Uva12113-overlapping squares (Binary enumeration)

Problem uva12113-overlapping Squares

accept:116 submit:596time limit:3000 mSec problem Description

Input

The input consists of several test cases. Each test case is a contained in-ve lines and each line contains nine characters. If the horizontal border of a? lled square is visible it's denoted with "(ASCII value) sign and if vertical border of a? lled square is visible then it's denoted with ' | ' (ASCII value 124) character. The board contains no other character than ', ' | ' and of course ' (ASCII Value 32). The border lines of the squares can only be along the grid lines. Each board lines end with a ' # ' (Hash character) which denotes the end of line. This character isn't a part of the grid or square. The last test case was followed by a single zero, which should wasn't be processed.

Output

For each test case, print the case number and ' Yes ' or ' No ', depending on whether it's possible to form the target.

Sample Input

Sample ouput

Case 1:yes
Case 2:yes
Case 3:no
Case 4:yes

Solution: Feel the recent problems are very test code ability (but I am very water), think of only nine kinds of layout after the thinking of the problem is basically over, all the selection scheme only 2^9, direct binary enumeration, for the same selection scheme, different placement order will also bring different coverage results, The solution is next_permutation (), pre-processing the different small squares of the overlay lattice label, the next violence is good.

1#include <bits/stdc++.h>2 3 using namespacestd;4 5 Const intMAXN = (1<<9);6 Const intN =5, M =9;7 Const intKind =9;8 9 intedge[9][8] = {{1,3,9, -, -, +, +, A}};Ten intcore[9][4] = {{Ten, One, A, -}}; One intBits[kind+1],target[n*M]; A  - intRead () { -     Charstr[ -]; the     intCNT =0, edges =0; -      for(inti =0; I < n;i++){ - gets (str); -         if(str[0] =='0')return-1; +          for(intj =0; J < m;j++){ -             if(Str[j] = =' ') target[cnt++] =0; +             Elsetarget[cnt++] =1, edges++; A         } at     } -     returnedges; - } -  - voidinit () { -      for(inti =0; I <3; i++){ in          for(intj =0; J <3; j + +){ -             if(!i &&!j)Continue; to             intPlus,minus; +             if(J = =0) plus =9, minus =3; -             ElsePlus =2, minus =1; the              for(intK =0; k <8; k++){ *edge[i*3+j][k] = edge[i*3+j-minus][k]+Plus; $             }Panax Notoginseng              for(intK =0; k <4; k++){ -core[i*3+j][k] = core[i*3+j-minus][k]+Plus; the             } +         } A     } the } +  - intBitcount (ints) { $     returns = =0?0: Bitcount (s>>1) + (s&1); $ } -  - voidBitpos (ints) { the     intCNT =0; -      for(inti =0; I <9; i++){Wuyi         if(s& (1<<i)) bits[cnt++] =i; the     } - } Wu  - intIcase =1; About  $ intMain () - { - #ifdef Geh -Freopen ("HELLOWORLD.01,INP","R", stdin); A #endif + init (); the     intedge_cnt; -      while(edge_cnt=Read ()) { $         if(edge_cnt = =-1) Break; the         inttmp[m*N]; the         BOOLOK =false; the          for(ints =0; s < maxn;s++){ the             intn =Bitcount (s); - Bitpos (s); in             if(n>6|| N8&LT;EDGE_CNT)Continue; the              Do{ thememset (TMP,0,sizeof(TMP)); About                  for(inti =0; I < n;i++){ the                      for(intj =0; J <8; j + +){ theTMP[EDGE[BITS[I]][J]] =1; the                     } +                      for(intj =0; J <4; j + +){ -TMP[CORE[BITS[I]][J]] =0; the                     }Bayi                 } the  the                 if(MEMCMP (Tmp,target,sizeof(target)) ==0){ -OK =true; -                      Break; the                 } the} while(Next_permutation (bits,bits+n)); the             if(OK) Break; the         } -printf"Case %d:", icase++); the         if(OK) printf ("yes\n"); the         Elseprintf"no\n"); the     }94     return 0; the}

Uva12113-overlapping squares (Binary enumeration)