1.1 transform 简单模拟

自己写的代码很丑。

要注意旋转和镜像对应到二维数组时对应的操作

 

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

//顺时针旋转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?因为旋转过程可以分解为先转秩，再镜像
			//a是原数组，b是a的转秩，c是b的镜像
			//a[i][j]=b[j][i],b[i][j]=c[n-1-i][j],由此推得a和c的关系
			result[i][j] = copy[n-1-j][i];
		}
	}
}

//镜像操作
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];
		}
	}
}

//检查是否到达目标状态
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];	//矩阵的初始化状态
	char copy[N][N];	//初始化状态的拷贝
	char dest[N][N];	//目标状态
	int n,i,j;
	in>>n;	//矩阵的尺寸

	for(i=0; i<n; i++)
	{
		for(j=0; j<n; j++)
		{
			in>>square[i][j];	//读入初始化状态
			copy[i][j] = square[i][j];
		}
	}
	for(i=0; i<n; i++)
	{
		for(j=0; j<n; j++)
		{
			in>>dest[i][j];	//读入目标状态
		}
	}

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

	rotate(temp,copy,n);//copy是原数组，temp是顺时针旋转90度的
	if(check(temp,dest,n))
	{
		operation = 1;
	}
	rotate(copy,temp,n);//copy是旋转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现在是原数组的镜像
	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;
}