UVA-253 Cube painting

最新推荐文章于 2019-04-29 13:34:38 发布

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本文介绍了一个编程问题的解决方法，该问题要求判断两个六面体是否可以通过旋转变为相同涂色状态。文章提供了两种解决方案，一种是通过检查相对面的颜色进行简化判断，另一种则是采用暴力模拟所有可能的旋转状态。

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题目来源：https://cn.vjudge.net/problem/UVA-253

UVA-253 Cube painting

We have a machine for painting cubes. It is supplied with threedifferent colors: blue, red and green. Each face of the cube gets oneof these colors. The cube’s faces are numbered as in Figure 1.Since a cube has 6 faces, our machine can paint a face-numberedcube in 36 = 729 different ways. When ignoring the face-numbers,the number of different paintings is much less, because a cube canbe rotated. See example below.We denote a painted cube by a string of 6 characters, where eachcharacter is a ‘b’, ‘r’, or ‘g’. The i-th character (1 ≤ i ≤ 6) from theleft gives the color of face i. For example, Figure 2 is a picture of“rbgggr” and Figure 3 corresponds to “rggbgr”. Notice that bothcubes are painted in the same way: by rotating it around the verticalaxis by 90°, the one changes into the other.

Input

The input of your program is a textfile that ends with the standard end-of-file marker. Each line is astring of 12 characters. The first 6 characters of this string are the representation of a painted cube, theremaining 6 characters give you the representation of another cube. Your program determines whetherthese two cubes are painted in the same way, that is, whether by any combination of rotations one canbe turned into the other. (Reflections are not allowed.)

Output

The output is a file of boolean. For each line of input, output contains ‘TRUE’ if the second half can beobtained from the first half by rotation as describes above, ‘FALSE’ otherwise.

Sample Input

rbgggrrggbgr

rrrbbbrrbbbr

rbgrbgrrrrrg

Sample Output

TRUE

FALSE

题意：1个骰子有六个面，用三种颜色上色，现有两个这题的骰子，问是否是一样的？

分析：两个对面匹配？开始是这么想的，然后抱着一试的心态交了一次，竟然AC了。

当后来队友和我说这样是不行，并且给出了一组反例，说只能24中情况模拟，自己懒就没有重打了

反例：

rbbgrggbgrrb(这个是FALSE)

“AC”的代码：

#include <bits/stdc++.h>
using namespace std;
int main()
{
	string s;
	bool v[4];
	vector<pair<char,char> >a,b,c;
	while(cin>>s)
	{
		a.clear();b.clear();c.clear();
		memset(v,false,sizeof(v));
		//将对边存入 
		a.push_back(make_pair(s[0],s[5]));
		a.push_back(make_pair(s[1],s[4]));
		a.push_back(make_pair(s[2],s[3]));
		b.push_back(make_pair(s[6],s[11]));
		b.push_back(make_pair(s[7],s[10]));
		b.push_back(make_pair(s[8],s[9]));
		c.push_back(make_pair(s[11],s[6]));
		c.push_back(make_pair(s[10],s[7]));
		c.push_back(make_pair(s[9],s[8]));
		/*
for(int i=0;i<3;i++)
		{
			cout<<a[i].first<<a[i].second<<endl;	
		}
		for(int i=0;i<3;i++)
		{
			cout<<b[i].first<<b[i].second<<endl;
			
		}	*/
		int f=0;
		//寻找对边是否匹配 
		for(int i=0;i<3;i++)
		{
			f=1;
			for(int j=0;j<3;j++)
			{
				if((a[i]==b[j]||a[i]==c[j])&&!v[j])
				{
					f=0;
					v[j]=true;
					break;	
				}
			}
			if(f)break;
		}
		if(f) puts("FALSE");
		else puts("TRUE");
		
	}
}

队友的代码（直接暴力）：

#include<stdio.h>
#include<string.h>
int a[3][6]={5,1,3,4,6,2,4,2,1,6,5,3,1,3,5,2,4,6};
char b[100][6];
char c[14],d[10];
int s;
int flag;
void zhuan(int r)
{
	int j;
	for(j=0;j<6;j++)
	{
		if(s)
		b[s][j]=b[s-1][a[r][j]-1];
		else
		b[s][j]=c[a[r][j]-1];
	}
	b[s][6]='\0';
	if(strcmp(b[s],d)==0)
	{
		flag=1;
	}
	s++;
}
int main()
{
	int i,j,k,l;
	while(~scanf("%s",c))
	{
		flag=0;
		for(i=6;i<12;i++)
		{
			d[i-6]=c[i];
		}
		d[6]='\0';
		s=0;
		for(i=0;i<4;i++)
		{
			for(k=0;k<4;k++)
			{
				for(l=0;l<4;l++)
				{
					if(flag)break;
					zhuan(2);
				}
				if(flag)break;
				zhuan(1);
			}
			if(flag)break;
			zhuan(0);
		}
		if(flag)puts("TRUE");
		else puts("FALSE");
	}
}