&
        c[21] == c[22] && c[22] == c[23] && c[23] == c[24];
}

void dfs(int dep, int flag)
{
    if (check())
    {
        cnt = dep;
        stop = 1;
        return;
    }
    if (dep >= n)
        return;
    if (flag != 1)
    {
        for (int i = 1; i <= 3; i++)
        {
            turn1();
            ans[dep + 1] = i;
            dfs(dep + 1, 1);
            if (stop)
                return;
        }
        turn1();
    }
    if (flag != 2)
    {
        for (int i = 1; i <= 3; i++)
        {
            turn2();
            ans[dep + 1] = 3 + i;
            dfs(dep + 1, 2);
            if (stop)
                return;
        }
        turn2();
    }
    if (flag != 3)
    {
        for (int i = 1; i <= 3; i++)
        {
            turn3();
            ans[dep + 1] = 6 + i;
            dfs(dep + 1, 3);
            if (stop)
                return;
        }
        turn3();
    }
    
    if (flag != 3)
    {
        for (int i = 1; i <= 3; i++)
        {
            turn4();
            ans[dep + 1] = 9 + i;
            dfs(dep + 1, 3);
            if (stop)
                return;
        }
        turn4();
    }
    if (flag != 2)
    {
        for (int i = 1; i <= 3; i++)
        {
            turn5();
            ans[dep + 1] = 12 + i;
            dfs(dep + 1, 2);
            if (stop)
                return;
        }
        turn5();
    }
    if (flag != 1)
    {
        for (int i = 1; i <= 3; i++)
        {
            turn6();
            ans[dep + 1] = 15 + i;
            dfs(dep + 1, 1);
            if (stop)
                return;
        }
        turn6();
    }
    
}

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= 24; i++)
        scanf("%d", &c[i]);
    dfs(0, 0);
    for (int i = 1; i <= cnt; i++)
        printf("%d ", ans[i]);

    return 0;
}

 

清北学堂模拟赛d5t6 cube

标签:col   names   printf   code   scanf   font   mes   emc   memcpy   

原文地址:http://www.cnblogs.com/zbtrs/p/7652008.html

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