• UVa 227 Puzzle


    Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu

    Status

    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 }
  • 相关阅读:
    2017浙江工业大学-校赛决赛 猜猜谁是我
    2017浙江工业大学-校赛决赛 竹之书
    2017浙江工业大学-校赛决赛 小M和天平
    2017"百度之星"程序设计大赛
    2017"百度之星"程序设计大赛
    2017"百度之星"程序设计大赛
    2017"百度之星"程序设计大赛
    2015-2016机器人操作系统(ROS)及其应用暑期学校资料汇总 ROS Summer School 持续更新
    2016“智能无人系统”暑期学校总结
    ROS_Kinetic_20 ROS基础补充
  • 原文地址:https://www.cnblogs.com/caterpillarofharvard/p/4239536.html
Copyright © 2020-2023  润新知