[Usaco1.2 _ 2]★Transformations block Conversion

The black-and-white tile pattern of an n x n (1 <= n <= 10) Square is converted into a new square pattern. Write a program to find the minimum way to convert the original pattern into a new pattern using the following column conversion method:

1: Turn 90 degrees: The pattern turns 90 degrees clockwise.

2: turn 180 degrees: The pattern turns clockwise to 180 degrees.

3: Turn 270 degrees: The pattern turns clockwise to 270 degrees.

4: Reflection: The image is flipped horizontally (the image with the central vertical line as the heart to form the original pattern ).

5: combination: The pattern is flipped horizontally, and then converted again according to one to three.

6: Do not change: the original pattern does not change.

7: Invalid conversion: the new pattern cannot be obtained using the above method.

If you have multiple available conversion methods, select the one with the smallest serial number.

Only one step from 1 to 7 is used to complete the conversion.

[Edit] format

Program name: Transform

Input Format:

(File transform. In)

The first line is a separate integer n.

Line 2 to line n + 1: line N contains n characters (either "@" or "-"); this is the square before conversion.

Line n + 2 to line 2nd * n + 1: line N contains n characters ("@" or "-"). This is the converted square.

Output Format:

(File transform. out)

A separate row contains a number ranging from 1 to 7 (as described above), indicating that the square before conversion needs to be converted into the transformation method of the square after conversion.

[Edit] sample input

3@[email protected]---@@-@[email protected]@--[email protected]

[Edit] sample output

1

/* ID: zwcwu52prog: transformlang: C ++ */# include <iostream> # include <fstream> # include <string> using namespace STD; char blksrc [10] [10], blkdst [10] [10]; // source array, target matrix int main () {ofstream fout ("transform. out "); ifstream fin (" transform. in "); unsigned N; // order unsigned num = 7; // solution number fin> N; // read the source array for (unsigned I = 0; I <n; I ++) {for (unsigned J = 0; j <n; j ++) {fin> blksrc [I] [J];} // read the target phalanx for (unsigned I = 0; I <n; I ++) {for (unsigned J = 0; j <n; j ++) {fin> blkdst [I] [J];} do {bool B = true; // check whether the do {bool B = true; // check by solution 1 for (unsigned I = 0; I <n; I ++) {for (unsigned J = 0; j <n; j ++) {If (blksrc [I] [J]! = Blkdst [J] [n-I-1]) {B = false; break ;}}if (true = B) {num = 1; break ;} // detect B = true by solution 2; for (unsigned I = 0; I <n; I ++) {for (unsigned J = 0; j <N; j ++) {If (blksrc [I] [J]! = Blkdst [n-I-1] [n-J-1]) {B = false; break ;}} if (true = B) {num = 2; break;} // detect B = true by solution 3; for (unsigned I = 0; I <n; I ++) {for (unsigned J = 0; j <N; j ++) {If (blksrc [I] [J]! = Blkdst [n-J-1] [I]) {B = false; break ;}}if (true = B) {num = 3; break ;} // check according to solution 4 B = true; For (unsigned I = 0; I <n; I ++) {for (unsigned J = 0; j <N; j ++) {If (blksrc [I] [J]! = Blkdst [I] [n-J-1]) {B = false; break ;}}if (true = B) {num = 4; break ;} // check according to solution 5 B = true; For (unsigned I = 0; I <n; I ++) {for (unsigned J = 0; j <N; j ++) {If (blksrc [I] [J]! = Blkdst [n-J-1] [n-I-1]) {B = false; break ;}} if (true = B) {num = 5; break;} // according to solution 5 detection B = true; For (unsigned I = 0; I <n; I ++) {for (unsigned J = 0; j <N; j ++) {If (blksrc [I] [J]! = Blkdst [n-I-1] [J]) {B = false; break ;}}if (true = B) {num = 5; break ;} // check according to solution 5 B = true; For (unsigned I = 0; I <n; I ++) {for (unsigned J = 0; j <N; j ++) {If (blksrc [I] [J]! = Blkdst [J] [I]) {B = false; break ;}}if (true = B) {num = 5; break ;} // check B = true according to solution 6; For (unsigned I = 0; I <n; I ++) {for (unsigned J = 0; j <N; j ++) {If (blksrc [I] [J]! = Blkdst [I] [J]) {B = false; break ;}}if (true = B) {num = 6; break ;}} while (0 ); fout <num <Endl; return 0 ;}