Transformations-优快云博客

本文介绍了一种用于检测图像是否通过旋转或翻转等变换得到另一图像的算法。该算法通过比较原始图像与目标图像的不同变换状态来判断两者间的关系，支持多种变换模式，包括旋转90度、180度、270度及翻转。

这种实现方法太蠢，实际上根本不需要自己算出坐标变换，只需要实现两个操作：旋转，以及翻转。然后叠加即可（参见ANALYSIS）
transform.cpp

/*
ID: america6
PROG: transform
LANG: C++11
*/
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;
int n;

bool trans(const int& mode, const string& a, const string& b) {
    //bool flag = true;
    switch (mode) {
        case 1: {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (a[i*n+j] != b[j*n+n-i-1]) {
                        return false;
                    }
                }
            }
            break;
        }
        case 2: {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (a[i*n+j] != b[(n-i-1)*n+n-j-1]) {
                        return false;
                    }
                }
            }
            break;
        }
        case 3: {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (a[i*n+j] != b[(n-j-1)*n+i]) {
                        return false;
                    }
                }
            }
            break;
        }
        case 4: {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (a[i*n+j] != b[i*n+n-j-1]) {
                        return false;
                    }
                }
            }
            break;
        }
        case 5: {
            bool f1 = true;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (a[i*n+j] != b[(n-j-1)*n+n-i-1]
                     /*&& a[i*n+j] != b[(n-i-1)*n+j]
                     && a[i*n+j] != b[j*n+i]*/) {
                        f1 = false;
                    }
                }
            }
            bool f2 = true;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (a[i*n+j] != b[(n-i-1)*n+j]
                    /*a[i*n+j] != b[(n-j-1)*n+n-i-1]
                     && a[i*n+j] != b[(n-i-1)*n+j]
                     && a[i*n+j] != b[j*n+i]*/) {
                        f2 = false;
                    }
                }
            }
            bool f3 = true;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (a[i*n+j] != b[j*n+i]
                    /*a[i*n+j] != b[(n-j-1)*n+n-i-1]
                     && a[i*n+j] != b[(n-i-1)*n+j]
                     && a[i*n+j] != b[j*n+i]*/) {
                        f3 = false;
                    }
                }
            }
            return f1 || f2 || f3;
        }
        case 6: {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (a[i*n+j] != b[i*n+j]) {
                        return false;
                    }
                }
            }
            break;
        }
    }
    return true;
}

int main() {
    ofstream fout ("transform.out");
    ifstream fin ("transform.in");
    fin >> n;
    string before, after;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            char temp;
            fin >> temp;
            before.push_back(temp);
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            char temp;
            fin >> temp;
            after.push_back(temp);
        }
    }
    int mode = 7;
    for (int i = 1; i <= 6; i++) {
        if (trans(i, before, after) == true) {
            mode = i;
            break;
        }
    }
    fout << mode << endl;
    return 0;
}