ZOJ2477 Magic Cube
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
- 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 ...
- 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 ...
- 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 ...
- 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$a_i ...
- sdutoj 2606 Rubik’s cube
http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2606 Rubik’s cube Time Li ...
- USACO 3.2 Magic Squares
Magic SquaresIOI'96 Following the success of the magic cube, Mr. Rubik invented its planar version, ...
- 用DirectX实现魔方（一）
关于魔方 魔方英文名字叫做Rubik's Cube,是由匈牙利建筑学教授和雕塑家Ernő Rubik于1974年发明,最初叫做Magic Cube(这大概也是中文名字的来历吧),1980年Ideal ...
- 一位学长的ACM总结（感触颇深）
发信人: fennec (fennec), 信区: Algorithm 标 题: acm 总结 by fennec 发信站: 吉林大学牡丹园站 (Wed Dec 8 16:27:55 2004) AC ...
- ZOJ - 2477 dfs [kuangbin带你飞]专题二
注意输入的处理,旋转操作打表.递增枚举可能步数,作为限制方便找到最短路. AC代码:90ms #include<cstdio> #include<cstring> char m ...
随机推荐
- (转）Struts2的标签库
http://blog.csdn.net/yerenyuan_pku/article/details/68638679 Struts2的标签库 对于一个MVC框架而言,重点是实现两部分:业务逻辑控制器 ...
- makefile 与 make
所要完成的Makefile 文件描述了整个工程的编译.连接等规则.其中包括:工程中的哪些源文件需要编译以及如何编译.需要创建那些库文件以及如何创建这些库文件.如何最后产生我们想要的可执行文件.尽管看起 ...
- mvc 类中对应数据库属性
[StringLength()] //可空 对应数据库可空 [DefaultValue("")] [DisplayName("添加人用户名")] public ...
- R语言数据重塑
使用cbind()函数连接多个向量来创建数据帧.此外,使用rbind()函数合并两个数据帧 使用merge()函数合并两个数据帧.数据帧必须具有相同的列名称,在其上进行合并 melt()拆分数 ...
- ionic4封装样式原理
查看文档: https://www.cnblogs.com/WhiteCusp/p/4342502.html https://www.jianshu.com/p/bb291f9678e1 https: ...
- Navicat 导出为 Excel 文件
1:先选择一个表 再点击导出向导 2:看图 3:选择保存的路径 4:选择字段 5:最后执行吧
- 本地读取服务器Xml文件及本地读本地的xml
updateUrl="ServerUrl"(服务器路径) WebClient wc = new WebClient(); Stream stream = wc.OpenRead(u ...
- 原生Ajax的使用——含开放API接口
看了两天关于Ajax的使用,总感觉云里雾里的. 故在此总结梳理一下,如果疏漏错误还请纠正支出. Ajax能够在向服务器请求额外的数据时,不必重新加载/卸载整个页面,实现一小块区域性的刷新,也是常说的异 ...
- API开发管理平台eoLinker AMS 4.1版本发布：加入聚合空间，发布AMS专业版等
eoLinker AMS是集API文档管理.API自动化测试.开发协作三位一体的综合API开发管理平台,是中国最大的在线API管理平台. eoLinker AMS 4.1更新内容: 1.新增" ...
- api 签名算法
<?php define('token', 'tokensecret'); // 定义私钥token /** * 哈希验证签名 */ function hmacSign($array, $tok ...