uva 101 - The Blocks Problem

题目就是根据题意进行四个操作  。。复制一下别人的翻译：

The valid commands for the robot arm that manipulates blocks are:
机械臂操作积木的有效指令列举如下：

  • move a onto b
    • where a and b are block numbers, puts block a onto block b after returning any blocks that are stacked on top of blocks a and b to their initial positions.
    • a和b都是积木的编号，先将a和b上面所有的积木都放回原处，再将a放在b上。
  • move a over b
    • where a and b are block numbers, puts block a onto the top of the stack containing block b, after returning any blocks that are stacked on top of block a to their initial positions.
    • a和b都是积木的编号，先将a上面所有的积木放回原处，再将a放在b上。（b上原有积木不动）
  • pile a onto b
    • where a and b are block numbers, moves the pile of blocks consisting of block a, and any blocks that are stacked above block a, onto block b. All blocks on top of block b are moved to their initial positions prior to the pile taking place. The blocks stacked above block a retain their order when moved.
    • a和b都是积木的编号，将a和其上面所有的积极组成的一摞整体移动到b上。在移动前要先将b上面所有的积极都放回原处。移动的一摞积木要保持原来的顺序不变。
  • pile a over b
    • where a and b are block numbers, puts the pile of blocks consisting of block a, and any blocks that are stacked above block a, onto the top of the stack containing block b. The blocks stacked above block a retain their original order when moved.
    • a和b都是积木的编号，将a和其上面所有的积极组成的一摞整体移动到b所在一摞积木的最上面一个积木上。移动的一摞积木要保持原来的顺序不变。
  • quit
    • terminates manipulations in the block world.
    • 结束积木世界的操纵。

Any command in which a = b or in which a and b are in the same stack of blocks is an illegal command. All illegal commands should be ignored and should have no affect on the configuration of blocks.
当a = b或a和b处在同一摞时，任何企图操作a和b的命令都是非法的。所有非法的命令都要忽略，且不能对当前积木的状态产生作用。

 

 

 

代码比较恶心。。。就是纯粹的模拟

 

题目：

The Blocks Problem 

 

 

 

Background 

Many areas of Computer Science use simple, abstract domains for both analytical and empirical studies. For example, an early AI study of planning and robotics (STRIPS) used a block world in which a robot arm performed tasks involving the manipulation of blocks.

In this problem you will model a simple block world under certain rules and constraints. Rather than determine how to achieve a specified state, you will ``program'' a robotic arm to respond to a limited set of commands.

 

 

 

The Problem 

The problem is to parse a series of commands that instruct a robot arm in how to manipulate blocks that lie on a flat table. Initially there are n blocks on the table (numbered from 0 to n-1) with block bi adjacent to block bi+1 for all $0 \leq i < n-1$ as shown in the diagram below:

 
\begin{figure} \centering \setlength{\unitlength}{0.0125in} % \begin{picture} (2... ...raisebox{0pt}[0pt][0pt]{$\bullet \bullet \bullet$ }}} \end{picture} \end{figure}
Figure: Initial Blocks World

 

The valid commands for the robot arm that manipulates blocks are:

 

  • move a onto b

    where a and b are block numbers, puts block a onto block b after returning any blocks that are stacked on top of blocks a and b to their initial positions.

     

  • move a over b

    where a and b are block numbers, puts block a onto the top of the stack containing block b, after returning any blocks that are stacked on top of block a to their initial positions.

     

  • pile a onto b

    where a and b are block numbers, moves the pile of blocks consisting of block a, and any blocks that are stacked above block a, onto block b. All blocks on top of block b are moved to their initial positions prior to the pile taking place. The blocks stacked above block a retain their order when moved.

     

  • pile a over b

    where a and b are block numbers, puts the pile of blocks consisting of block a, and any blocks that are stacked above block a, onto the top of the stack containing block b. The blocks stacked above block a retain their original order when moved.

     

  • quit

    terminates manipulations in the block world.

 

Any command in which a = b or in which a and b are in the same stack of blocks is an illegal command. All illegal commands should be ignored and should have no affect on the configuration of blocks.

 

 

 

The Input 

The input begins with an integer n on a line by itself representing the number of blocks in the block world. You may assume that 0 < n < 25.

The number of blocks is followed by a sequence of block commands, one command per line. Your program should process all commands until the quit command is encountered.

 

You may assume that all commands will be of the form specified above. There will be no syntactically incorrect commands.

 

 

 

The Output 

 

The output should consist of the final state of the blocks world. Each original block position numbered i ( $0 \leq i < n$ where n is the number of blocks) should appear followed immediately by a colon. If there is at least a block on it, the colon must be followed by one space, followed by a list of blocks that appear stacked in that position with each block number separated from other block numbers by a space. Don't put any trailing spaces on a line.

 

There should be one line of output for each block position (i.e., n lines of output where n is the integer on the first line of input).

 

 

 

Sample Input 

 

10
move 9 onto 1
move 8 over 1
move 7 over 1
move 6 over 1
pile 8 over 6
pile 8 over 5
move 2 over 1
move 4 over 9
quit

 

 

 

Sample Output 

 

 0: 0
 1: 1 9 2 4
 2:
 3: 3
 4:
 5: 5 8 7 6
 6:
 7:
 8:
 9:



代码：
  1 #include <iostream>
  2 #include <vector>
  3 #include <deque>
  4 #include <string>
  5 #include <string.h>
  6 #include <cstdio>
  7 using namespace std;
  8 
  9 int blocks[30][30];
 10 int height[30];
 11 int bheight[30];
 12 int pos[30];
 13 int num;
 14 
 15 int comprehend(char* cmd)
 16 {
 17     if(strstr(cmd,"move"))
 18     {
 19         if(strstr(cmd,"onto"))return 1;
 20         else if(strstr(cmd,"over"))return 2;
 21     }
 22     else
 23     {
 24         if(strstr(cmd,"onto")) return 3;
 25         else return 4;
 26     }
 27 
 28 }
 29 
 30 void run(char* cmd)
 31 {
 32     int cmdkind = comprehend(cmd);
 33     int from,to;
 34     char rub[100];
 35     if(cmdkind==1)
 36     {
 37         sscanf(cmd,"%s %d %s %d",rub,&from,rub,&to);
 38 
 39 
 40 
 41         int px = pos[from];
 42         int py = height[from];
 43         int ptx = pos[to];
 44         int pty = height[to];
 45          if(from==to||px==ptx)return;
 46         for(int i= py+1;i<=bheight[px];i++)
 47         {
 48             int now = blocks[px][i];
 49             blocks[now][0]=now;
 50             bheight[now]=0;
 51             pos[now] = now;
 52             height[now]=0;
 53         }
 54         bheight[px]=py;
 55 
 56 
 57 
 58         for(int i=pty+1;i<=bheight[ptx];i++)
 59         {
 60             int now = blocks[ptx][i];
 61             blocks[now][0]=now;
 62             bheight[now]=0;
 63 
 64             pos[now] = now;
 65             height[now]=0;
 66         }
 67 
 68        // cout<<"px"<<px<<"py"<<py<<"ptx"<<ptx<<"pty"<<pty<<endl;
 69         bheight[ptx]=pty+1;
 70         blocks[ptx][pty+1] = from;
 71 
 72         bheight[px]--;
 73         pos[from]=ptx;
 74         height[from]=pty+1;
 75 
 76     }
 77     if(cmdkind==2)
 78     {
 79         sscanf(cmd,"%s %d %s %d",rub,&from,rub,&to);
 80 
 81 
 82         int px = pos[from];
 83         int py = height[from];
 84         int ptx = pos[to];
 85         int pty = bheight[ptx];
 86          if(from==to||px==ptx)return ;
 87         for(int i= py+1;i<=bheight[px];i++)
 88         {
 89             int now = blocks[px][i];
 90             blocks[now][0]=now;
 91             bheight[now]=0;
 92             pos[now] = now;
 93             height[now]=0;
 94         }
 95         bheight[px]=py;
 96 
 97 
 98 
 99         //cout<<"px"<<px<<"py"<<py<<"ptx"<<ptx<<"pty"<<pty<<endl;
100         bheight[ptx]=pty+1;
101         blocks[ptx][pty+1] = from;
102 
103         bheight[px]--;
104         pos[from]=ptx;
105         height[from]=pty+1;
106 
107 
108 
109     }
110 
111     if(cmdkind==3)
112     {
113         sscanf(cmd,"%s %d %s %d",rub,&from,rub,&to);
114 
115         int px = pos[from];
116         int py = height[from];
117 
118 
119         int ptx = pos[to];
120         int pty = height[to];
121 
122         if(from==to||px==ptx)return;
123 
124         for(int i=pty+1;i<=bheight[ptx];i++)
125         {
126             int now = blocks[ptx][i];
127             blocks[now][0]=now;
128             bheight[now]=0;
129 
130             pos[now] = now;
131             height[now]=0;
132         }
133         bheight[ptx] = pty;
134 
135         for(int i=py;i<=bheight[px];i++)
136         {
137             int now = blocks[px][i];
138             blocks[ptx][++pty] = now;
139             pos[now] = ptx;
140             height[now] = pty;
141         }
142         bheight[px] = py-1;
143 
144        // cout<<"px"<<px<<"py"<<py<<"ptx"<<ptx<<"pty"<<pty<<endl;
145         bheight[ptx]=pty;
146     }
147 
148 
149 
150     if(cmdkind==4)
151     {
152         sscanf(cmd,"%s %d %s %d",rub,&from,rub,&to);
153 
154 
155         int px = pos[from];
156         int py = height[from];
157          int ptx = pos[to];
158         int pty = bheight[ptx];
159           if(from==to||px==ptx)return ;
160         for(int i= py;i<=bheight[px];i++)
161         {
162             int now = blocks[px][i];
163             blocks[ptx][++pty]=now;
164             pos[now] = ptx;
165             height[now]=pty;
166         }
167         bheight[px]=py-1;
168 
169        // cout<<"px"<<px<<"py"<<py<<"ptx"<<ptx<<"pty"<<pty<<endl;
170         bheight[ptx]=pty;
171     }
172 
173 }
174 void print()
175 {
176     for(int i=0;i<num;i++)
177     {
178         printf("%d:",i);
179         for(int j=0;j<=bheight[i];j++)
180         {
181             printf(" %d",blocks[i][j]);
182         }
183         printf("\n");
184     }
185 
186 }
187 
188 
189 int main()
190 {
191     cin>>num;
192 
193     char cmd[100];
194 
195     memset(height,0,sizeof(height));
196     for(int i=0;i<num;i++)
197     {
198         blocks[i][0]=i;
199         bheight[i]=0;
200         height[i]=0;
201         pos[i]=i;
202     }
203 
204 
205     getchar();
206    // print();
207     while(gets(cmd) && !strstr(cmd,"quit"))
208     {
209         run(cmd);
210      
211     }
212        print();
213 
214     return 0;
215 }

 

 

posted @ 2013-11-21 23:27  doubleshik  阅读(402)  评论(0)  编辑  收藏  举报