Section 1.2 Transformations

最新推荐文章于 2025-12-31 19:32:08 发布

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本文介绍了一种用于检测矩阵经过特定变换后的状态匹配算法。通过旋转和平面对称等操作，该算法能够判断两个矩阵是否可以通过预定义的变化方式从一个转换到另一个。支持90度、180度和270度旋转，以及水平翻转和组合变换等多种情况。

这道搜索题没什么绕弯的地方，把transformation用矩阵位置变换表示，然后一一验证即可。

对于#5 combination我写得比较啰嗦: )

/*
ID: alexyua2
PROG: transform
LANG: C++
*/

#include<fstream>
using namespace std;

ifstream fin("transform.in");
ofstream fout("transform.out");

//ifstream fin("in.txt");
//ofstream fout("out.txt");

int MAX = 11;

int main()
{
    //input
    int N;
    fin>>N;
    char before[MAX][MAX],after[MAX][MAX];
    int i,j;
    for(i=1;i<=N;i++)
        for(j=1;j<=N;j++)
            fin>>before[i][j];
    for(i=1;i<=N;i++)
        for(j=1;j<=N;j++)
            fin>>after[i][j];

    //search transformation pattern
        //#1:90 Degree Rotation
    int state = 1;
    for(i=1;i<=N;i++)
        for(j=1;j<=N;j++)
            if(before[i][j] != after[j][N+1-i])
            {
                state = 0;
                break;
            }
    if(state == 1)
    {
        fout<<"1" <<endl;
        return 0;
    }
        //#2:180 Degree Rotation
    state = 1;
    for(i=1;i<=N;i++)
        for(j=1;j<=N;j++)
            if(before[i][j] != after[N+1-i][N+1-j])
            {
                state = 0;
                break;
            }
    if(state == 1)
    {
        fout<<"2" <<endl;
        return 0;
    }
        //#3:270 Degree Rotation
    state = 1;
    for(i=1;i<=N;i++)
        for(j=1;j<=N;j++)
            if(before[i][j] != after[N+1-j][i])
            {
                state = 0;
                break;
            }
    if(state == 1)
    {
        fout<<"3" <<endl;
        return 0;
    }
        //#4:Reflection
    state = 1;
    for(i=1;i<=N;i++)
        for(j=1;j<=N;j++)
            if(before[i][j] != after[i][N+1-j])
            {
                state = 0;
                break;
            }
    if(state == 1)
    {
        fout<<"4" <<endl;
        return 0;
    }
        //#5:Combination
    char temp[MAX][MAX];
    for(i=1;i<=N;i++)
        for(j=1;j<=N;j++)
            temp[i][j] = before[i][N+1-j];
    state = 1;
    for(i=1;i<=N;i++)
        for(j=1;j<=N;j++)
            if(temp[i][j] != after[j][N+1-i])
            {
                state = 0;
                break;
            }
    if(state == 1)
    {
        fout<<"5" <<endl;
        return 0;
    }
    state = 1;
    for(i=1;i<=N;i++)
        for(j=1;j<=N;j++)
            if(temp[i][j] != after[N+1-i][N+1-j])
            {
                state = 0;
                break;
            }
    if(state == 1)
    {
        fout<<"5" <<endl;
        return 0;
    }
    state = 1;
    for(i=1;i<=N;i++)
        for(j=1;j<=N;j++)
            if(temp[i][j] != after[N+1-j][i])
            {
                state = 0;
                break;
            }
    if(state == 1)
    {
        fout<<"5" <<endl;
        return 0;
    }
        //#6:No Change
    state = 1;
    for(i=1;i<=N;i++)
        for(j=1;j<=N;j++)
            if(before[i][j] != after[i][j])
            {
                state = 0;
                break;
            }
    if(state == 1)
    {
        fout<<"6" <<endl;
        return 0;
    }
        //#7:No Answer
    fout<<"7" <<endl;
    //close files
    fin.close();
    fout.close();
    return 0;
}