Description
A fancy square image encryption algorithm works as follow:
0. consider the image as an N x N matrix
1. choose an integer k∈ {0, 1, 2, 3}
2. rotate the square image k * 90 degree clockwise
3. if N is odd stop the encryption process
4. if N is even split the image into four equal sub-squares whose length is N / 2 and encrypt them recursively starting from step 0
Apparently different choices of the k serie result in different encrypted images. Given two images A and B, your task is to find out whether it is POSSIBLE that B is encrypted from A. B is possibly encrypted from A if there is a choice of k serie that encrypt A into B.
Input
Input may contains multiple testcases.
The first line of the input contains an integer T(1 <= T <= 10) which is the number of testcases.
The first line of each testcase is an integer N, the length of the side of the images A and B.
The following N lines each contain N integers, indicating the image A.
The next following N lines each contain N integers, indicating the image B.
For 20% of the data, 1 <= n <= 15
For 100% of the data, 1 <= n <= 100, 0 <= Aij, Bij <= 100000000
Output
For each testcase output Yes or No according to whether it is possible that B is encrypted from A.
- Sample Input
-
3 2 1 2 3 4 3 1 4 2 2 1 2 4 3 3 1 4 2 4 4 1 2 3 1 2 3 4 2 3 4 1 3 4 1 2 3 4 4 1 2 3 1 2 1 4 4 3 2 1 3 2
- Sample Output
-
Yes No Yes
Solution:
1 #include <cstdlib>
2 #include <cstdio>
3 #include <vector>
4 using namespace std;
5
6
7 void rotateImg(vector<vector<int> > &A) {
8
9 int N = A.size();
10 int r = N >> 1;
11 int c = (N + 1) >> 1;
12 for (int i = 0; i < r; ++i) {
13 for (int j = 0; j < c; ++j) {
14 int tmp = A[i][j];
15 A[i][j] = A[N - 1 - j][i];
16 A[N - 1 - j][i] = A[N - 1 - i][N - 1 - j];
17 A[N - 1 - i][N - 1 - j] = A[j][N - 1 - i];
18 A[j][N - 1 - i] = tmp;
19 }
20 }
21
22 }
23
24 bool isEqual(vector<vector<int> > A, vector<vector<int> > B) {
25 int N = A.size();
26 for (int i = 0; i < N; ++i) {
27 for (int j = 0; j < N; ++j) {
28 if (A[i][j] != B[i][j]) return false;
29 }
30 }
31 return true;
32 }
33
34 vector<vector<int> > split(vector<vector<int> > A, int x, int y) {
35 int N = A.size() >> 1;
36 vector<vector<int> > v(N, vector<int>(N, 0));
37 for (int i = 0; i < N; ++i) {
38 for (int j = 0; j < N; ++j) {
39 v[i][j] = A[i + x][j + y];
40 }
41 }
42 return v;
43 }
44
45 bool isEncrypted(vector<vector<int> > A, vector<vector<int> > B) {
46 int N = A.size();
47 if (N == 1) return A[0][0] == B[0][0];
48 if (N & 0x01) {
49 vector<vector<int> > AA = A;
50 for (int k = 0; k < 4; ++k) {
51 if (isEqual(AA, B)) return true;
52 rotateImg(AA);
53 }
54 return false;
55 }
56 else {
57 int N_2 = N >> 1;
58 vector<vector<int> > AA = A;
59 int x[] = { 0, N_2, 0, N_2 };
60 int y[] = { 0, 0, N_2, N_2 };
61 for (int k = 0; k < 4; ++k) {
62 bool flag = true;
63 for (int kk = 0; kk < 4; ++kk) {
64 if (!isEncrypted(split(AA, x[kk], y[kk]), split(B, x[kk], y[kk]))) {
65 flag = false;
66 break;
67 }
68 }
69 if (flag) return true;
70 rotateImg(AA);
71 }
72 return false;
73 }
74 }
75
76
77
78
79 int main() {
80 int T;
81 scanf("%d", &T);
82 while (T--) {
83 int N;
84 scanf("%d", &N);
85 vector<vector<int> > A(N, vector<int>(N, 0));
86 vector<vector<int> > B(N, vector<int>(N, 0));
87 for (int i = 0; i < N; ++i) {
88 for (int j = 0; j < N; ++j) {
89 int v;
90 scanf("%d", &v);
91 A[i][j] = v;
92 }
93 }
94
95 for (int i = 0; i < N; ++i) {
96 for (int j = 0; j < N; ++j) {
97 int v;
98 scanf("%d", &v);
99 B[i][j] = v;
100 }
101 }
102
103 if (isEncrypted(A, B)) printf("Yes\n");
104 else printf("No\n");
105 }
106 }