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:
In the case that more than one transform could have been used, choose the one with the minimum number above.
| 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 |
3 @-@ --- @@- @-@ @-- --@
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.
1
题解：模拟题
写一个旋转90度函数，和一个投影函数 其他几种情况可运用这两个函数实现
旋转90°：目标矩阵 after[j][N-i-1]=原矩阵 before[i][j]
投影： 目标矩阵 after[i][N-j-1]=原矩阵 before[i][j]
/*
ID: cxq_xia1
PROG: transform
LANG: C++
*/
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=11;
int N;
char before1[maxn][maxn],after1[maxn][maxn];
bool isSame(char before[maxn][maxn],char after[maxn][maxn])
{
for(int i=0;i<N;i++)
{
for(int j=0;j<N;j++)
{
if(before[i][j]!=after[i][j])
{
return false;
}
}
}
return true;
}
void clockwise_90(char before[maxn][maxn],char after[maxn][maxn])
{
for(int i=0;i<N;i++)
{
for(int j=0;j<N;j++)
{
after[j][N-i-1]=before[i][j];
}
}
}
void Reflection(char before[maxn][maxn],char after[maxn][maxn])
{
for(int i=0;i<N;i++)
{
for(int j=0;j<N;j++)
{
after[i][N-j-1]=before[i][j];
}
}
}
int solve()
{
char wise_90[maxn][maxn],wise_180[maxn][maxn],wise_270[maxn][maxn],reflect[maxn][maxn];
clockwise_90(before1,wise_90);
if(isSame(after1,wise_90))
return 1;
clockwise_90(wise_90,wise_180);
if(isSame(after1,wise_180))
return 2;
clockwise_90(wise_180,wise_270);
if(isSame(after1,wise_270))
return 3;
Reflection(before1,reflect);
if(isSame(after1,reflect))
return 4;
Reflection(after1,reflect);
if(isSame(wise_90,reflect)||isSame(wise_180,reflect)||isSame(wise_270,reflect))
return 5;
if(isSame(before1,after1))
return 6;
return 7;
}
int main()
{
freopen("transform.in","r",stdin);
freopen("transform.out","w",stdout);
while(cin >> N)
{
for(int i=0;i<N;i++)
{
for(int j=0;j<N;j++)
{
cin >> before1[i][j];
}
}
for(int i=0;i<N;i++)
{
for(int j=0;j<N;j++)
{
cin >> after1[i][j];
}
}
cout <<solve()<<endl;
}
return 0;
}
原文：http://www.cnblogs.com/WillsCheng/p/4768356.html