hihoCoder #1240 Image Encryption

时间：2015-10-11 21:30:12 阅读：266 评论：0 收藏：0 [点我收藏+]

标签：

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

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 }

hihoCoder #1240 Image Encryption

标签：

原文地址：http://www.cnblogs.com/liew/p/4869885.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行