USACO Section 1.2.3 Transformations

Mar 11, 2013 2 分钟阅读时间 usc, csdn

此页面通过工具从 csdn 导出，格式可能有问题。

题目

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)

思路

思路很简单，就是枚举各种转换方法，然后与莫状态进行比较。

繁琐即需要注意的是，对于每种转换方式 i和j的关系，稍有疏忽就会全军覆没

另外注意的就是：如果初末状态相同，不能直接输出6，要先看看之前的5种方法能不能得到。因为题目要求从小号优先。我就被坑了一次。

代码

/*
ID:zhrln1
PROG:transform
LANG:C++
*/
#include <cstdio>
int n;
typedef struct board{
	char v[11][11];
} Board;
Board a,b;
Board rdboard(int n){
	Board c;
	for (int i(1);i<=n;i++){
		for (int j(1);j<=n;j++){
			scanf("%c",&c.v[i][j]);
		}
		scanf("\n");
	}
	return c;
}
Board change(Board a,int k,int n){
	Board c;
	for (int i(1);i<=n;i++){
		for (int j(1);j<=n;j++){
			switch(k){
				case 1: c.v[i][j]=a.v[n-j+1][i];break;
				case 2: c.v[i][j]=a.v[n+1-i][n+1-j];break;
				case 3: c.v[i][j]=a.v[j][n+1-i];break;
				case 4: c.v[i][j]=a.v[i][n+1-j];break;
			}
		}
	}
	return c;
}
int isequel(Board a,Board b,int n){
	for (int i(1);i<=n;i++){
		for (int j(1);j<=n;j++){
			if (a.v[i][j]!=b.v[i][j]) return 0;
		}
	}
	return 1;
}
int main(){
	freopen("transform.in","r",stdin);
	freopen("transform.out","w",stdout);
	scanf("%d\n",&n);
	a=rdboard(n);
	b=rdboard(n);
	for (int i(1);i<=4;i++){
		Board t=change(a,i,n);
		if (isequel(t,b,n)){
			printf("%d\n",i);
			return 0;
		}
		if (i==4){
			for (int j(1);j<=3;j++){
				if (isequel(change(t,j,n),b,n)){
					printf("5\n");
					return 0;
				}
			}
		}
	}
	if (isequel(a,b,n)){
		printf("6\n");
		return 0;
	}
	printf("7\n");
	return 0;
}

USACO Algorithm ACM

huiren

Code Artisan

问渠那得清如许，为有源头活水来