CROC 2016 - Elimination Round (Rated Unofficial Edition) A. Amity Assessment 水题

A. Amity Assessment

题目连接：

http://www.codeforces.com/contest/655/problem/A

Description

Bessie the cow and her best friend Elsie each received a sliding puzzle on Pi Day. Their puzzles consist of a 2 × 2 grid and three tiles labeled 'A', 'B', and 'C'. The three tiles sit on top of the grid, leaving one grid cell empty. To make a move, Bessie or Elsie can slide a tile adjacent to the empty cell into the empty cell as shown below:

In order to determine if they are truly Best Friends For Life (BFFLs), Bessie and Elsie would like to know if there exists a sequence of moves that takes their puzzles to the same configuration (moves can be performed in both puzzles). Two puzzles are considered to be in the same configuration if each tile is on top of the same grid cell in both puzzles. Since the tiles are labeled with letters, rotations and reflections are not allowed.

Input

The first two lines of the input consist of a 2 × 2 grid describing the initial configuration of Bessie's puzzle. The next two lines contain a 2 × 2 grid describing the initial configuration of Elsie's puzzle. The positions of the tiles are labeled 'A', 'B', and 'C', while the empty cell is labeled 'X'. It's guaranteed that both puzzles contain exactly one tile with each letter and exactly one empty position.

Output

Output "YES"(without quotes) if the puzzles can reach the same configuration (and Bessie and Elsie are truly BFFLs). Otherwise, print "NO" (without quotes).

Sample Input

AB
XC
XB
AC

Sample Output

YES

Hint

题意

类似于华容道一样，一个2*2的格子，然后问你能不能从第一个推到第二个

题解：

由于是2*2的，我们比较简单的就可以发现，其实他按照顺时针\逆时针的顺序是不会改变的。

所以只用check一下顺序就好了。

代码

#include<bits/stdc++.h>
using namespace std;

string s1,s2;

int main()
{
    string a,b,c,d;
    cin>>a>>b>>c>>d;
    for(int i=0;i<2;i++)
        if(a[i]!='X')s1+=a[i];
    for(int i=1;i>=0;i--)
        if(b[i]!='X')s1+=b[i];
    for(int i=0;i<2;i++)
        if(c[i]!='X')s2+=c[i];
    for(int i=1;i>=0;i--)
        if(d[i]!='X')s2+=d[i];
    s1+=s1;
    if(s1.find(s2)!=string::npos)cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
}