UVa 227 Puzzle
| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
A children's puzzle that was popular 30 years ago consisted of a 5x5 frame which contained 24 small squares of equal size. A unique letter of the alphabet was printed on each small square. Since there were only 24 squares within the frame, the frame also contained an empty position which was the same size as a small square. A square could be moved into that empty position if it were immediately to the right, to the left, above, or below the empty position. The object of the puzzle was to slide squares into the empty position so that the frame displayed the letters in alphabetical order.
The illustration below represents a puzzle in its original configuration and in its configuration after the following sequence of 6 moves:
1) The square above the empty position moves.
2) The square to the right of the empty position moves.
3) The square to the right of the empty position moves.
4) The square below the empty position moves.
5) The square below the empty position moves.
6) The square to the left of the empty position moves.
Write a program to display resulting frames given their initial configurations and sequences of moves.
Input
Input for your program consists of several puzzles. Each is described by its initial configuration and the sequence of moves on the puzzle. The first 5 lines of each puzzle description are the starting configuration. Subsequent lines give the sequence of moves.
The first line of the frame display corresponds to the top line of squares in the puzzle. The other lines follow in order. The empty position in a frame is indicated by a blank. Each display line contains exactly 5 characters, beginning with the character on the leftmost square (or a blank if the leftmost square is actually the empty frame position). The display lines will correspond to a legitimate puzzle.
The sequence of moves is represented by a sequence of As, Bs, Rs, and Ls to denote which square moves into the empty position. A denotes that the square above the empty position moves; B denotes that the square below the empty position moves; L denotes that the square to the left of the empty position moves; R denotes that the square to the right of the empty position moves. It is possible that there is an illegal move, even when it is represented by one of the 4 move characters. If an illegal move occurs, the puzzle is considered to have no final configuration. This sequence of moves may be spread over several lines, but it always ends in the digit 0. The end of data is denoted by the character Z.
Output
Output for each puzzle begins with an appropriately labeled number (Puzzle #1, Puzzle #2, etc.). If the puzzle has no final configuration, then a message to that effect should follow. Otherwise that final configuration should be displayed.
Format each line for a final configuration so that there is a single blank character between two adjacent letters. Treat the empty square the same as a letter. For example, if the blank is an interior position, then it will appear as a sequence of 3 blanks - one to separate it from the square to the left, one for the empty position itself, and one to separate it from the square to the right.
Separate output from different puzzle records by one blank line.
Note: The first record of the sample input corresponds to the puzzle illustrated above.
Sample Input
TRGSJ XDOKI M VLN WPABE UQHCF ARRBBL0 ABCDE FGHIJ KLMNO PQRS TUVWX AAA LLLL0 ABCDE FGHIJ KLMNO PQRS TUVWX AAAAABBRRRLL0 Z
Sample Output
Puzzle #1: T R G S J X O K L I M D V B N W P A E U Q H C F Puzzle #2: A B C D F G H I E K L M N J P Q R S O T U V W X Puzzle #3: This puzzle has no final configuration.
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <algorithm> 5 using namespace std; 6 7 struct zb 8 { 9 int x,y; 10 } xy; 11 12 void findxy( string s, int x ) 13 { 14 int len=s.size(); 15 for( int i=0; i!=len; ++i ) 16 { 17 if(s[i]==' ') 18 { 19 xy.x=x; 20 xy.y=i; 21 return; 22 } 23 } 24 return; 25 } 26 27 int main() 28 { 29 string s,st; 30 vector<string> v; 31 int ti(1); 32 while(true) 33 { 34 v.clear(); 35 getline(cin,s); 36 if(s[0]=='Z') 37 break; 38 if(ti>1) 39 cout<<endl; // 输出中间加空行 40 v.push_back(s); 41 findxy(s,0); 42 for(int i=1; i!=5; ++i) 43 { 44 getline(cin,s); 45 v.push_back(s); 46 findxy(s,i); 47 } 48 // 读完矩阵，并找到空格坐标。 49 s=""; 50 while(true) 51 { 52 getline(cin,st); 53 s+=st; 54 if(st[st.size()-1]=='0') break; 55 } 56 // 读完判断序列 ABLR 57 bool cando=true; 58 int len=s.size(); 59 for( int i=0; i!=len; ++i ) 60 { 61 char c=s[i]; 62 if(c=='0') 63 break; 64 int line=xy.x,row=xy.y; 65 if(c=='A') 66 { 67 if(line==0) 68 { 69 cando=false; 70 break; 71 } 72 else 73 { 74 --xy.x; 75 swap(v[line][row],v[line-1][row]); 76 } 77 } 78 if(c=='B') 79 { 80 if(line==4) 81 { 82 cando=false; 83 break; 84 } 85 else 86 { 87 ++xy.x; 88 swap(v[line][row],v[line+1][row]); 89 } 90 } 91 if(c=='L') 92 { 93 if(row==0) 94 { 95 cando=false; 96 break; 97 } 98 else 99 { 100 --xy.y; 101 swap(v[line][row],v[line][row-1]); 102 } 103 } 104 if(c=='R') 105 { 106 if(row==4) 107 { 108 cando=false; 109 break; 110 } 111 else 112 { 113 ++xy.y; 114 swap(v[line][row],v[line][row+1]); 115 } 116 } 117 } 118 // 走完 119 cout<<"Puzzle #"<<ti<<":"<<endl; 120 if(cando) 121 { 122 for( int i=0; i!=5; ++i ) 123 { 124 for( int j=0; j!=4; ++j ) 125 cout<<v[i][j]<<" "; 126 cout<<v[i][4]<<endl; 127 } 128 } 129 else cout<<"This puzzle has no final configuration."<<endl; 130 ++ti; 131 } 132 return 0; 133 }
