Section 1.2 Transformations

题目原文

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, NLine 2..N+1:N lines of N characters (each either `@' or `-'); this is the square before transformationLine 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

分析

本题首先给出了一个方阵（使用字符表示）的6种变换方式，分别包括旋转（90 180 270）、镜像反转、反转加旋转、保持原状，然后输入一个n*n的方阵以及变换后的结果，要求判断是由哪种变换得到的结果，如果不在上述变换种类中，则输出7.

要解决本题，首先要实现旋转和镜像反转两种基本的变换方式，这里只需要实现旋转90度以及镜像反转两种操作即可，其他操作都是这两种操作的组合。实现基本操作后只需要将原始状态和结果状态输入，依次判断七种变换方式即可。

提交代码

/*ID: PROG: transformLANG: C++*/#include <fstream>#include <algorithm>#include <vector>using namespace std;typedef vector<vector<char> > pattern;pattern rotate90(pattern input,int n){pattern output;output.resize(n);for (int i=0;i!=n;i++){output[i].resize(n);}for (int i=0;i!=n;i++){for (int j=0;j!=n;j++){output[i][j] = input[n-1-j][i];}}return output;}pattern reflect(pattern input,int n){pattern output;output.resize(n);for (int i=0;i!=n;i++){output[i].resize(n);}for (int i=0;i!=n;i++){for (int j=0;j!=n;j++){output[i][j] = input[i][n-1-j];}}return output;}int transform(pattern source,pattern target,int n){if (target == rotate90(source,n)){return 1;}else if(target == rotate90(rotate90(source,n),n)){return 2;}else if (target == rotate90(rotate90(rotate90(source,n),n),n)){return 3;}else if (target == reflect(source,n)){return 4;}else if (target == source){return 6;}else{if (target == rotate90(reflect(source,n),n)|| target == rotate90(rotate90(reflect(source,n),n),n)|| target == rotate90(rotate90(rotate90(source,n),n),n)){return 5;}elsereturn 7;}}int main(){ifstream fin("transform.in");int n;fin >> n;pattern t1(n);pattern t2(n);for (int i=0;i!=2*n;i++){for (int j=0;j!=n;j++){char a;fin >> a;if(i <n){t1[i%n].push_back(a);}else{t2[i%n].push_back(a);}}}ofstream fout("transform.out");fout << transform(t1,t2,n) << endl;return 0;}

提交结果

TASK: transformLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.008 secs, 3504 KB]   Test 2: TEST OK [0.008 secs, 3504 KB]   Test 3: TEST OK [0.008 secs, 3504 KB]   Test 4: TEST OK [0.005 secs, 3504 KB]   Test 5: TEST OK [0.008 secs, 3504 KB]   Test 6: TEST OK [0.005 secs, 3504 KB]   Test 7: TEST OK [0.003 secs, 3504 KB]   Test 8: TEST OK [0.003 secs, 3504 KB]All tests OK.

官方参考答案

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <assert.h>#define MAXN 10typedef struct Board Board;struct Board {    int n;    char b[MAXN][MAXN];};/* rotate 90 degree clockwise: [r, c] -> [c, n - r] */Boardrotate(Board b){    Board nb;    int r, c;    nb.n = b.n;    for(r=0; r<b.n; r++)        for(c=0; c<b.n; c++)            nb.b[c][b.n-1 - r] = b.b[r][c];    return nb;}/* reflect board horizontally: [r, c] -> [r, n-1 -c] */Boardreflect(Board b){    Board nb;    int r, c;    nb.n = b.n;    for(r=0; r<b.n; r++)        for(c=0; c<b.n; c++)            nb.b[r][b.n-1 - c] = b.b[r][c];    return nb;}/* return non-zero if and only if boards are equal */inteqboard(Board b, Board bb){    int r, c;    if(b.n != bb.n)        return 0;    for(r=0; r<b.n; r++)        for(c=0; c<b.n; c++)            if(b.b[r][c] != bb.b[r][c])                return 0;    return 1;}Boardrdboard(FILE *fin, int n){    Board b;    int r, c;    b.n = n;    for(r=0; r<n; r++) {        for(c=0; c<n; c++)            b.b[r][c] = getc(fin);        assert(getc(fin) == '\n');    }    return b;}voidmain(void){    FILE *fin, *fout;    Board b, nb;    int n, change;    fin = fopen("transform.in", "r");    fout = fopen("transform.out", "w");    assert(fin != NULL && fout != NULL);    fscanf(fin, "%d\n", &n);    b = rdboard(fin, n);    nb = rdboard(fin, n);    if(eqboard(nb, rotate(b)))        change = 1;    else if(eqboard(nb, rotate(rotate(b))))        change = 2;    else if(eqboard(nb, rotate(rotate(rotate(b)))))        change = 3;    else if(eqboard(nb, reflect(b)))        change = 4;    else if(eqboard(nb, rotate(reflect(b)))         || eqboard(nb, rotate(rotate(reflect(b))))         || eqboard(nb, rotate(rotate(rotate(reflect(b))))))        change = 5;    else if(eqboard(nb, b))        change = 6;    else        change = 7;    fprintf(fout, "%d\n", change);    exit(0);}

THE END