The Blocks Problem

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5004		Accepted: 2119

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 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 <= i < n-1 as shown in the
diagram below:



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 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 should consist of the final state of the blocks world. Each
original block position numbered i ( 0 <= 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).

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

0: 0

1: 1 9 2 4

2:

3: 3

4:

5: 5 8 7 6

6:

7:

8:

9:

题解：模拟题，有一个值得学习的小技巧，输入一共有4种指令，但是如果完全独立地处理各种指令，代码会变得冗长而且容易错，更好的方法是提取出指令之间的共同点，编写函数
减少重复代码
因为这个题木块的高度不确定，所以用vector很好

提示：vector之间可以之间赋值或者作为函数的返回值
代码:

 #include<cstdio>

 #include<string>

 #include<vector>

 #include<iostream>

 using namespace std;

 const int maxn = ;

 int n;

 vector<int> pile[maxn];

 //找木块a所在的pile 和height， 以引用的形式返回调用者

 void find_block(int a , int & p , int & h)

 {

     for(p =  ; p < n ;p++)

         for( h =  ; h < pile[p].size() ; h++)

             if(pile[p][h]==a) return;

 }

 //把第p堆高度为h的木块上方的所有木块移回原位

 void clear_above(int p , int h )

 {

     for(int i = h+; i < pile[p].size(); i++)

     {

         int b = pile[p][i];

         pile[b].push_back(b);//把木块b放回原位

     }

     pile[p].resize(h+);//pile只保留下标为0～h的元素

 }

 //把第p堆高度为h及其上方的木块整体移动到p2堆的顶部

 void pile_onto(int p , int h , int p2)

 {

     for(int i = h ; i < pile[p].size() ; i++)

     pile[p2].push_back(pile[p][i]);

     pile[p].resize(h);

 }

 void print()

 {

     for(int i =  ; i < n ;i++)

     {

         printf("%d:",i);

         for(int j =  ;j < pile[i].size();j++) printf(" %d",pile[i][j]);

         printf("\n");

     }

 }

 int main()

 {

     int a , b;

     cin>>n;

     string s1 , s2;

     for(int i =  ; i < n ;i++) pile[i].push_back(i);

     while(cin>>s1>>a>>s2>>b)

     {

         int pa , pb , ha, hb ;

         find_block(a,pa,ha);

         find_block(b,pb,hb);

         if(pa==pb) continue;//非法指令

         if(s2=="onto") clear_above(pb,hb);

         if(s1=="move") clear_above(pa,ha);

         pile_onto(pa,ha,pb);

     }

     print();

     return ;

 }