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 @-@ --- @@- @-@ @-- --@
1
/* ID: jszhais1 PROG: transform LANG: C++ */ #include <iostream> #include <fstream> #include <string> using namespace std; bool rotation(int length,char (*map)[10],char (*rotationMap)[10],int degree) { bool judger = true; switch (degree) { case 450: for (int i=0; i<length&&judger; i++) for(int j=0; j<length; j++) if(map[i][j]!=rotationMap[i][j]) { judger=false; break; } break; case 90: for (int j=0,ii=0; j<length; j++,ii++) for(int i=length-1,jj=0; i>=0&&judger; i--,jj++) if(map[i][j] != rotationMap[ii][jj]) { judger=false; break; } break; case 180: for (int i=length-1,ii=0; i>=0&&judger; i--,ii++) for(int j=length-1,jj=0; j>=0; j--,jj++) if(map[i][j]!=rotationMap[ii][jj]) { judger=false; break; } break; case 270: for (int j=length-1,ii=0; j>=0&&judger; j--,ii++) for(int i=0,jj=0; i<length; i++,jj++) if(map[i][j]!=rotationMap[ii][jj]) { judger=false; break; } break; case 360: for (int i=0; i<length&&judger; i++) for(int j=length-1,jj=0; j>=0; j--,jj++) if(map[i][j]!=rotationMap[i][jj]) { judger=false; break; } break; default: break; } return judger; } void getReflectionMap(int length,char (*map)[10],char (*reflectionMap)[10]) { for (int i=0; i<length; i++) { for(int j=length-1,jj=0; j>=0; j--,jj++) reflectionMap[i][jj]=map[i][j]; } } int main(int argc, const char * argv[]) { ofstream fout("transform.out"); ifstream fin("transform.in"); int length; while(fin>>length){ char map[10][10]={' '}; char rotationMap[10][10] = {' '}; char reflectionMap[10][10] = {' '}; for(int i=0 ; i<length ;i++) for(int j=0; j< length; j++) fin>>map[i][j]; // getReflectionMap(3, &map[0], &reflectionMap[0]); for(int i=0 ; i<length ;i++) for(int j=0; j< length; j++) fin>>rotationMap[i][j]; for(int degree=90 ;degree<=450;degree+=90) { if(rotation(length,&map[0], &rotationMap[0], degree)) { switch (degree) { case 450: fout<<6<<endl; break; case 90: fout<<1<<endl; break; case 180: fout<<2<<endl; break; case 270: fout<<3<<endl; break; case 360: fout<<4<<endl; break; default: break; } return 0; } } getReflectionMap(length, &map[0], &reflectionMap[0]); for(int degree=90 ;degree<=450;degree+=90) { if(rotation(length,&reflectionMap[0], &rotationMap[0], degree)) { fout<<5<<endl; return 0; } } fout<<7<<endl; } return 0; }结果:
USER: jim zhai [jszhais1] TASK: transform LANG: C++ Compiling... Compile: OK Executing... Test 1: TEST OK [0.000 secs, 3360 KB] Test 2: TEST OK [0.000 secs, 3360 KB] Test 3: TEST OK [0.000 secs, 3360 KB] Test 4: TEST OK [0.000 secs, 3360 KB] Test 5: TEST OK [0.000 secs, 3360 KB] Test 6: TEST OK [0.000 secs, 3360 KB] Test 7: TEST OK [0.000 secs, 3360 KB] Test 8: TEST OK [0.000 secs, 3360 KB] All tests OK
Here are the test data inputs:
------- test 1 ---- 3 --- --- --- --- -@- --- ------- test 2 ---- 5 -@@@- -@@-- -@--- ----- ----- ----- ----@ ---@@ --@@@ ----- ------- test 3 ---- 5 @@@@@ @---@ @@@@@ @@@@@ @@@@@ @@@@@ @@@@@ @@@@@ @---@ @@@@@ ------- test 4 ---- 6 -@-@-@ @-@-@- -@-@-@ @-@-@- -@-@-@ @-@-@- @-@-@- -@-@-@ @-@-@- -@-@-@ @-@-@- -@-@-@ ------- test 5 ---- 3 @@@ --- @@@ @@@ --- @@@ ------- test 6 ---- 4 @@@@ @@@@ -@@@ @@@@ @@@@ @@@@ @@@- @@@@ ------- test 7 ---- 4 @-@@ @@@@ @@@@ @@@@ @@@@ @@@@ @@@@ @-@@ ------- test 8 ---- 10 @--------@ ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- @--------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------@
We represent a board as a data structure containing the dimension and the contents. We pass around the data structure itself, not a reference to it, so that we can return new boards, and so on.
This makes it easy to define reflect and rotate operations that return reflected and rotated boards.
Once we have these, we just check to see what combination of transformations makes the old board into the new board.
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <assert.h> #define MAXN 10 typedef struct Board Board; struct Board { int n; char b[MAXN][MAXN]; }; /* rotate 90 degree clockwise: [r, c] -> [c, n+1 - r] */ Board rotate(Board b) { Board nb; int r, c; nb = b; 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] */ Board reflect(Board b) { Board nb; int r, c; nb = b; 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 */ int eqboard(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; } Board rdboard(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; } void main(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); }