1.1 transform 簡單模擬

自己寫的代碼很醜。

要注意旋轉和鏡像對應到二維數組時對應的操作

 

Transformations

A square pattern of size N x N (1 <= N <= 10) black and white square tiles is transformed into another square pattern. Write a program that will recognize the minimum transformation that has been applied to the original pattern given the following list of possible transformations:

• #1: 90 Degree Rotation: The pattern was rotated clockwise 90 degrees.
• #2: 180 Degree Rotation: The pattern was rotated clockwise 180 degrees.
• #3: 270 Degree Rotation: The pattern was rotated clockwise 270 degrees.
• #4: Reflection: The pattern was reflected horizontally (turned into a mirror image of itself by reflecting around a vertical line in the middle of the image).
• #5: Combination: The pattern was reflected horizontally and then subjected to one of the rotations (#1-#3).
• #6: No Change: The original pattern was not changed.
• #7: Invalid Transformation: The new pattern was not obtained by any of the above methods.

In the case that more than one transform could have been used, choose the one with the minimum number above.

PROGRAM NAME: transform

INPUT FORMAT

Line 1:	A single integer, N
Line 2..N+1:	N lines of N characters (each either `@' or `-'); this is the square before transformation
Line N+2..2*N+1:	N lines of N characters (each either `@' or `-'); this is the square after transformation

SAMPLE INPUT (file transform.in)

3


@-@


---


@@-


@-@


@--


--@

OUTPUT FORMAT

A single line containing the the number from 1 through 7 (described above) that categorizes the transformation required to change from the `before' representation to the `after' representation.

SAMPLE OUTPUT (file transform.out)

1



/*
ID: xdzhbin1
LANG: C++
TASK:transform
*/

#include <fstream>
using namespace std;

#define N 10

//順時針旋轉90度
void rotate(char result[][N],const char copy[][N], int n)
{
	int i,j;
	for(i=0; i<n; i++)
	{
		for(j=0; j<n; j++)
		{
			//why this?因爲旋轉過程可以分解爲先轉秩，再鏡像
			//a是原數組，b是a的轉秩，c是b的鏡像
			//a[i][j]=b[j][i],b[i][j]=c[n-1-i][j],由此推得a和c的關係
			result[i][j] = copy[n-1-j][i];
		}
	}
}

//鏡像操作
void mirror(char result[][N],const char copy[][N], int n)
{
	int i,j;
	for(i=0; i<n; i++)
	{
		for(j=0; j<n; j++)
		{
			result[i][j] = copy[i][n-1-j];
		}
	}
}

//檢查是否到達目標狀態
bool check(const char src[][N], const char dest[][N], int n)
{
	int i,j;
	for(i=0; i<n; i++)
		for(j=0; j<n; j++)
			if(src[i][j]!=dest[i][j]) return false;
	return true;
}

int main()
{
	ifstream in("transform.in",ios_base::in);
	ofstream out("transform.out",ios_base::out);

	char square[N][N];	//矩陣的初始化狀態
	char copy[N][N];	//初始化狀態的拷貝
	char dest[N][N];	//目標狀態
	int n,i,j;
	in>>n;	//矩陣的尺寸

	for(i=0; i<n; i++)
	{
		for(j=0; j<n; j++)
		{
			in>>square[i][j];	//讀入初始化狀態
			copy[i][j] = square[i][j];
		}
	}
	for(i=0; i<n; i++)
	{
		for(j=0; j<n; j++)
		{
			in>>dest[i][j];	//讀入目標狀態
		}
	}

	char temp[N][N];
	int operation = 0;

	rotate(temp,copy,n);//copy是原數組，temp是順時針旋轉90度的
	if(check(temp,dest,n))
	{
		operation = 1;
	}
	rotate(copy,temp,n);//copy是旋轉180度的
	if(!operation && check(copy,dest,n))
	{
		operation = 2;
	}
	rotate(temp,copy,n);
	if(!operation && check(temp,dest,n))
	{
		operation = 3;
	}

	mirror(temp,square,n);//temp現在是原數組的鏡像
	if(!operation && check(temp,dest,n))
	{
		operation = 4;
	}

	rotate(copy,temp,n);
	if(!operation && check(copy,dest,n))
		operation = 5;
	rotate(temp,copy,n);
	if(!operation && check(temp,dest,n))
		operation = 5;
	rotate(copy,temp,n);
	if(!operation && check(copy,dest,n))
		operation = 5;

	if(!operation && check(square,dest,n))
		operation = 6;
	if(!operation)
		operation = 7;

	out<<operation<<endl;
	in.close();
	out.close();
	return 0;
}