LA 3510 (displacement cycle decomposition) Pixel Shuffle

The idea is quite simple, each command in the title (including the inverse of the command) is equivalent to a permutation.

The time complexity of O (n2k) is used to find the product A of these permutations from right to left, and then m to make AM = I (i is congruent permutation)

First, the A decomposition cycle, M is equal to all the length of the Loop section least common multiple.

It is important to note that:

The execution command is executed from right to left, which is said in the title =_=

The other commands are OK, mix that command to get me confused, the title is reversed, we want to reverse the original image (I, j) in the new image position.

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <string>5#include <sstream>6#include <vector>7 using namespacestd;8 9 intgcdintAintb)Ten{returnb = =0? A:GCD (b, a%b); } One  A intLcmintAintb) -{returnA/GCD (A, b) *b;} -  the Const intMAXN =1025; - intN; - Chars[ -], op[ *][Ten]; -  +InlineintID (intIintj) -{returnI*n +J;} +  A intNewpos (intIintJConst Char*op) at { -      if(op[0] =='R')returnID (n-j-1, i); -      if(op[0] =='s')returnID (i, n-j-1); -      if(op[0] =='b'&& op[1] =='h') -      { -          if(I >= n/2)returnID (i, n-j-1); in          returnID (i, j); -      } to      if(op[0] =='b'&& op[1] =='v') +      { -          if(I >= n/2)returnID (N-(i-n/2)-1, j); the          returnID (i, j); *      } $      if(op[0] =='D')Panax Notoginseng      { -          if(I &1)returnID (n/2+ i/2, j); the          returnID (i/2, j); +      } A      if(op[0] =='m') the      { +          intK = i/2; -          if(J < n/2)returnI2==0? ID (k*2, j*2): ID (k*2, j*2+1); $          Else returnI2==0? ID (2*k+1,2* (j-n/2)): ID (2*k+1,2* (j-n/2)+1); $      } -      returnID (i, j); - } the  - intCUR[MAXN * MAXN], ORIGIN[MAXN *MAXN];Wuyi  the voidApplyConst Char*op) - { Wu      for(inti =0; i < n*n; i++) Origin[i] =Cur[i]; -     BOOLINV = Op[strlen (OP)-1] =='-'?true:false; About      for(inti =0; I < n; i++) $          for(intj =0; J < N; J + +) -         { -             intp = ID (i, j), p2 =Newpos (I, J, op); -             if(INV) Cur[p] =ORIGIN[P2]; A             ElseCUR[P2] =Origin[p]; +         } the } -  $ BOOLVIS[MAXN *MAXN]; the  the intSolve () the { thememset (Vis,false,sizeof(Vis)); -     intAns =1; in      for(inti =0; i < n*n; i++)if(!Vis[i]) the     { the         intCNT =0, j =i; About          Do the         { theVIS[J] =1; thecnt++; +j =Cur[j]; -} while(J! =i); theAns =LCM (cnt, ans);Bayi     } the     returnans; the } -  - intMain () the { the     //freopen ("In.txt", "R", stdin); the  the     intT; -scanf"%d", &T); the      for(intKase =0; Kase < T; kase++) the     { the         if(Kase) puts ("");94  thescanf"%d", &N); GetChar (); the         stringLine , temp; the getline (CIN, line);98 StringStream SS (line); Aboutvector<string>op; -          while(SS >>temp) Op.push_back (temp);101 102          for(inti =0; i < n*n; i++) Cur[i] =i;103          for(inti = op.size ()-1; I >=0; i--) Apply (OP[I].C_STR ());104printf"%d\n", Solve ()); the     }106 107     return 0;108}

code June

LA 3510 (displacement cycle decomposition) Pixel Shuffle