题目:

This is a very popular game for children. In this game, there's a cube, which consists of 3 * 3 * 3 small cubes. We can unwrap the cube, it will become like this:

      w w w
      w w w
      w w w
r r r g g g b b b o o o
r r r g g g b b b o o o
r r r g g g b b b o o o
      y y y
      y y y
      y y y

The letters means the color on the small cubes. For example, 'r' means red, 'g' means green, 'y' means yellow....The goal for this game is to rotate the faces of the cube to make each of the faces contains only one color. Note there're exact 6 kind of colors on the cube and there're exact 9 small rectangles totally in any time in the game.

Do you know how to rotate the faces? I think most of you have known it. But I would like to show it again. When a face is rotated, the configuration of colors in all the adjacent faces changes. For the cube above, after we rotate the green face clock-wise, the last line of 'w' face will become the left column of 'b' face, the left column of 'b' face will become the top line of 'y' face, etc. As you may know, reaching the final position from a scrambled configuration can be quite challenging.

In this problem, you are given a configuration of the cube, and asked to give a way to reach the final position. To reduce the difficulty, the steps required will never be greater than 5.

输入:

The input contains an integer in the first line, which indicates the number of the test cases. In each test case, there're exact 10 lines. The first line is an empty line. The next 9 lines contain a configuration. The format can be seen in the sample input. For simplicity, we give an index to each face as follows:

 

    /---\
    |   |
    | 4 |
    |   |
/---+---+---+---\
|   |   |   |   |
| 0 | 1 | 2 | 3 |
|   |   |   |   |
\---+---+---+---/
    |   |
    | 5 |
    |   |
    \---/

Note that there's a space between two adjacent letters.

输出:

For each test case, the first line of the output is the smallest count N of the steps to reach the winning position. If the winning position can't be reached in 5 steps, print -1 in this line. Otherwise print each step in one line in the following N lines. A step contains two integers, the first one means the face index, and the second one means the direction. 1 means clock-wise and -1 means counter clock-wise. If the given position is the winning position, print 0 for such test case simply. If there're multiple solutions, any one is acceptable.

样例:

分析:我。。。自闭了两天竟然是没有写>5输出-1(°ཀ°),为什么我会默认测试点必然可以在5步内实现啊(#`Д´)ノ

只能说这个模拟太恶心了!!!

用IDA*优化,思路是计算每一个面有几个和中心方格颜色不一样的方格,6个面的总数除12并向上取整是最少剩余需要的步数(如果最理想的情况加上已走步数仍然大于5那肯定没戏啊)

模拟写得相当丑(灬ºωº灬)

 #include<iostream>
#include<sstream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<numeric>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<cctype>
#define PI acos(-1.0)
const int INF = 0x3f3f3f3f;
const int NINF = -INF - ;
typedef long long ll;
using namespace std;
char cube[][];
struct node
{
char maze[][];
};
struct cur
{
char maze[][];
};
node clockwise(node temp)
{
char gtm[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
memcpy(temp.maze, gtm, sizeof(temp.maze));
return temp;
}
node cclockwise(node temp)
{
char gtm[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
gtm[][] = temp.maze[][];
memcpy(temp.maze, gtm, sizeof(temp.maze));
return temp;
}
int bol, ans;
pair<int, int> P[];
void print()
{
for (int i = ; i < ans; ++i)
cout << P[i].first << ' ' << P[i].second << endl;
}
int hx(cur trans)
{
double sum = ;
for (int i = ; i < ; ++i)
{
for (int j = ; j < ; j += )
if (trans.maze[i][j] != trans.maze[][]) sum++;
}
for (int i = ; i < ; ++i)
{
for (int j = ; j < ; j += )
if (trans.maze[i][j] != trans.maze[][]) sum++;
}
for (int i = ; i < ; ++i)
{
for (int j = ; j < ; j += )
if (trans.maze[i][j] != trans.maze[][]) sum++;
}
for (int i = ; i < ; ++i)
{
for (int j = ; j < ; j += )
if (trans.maze[i][j] != trans.maze[][]) sum++;
}
for (int i = ; i < ; ++i)
{
for (int j = ; j < ; j += )
if (trans.maze[i][j] != trans.maze[][]) sum++;
}
for (int i = ; i < ; ++i)
{
for (int j = ; j < ; j += )
if (trans.maze[i][j] != trans.maze[][]) sum++;
}
return ceil(sum / );
}
void dfs(int rec, char pos[][], int deep)
{
if (rec > deep) return;
cur trans;
memcpy(trans.maze, pos, sizeof(trans.maze));
int h = hx(trans);
//cout << "test:" << h << endl;
if (h == )
{
ans = rec;
cout << rec << endl;
bol = ;
return;
}
if (h + rec > deep) return;
for (int i = ; i < ; ++i)
{
char tmp[][];
memcpy(tmp, pos, sizeof(tmp));
/*for (int j = 0; j < 9; ++j)
cout << tmp[j] << endl;
return;*/
node temp;
if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = clockwise(temp);
/*for (int m = 0; m < 3; ++m)
{
for (int n = 0; n < 3; ++n)
cout << temp.maze[m][n] << ' ';
cout << endl;
}*/
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
else if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = cclockwise(temp);
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
else if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = clockwise(temp);
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
else if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = cclockwise(temp);
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
else if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = clockwise(temp);
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
else if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = cclockwise(temp);
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
else if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = clockwise(temp);
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
else if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = cclockwise(temp);
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
else if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = clockwise(temp);
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
else if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = cclockwise(temp);
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
else if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = clockwise(temp);
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
else if (i == )
{
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
temp.maze[m][n] = tmp[k][j];
}
temp = cclockwise(temp);
for (int k = , m = ; k < , m < ; ++k, ++m)
{
for (int j = , n = ; j < , n < ; j += , ++n)
tmp[k][j] = temp.maze[m][n];
}
char cpy[][];
memcpy(cpy, tmp, sizeof(cpy));
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
cpy[][] = tmp[][];
memcpy(tmp, cpy, sizeof(tmp));
/*for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
return;*/
}
/*if (i == 1 && deep == 2)
{
cout << "first" << endl;
for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
}
if (i == 2 && deep == 2)
{
cout << "second" << endl;
for (int k = 0; k < 9; ++k)
cout << tmp[k] << endl;
}*/
dfs(rec + , tmp, deep);
//if (i == 1) return;
if (bol)
{
P[rec].first = i / ;
if (i % == ) P[rec].second = ;
else P[rec].second = -;
return;
}
}
}
int main()
{
int T;
cin >> T;
while (T--)
{
string nul;
getline(cin, nul);
for (int i = ; i < ; ++i)
{
char ss;
int num = ;
while ((ss = getchar()) != '\n')
{
if (ss != ' ') cube[i][num++] = ss;
else cube[i][num++] = ' ';
}
cube[i][num] = '\0';
}
/*for (int i = 0; i < 9; ++i)
cout << cube[i] << endl;*/
/*for (int i = 0; i < 3; ++i)
{
for (int j = 6; j < 11; j += 2)
{
cout << cube[i][j] << ' ';
}
cout << endl;
}*/
/*cur gat;
memcpy(gat.maze, cube, sizeof(gat.maze));
int p = judge(gat);
cout << p << endl;*/
bol = , ans = ;
for (int i = ; i <= ; ++i)
{
//cout << "deep:" << i << endl;
dfs(, cube, i);
if (bol)
{
print();
break;
}
}
if (!bol) cout << - << endl;
}
return ;
}

ZOJ2477 Magic Cube的更多相关文章

  1. ZOJ 2477 Magic Cube(魔方)

    ZOJ 2477 Magic Cube(魔方) Time Limit: 2 Seconds      Memory Limit: 65536 KB This is a very popular gam ...

  2. ZOJ 2477 Magic Cube 暴力,模拟 难度:0

    http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1477 用IDA*可能更好,但是既然时间宽裕数据简单,而且记录状态很麻烦,就直接 ...

  3. magic cube

    搜索题, 每个状态能扩展出12种状态,最多进行5次旋转12^5 要用到iddfs,或者我看到网上其他人用的ida* 我也是参考了别人的代码,而且这个题vj上有点问题,我看数据看了半天,愣是没看明白第二 ...

  4. 2019杭电多校二 F. Fantastic Magic Cube (FWT)

    大意: 给定$N^3$立方体, 每个单位立方体权值为三个坐标异或, 每次沿坐标轴切一刀, 得分为两半内权值和的乘积, 求切成$n^3$块的最大得分. 可以发现得分与切法无关, 假设每个点权值为$a_i ...

  5. sdutoj 2606 Rubik’s cube

    http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2606 Rubik’s cube Time Li ...

  6. USACO 3.2 Magic Squares

    Magic SquaresIOI'96 Following the success of the magic cube, Mr. Rubik invented its planar version, ...

  7. 用DirectX实现魔方(一)

    关于魔方 魔方英文名字叫做Rubik's Cube,是由匈牙利建筑学教授和雕塑家Ernő Rubik于1974年发明,最初叫做Magic Cube(这大概也是中文名字的来历吧),1980年Ideal ...

  8. 一位学长的ACM总结(感触颇深)

    发信人: fennec (fennec), 信区: Algorithm 标 题: acm 总结 by fennec 发信站: 吉林大学牡丹园站 (Wed Dec 8 16:27:55 2004) AC ...

  9. ZOJ - 2477 dfs [kuangbin带你飞]专题二

    注意输入的处理,旋转操作打表.递增枚举可能步数,作为限制方便找到最短路. AC代码:90ms #include<cstdio> #include<cstring> char m ...

随机推荐

  1. PHP 之转换excel表格中的经纬度

    <?php set_time_limit(0); include './plugin/PHPExcel/PHPExcel.php'; include './plugin/PHPExcel/PHP ...

  2. day2笔记

    今日内容: 1.常用数据类型即数据方法 2.文件处理 3.函数 一 列表: 在[]内,可以存放多个任意类型的值,并以逗号隔开. 一般用于存放学生的爱好,课堂的周期等等 优先掌握的操作: 1.按索引取值 ...

  3. webstrom常用键

    常用快捷键—Webstorm入门指南 提高代码编写效率,离不开快捷键的使用,Webstorm拥有丰富的代码快速编辑功能,你可以自由配置功能快捷键. 快捷键配置 点击“File”-> “setti ...

  4. CAD在网页中打印的图纸里面添加页眉及页脚

    1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 3 ...

  5. 实现动画之CSS与JavaScript对比

    曾经某个时期,大多数开发者使用 jQuery 给浏览器中的元素添加动画.让这个淡化,让那个扩大,很简单.随着互动的项目越来越复杂,移动设备的大量增加,表现性能变得越来越重要.Flash 被抛弃,有天赋 ...

  6. C++入职学习篇--代码规范(持续更新)

    C++入职学习篇--代码规范(持续更新) 一.头文件规范 在头文件中大家一般会定义宏.引入库函数.声明.定义全局变量等,在设计时最后进行分类,代码示范(自己瞎琢磨的,请多多指点): #ifndef T ...

  7. Linux文件、目录属性

    文件.目录属性 可能有人觉得,文件这东西这么普通有什么好讲的,,,其实不然,在linux中一切皆文件,搞清楚文件的本质,将对我们大有裨益. 文件.目录属性简介 举个例子 randolf@localho ...

  8. Django REST framework - 版本控制

    目录 Django REST framework 版本控制 为什么需要版本控制 DRF提供了5种版本控制方案 版本控制系统的使用 全局配置 局部配置 获取版本信息 Django REST framew ...

  9. ubuntu下Apache2配置

    Ubuntu下Apache2的CGI简单配置:http://blog.csdn.net/a623891391/article/details/47170355 Ubuntu Apache的安装和配置以 ...

  10. cxf 和 httpclient 客户端调用 webservice 接口

    一.cxf 生成 webservice 客户端 1.接口路径 http://ws.webxml.com.cn/WebServices/WeatherWS.asmx 2.进入你需要放置 webservi ...