HDU5540 Secrete Master Plan(模拟)

Secrete Master Plan

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1179    Accepted Submission(s): 694

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 matrix, but Fei didn't know the correct direction to hold the sheet. What a pity!

Given two secrete master plans. The first one is the master's original plan. The second one is the plan opened by Fei. As KongMing had many pockets to hand out, he might give Fei the wrong pocket. Determine if Fei receives the right pocket.

 

Input

The first line of the input gives the number of test cases, T(1≤T≤104). T test cases follow. Each test case contains 4 lines. Each line contains two integers ai0and ai1 (1≤ai0,ai1≤100). The first two lines stands for the original plan, the 3rd and 4th line stands for the plan Fei opened.

 

Output

For each test case, output one line containing "Case #x: y", where x is the test case number
(starting from 1) and y is either "POSSIBLE" or "IMPOSSIBLE" (quotes for clarity).

 

Sample Input

41 23 41 23 41 23 43 14 21 23 43 24 11 23 44 32 1

 

Sample Output

Case #1: POSSIBLECase #2: POSSIBLECase #3: IMPOSSIBLECase #4: POSSIBLE

/*思路：输入两组数据，每一组都是2*2的矩阵，是否可以通过第二个2*2的矩阵旋转得到第一个。注意输入的时候，要控制下，看这个例子1 23 4旋转一次以后为3 14 2可以发现，在输入的时候输入的顺序为1,2,4,3。即2*2矩阵对应的位置，然后旋转即可。注意旋转的最多次数，一个点只可能出现在四个位置，所以只需要旋转四次就可以了。*/#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define ll long long#define PI acos(-1)#define M(n, m) memset(n, m, sizeof(n));const int INF = 1e9 + 7;const int maxn = 1e5 + 100;using namespace std;int main(){    int a[22], b[22];    int num;    scanf("%d", &num);    for (int k = 1; k <= num; k ++)    {        // 按照圈输入        cin >> a[1] >> a[2] >> a[4] >> a[3];        cin >> b[1] >> b[2] >> b[4] >> b[3];        bool flag = false;        for (int i = 0;i <= 4;i ++)        {            if (a[1] == b[1] && a[2] == b[2] && a[3] == b[3] && a[4] == b[4])            {                flag = true;                printf("Case #%d: POSSIBLE\n", k);                break;            }            // 交换位置            else            {                int t = b[1];                b[1] = b[2];                b[2] = b[3];                b[3] = b[4];                b[4] = t;            }        }        if (!flag)            printf("Case #%d: IMPOSSIBLE\n", k);    }    return 0;}