The Blocks Problem

Background

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

In thisproblem you will model a simple block world under certain rules andconstraints. 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 armin how to manipulate blocks that lie on a flat table. Initially therearen blocks on the table (numbered from 0 to n-1)with block b_i adjacent to blockb_i+1for 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 blockb afterreturning any blocks that are stacked on top of blocks a andb totheir initial positions.
move a over b
where a and b are block numbers, puts block a onto the top of thestack containing blockb, after returning any blocks that are stackedon top of block a to their initial positions.
pile a onto b
where a and b are block numbers, moves the pile of blocks consistingof blocka, and any blocks that are stacked above block a, ontoblock b. All blocks on top of block b are moved to their initialpositions prior to the pile taking place. The blocks stacked above blocka retain their order when moved.
pile a over b
where a and b are block numbers, puts the pile of blocks consistingof blocka, and any blocks that are stacked above block a, ontothe top of the stack containing blockb. The blocks stacked above blocka retain their original order when moved.
quit
terminates manipulations in the block world.

Any command in which a = b or in which a and bare in the same stack of blocks is an illegal command. All illegalcommands should be ignored and should have noaffect on the configuration of blocks.

The Input

The input begins with an integer n on a line by itself representingthe 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, onecommand per line. Yourprogram should process all commands until thequit command isencountered.

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. Eachoriginal block position numberedi ( $0 \leq i < n$ wheren is the number of blocks) should appearfollowed immediately by a colon.If there is at least a blockon 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 wheren 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

Miguel Revilla
2000-04-06

题目的意思是：　有一排过去ｎ个木块，按指令移动木块；

move a over b:
在木块ａ所在的那一堆木块，把ａ上面的的木块全部放回原地，然后把ａ放到ｂ所在那堆木块上面。。

move a onto b:
把ａ所在的那堆木块，ａ上面的全部放回原处，ｂ所在的那堆也是一样，把ｂ上面的木块都放回原处，然后把ａ放到ｂ上面；

pile a over b:
把ａ所在的那一堆木块，ａ上面的全部木块包括ａ　，都移到ｂ所在的那堆木块上面。

pile a onto b:
把ｂ所在那堆木块，在ｂ上面的木块全都放回原处，并把ａ所在的那堆木块ａ上面的全部移到ｂ上面。。

总的来说就是碰到ｍｏｖｅ就要把ａ上面的全部放回原处。如果碰到　ｏｎｔｏ　就要把ｂ上面的全部放到原处。
因为ｍｏｖｅ是只移动ａ一个，所以ａ上面的要归位，而ｐｉｌｅ是移一堆，所以不用。
ｏｎｔｏ是要和ｂ贴在一起，所以ｂ上面的要归位，而ｏｖｅｒ是上方，不需要直接接触，所以不用。。

思路就是用栈来模拟，一开始就是ｎ个栈。每个栈里都是一个元素，然后按照指令移，在这个栈里ｐｏｐ（）掉它，在另一个栈里ｐｕｓｈ（）进去。。
分四种情况来做移动，每种情况处理方式不一样。要注意如果是一堆移过去，因为还是要按照这个顺序，多以要先把这一堆放到另一个数组，再按顺序ｐｕｓｈj进去。
模拟完输出即可。。

ＡＣ代码：

#include<iostream>
#include<stdio.h>
#include<string>
#include<stack>
using namespace std;

int main () {
	stack <int> sta[100];
	int t;
	int num[100];
	int res[100];
	string m1,m2;
	int p1,p2;
	cin >> t;
	getchar();
	for( int i = 0; i < t ;i++) {
		sta[i].push(i);
		num[i] = i;

	}
	while(1) {
		cin >> m1;
		if(m1 == "quit")
			break;
		cin >>p1 >>m2 >>p2;
		if (num[p1] == num[p2])
			continue;
		if (m1 == "move" && m2 == "over") {
			for (;sta[num[p1]].top() != p1; ) {
				sta[ sta[num[p1]].top() ].push(sta[num[p1]].top());	
				num[sta[num[p1]].top()] = sta[num[p1]].top();
				sta[num[p1]].pop();
			}
			sta[num[p2]].push(p1);
			sta[num[p1]].pop();
			num[p1] = num[p2];
		}
		if (m1 == "pile" && m2 == "over") {
			int k = 0;
			int temp[200];
			for (;sta[num[p1]].top() != p1; ) {
				temp[k++] = sta[num[p1]].top();
				num[sta[num[p1]].top()] = num[p2];
				sta[num[p1]].pop();
			}
			sta[num[p1]].pop();
			temp[k] = p1;
			num[p1] = num[p2];
			for(int w = k ;w >= 0; w--)
				sta[num[p2]].push(temp[w]);
		}
		if (m1 == "move" && m2 == "onto") {
			for (;sta[num[p1]].top() != p1; ) {
				sta[ sta[num[p1]].top() ].push(sta[num[p1]].top());	
				num[sta[num[p1]].top()] = sta[num[p1]].top();
				sta[num[p1]].pop();
			}
			for (;sta[num[p2]].top() != p2; ) {
				sta[ sta[num[p2]].top() ].push(sta[num[p2]].top());	
				num[sta[num[p2]].top()] = sta[num[p2]].top();
				sta[num[p2]].pop();
			}
			sta[num[p2]].push(sta[num[p1]].top());
			sta[num[p1]].pop();
			num[p1] = num[p2];
		}
		if (m1 == "pile" && m2 == "onto") {
			int k = 0;
			int temp[200];
			for (;sta[num[p1]].top() != p1; ) {
				temp[k++] = sta[num[p1]].top();
				num[sta[num[p1]].top()] = num[p2];
				sta[num[p1]].pop();
			}
			sta[num[p1]].pop();
			temp[k] = p1;
			num[p1] =num[p2];
			for (;sta[num[p2]].top() != p2; ) {
				sta[ sta[num[p2]].top() ].push(sta[num[p2]].top());	
				num[sta[num[p2]].top()] = sta[num[p2]].top();
				sta[num[p2]].pop();
			}
			for(int w = k ;w >= 0; w--)
				sta[num[p2]].push(temp[w]);
		
		}
	}
	int j;
	for(int i = 0;i < t;i++) {
		cout << i <<":";
		for( j = 0 ;!sta[i].empty();j++) {
			res[j] =  sta[i].top();
			sta[i].pop();
		}
		for (j = j -1; j >= 0;j--)
			cout<<" "<<res[j];
		cout << endl;
	}
	return 0;
}

UVA101 The Blocks Problem

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