ACM Secrete Master Plan

Problem Description

Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan instructs how to deploy soldiers on the four corners of the city wall. Unfortunately, when Fei opened the pocket he found there are only four numbers written in dots on a piece of sheet. The numbers form 2×2

 

Input

The first line of the input gives the number of test cases, T(1≤T≤104)Tai0ai11≤ai0,ai1≤1003rd4th

 

Output

For each test case, output one line containing "Case #x: y", where x is the test case number
(starting from 1) and y

 

Sample Input

4

1 2

3 4

1 2

3 4

 

1 2

3 4

3 1

4 2

 

1 2

3 4

3 2

4 1

 

1 2

3 4

4 3

2 1

 

Sample Output

Case #1: POSSIBLE

Case #2: POSSIBLE

Case #3: IMPOSSIBLE

Case #4: POSSIBLE

#include<bits/stdc++.h>
using namespace std;
/*
    方块旋转，两个矩形，如果能够通过旋转使两个一样则输出 possible
     否则则输出impossible 
*/
int  A[5][5],B[5][5];  //矩阵A，矩阵B 
//判断现两个矩形是否一样 
bool Judge()       
{
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
            if(A[i][j] != B[i][j])
                return false;
    return true;
}
//将一个矩形进行旋转，顺时针旋转，
//破点：将[0][0]先后进行向左、斜方向、向下交换 
void Swap()
{
    swap(A[0][0],A[0][1]);
    swap(A[0][0],A[1][1]);
    swap(A[0][0],A[1][0]);    
}
//通过旋转四次进行判定是否满足题意 
bool Solve()
{
    for(int i = 0; i < 4; i++)
    {
        if(Judge())
            return true;
        else
            Swap();
    }
    return false;
}

int main()
{
    int T,n;
    while(~scanf("%d",&T))
    {
        for(int p= 1; p <= T; p++)
        {
            
            scanf("%d %d %d %d",&A[0][0],&A[0][1],&A[1][0],&A[1][1]);
            scanf("%d %d %d %d",&B[0][0],&B[0][1],&B[1][0],&B[1][1]);
            printf("Case #%d: ",p);
            if(Solve())
            {
                printf("POSSIBLE
");
            }
            else
            {
                printf("IMPOSSIBLE
");    
            }
            
            
        }
    }
    return 0;
 }