101 - The Blocks Problem



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 b_i adjacent to block b_i+1 for all as shown in the diagram below:

  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. 

• a,b为块号，先将a,b块之上的所有块放回初始位置，然后把a块放到b块之上(要求a,b是相邻的，a,b之间没有别的块) 

• 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,b为块号，先将a块之上的所有块放回初始位置，然后把a块放到包含b块的一叠木块的最上方

• 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,b为块号，先将b块之上的所有块放回初始位置，然后把a及a之上的所有木块移动b之上

• 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,b为块号，把a及a之上的所有木块移动包含b块的一叠木块的最上方 

• 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 ( 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：

#include <iostream>
#include <stack>
#include <vector>
#include <string>

using namespace std;

vector<stack<int>> blocks;//下标代表位置从0-n,值代表各个位置上的块
vector<int> save_pos;//下标代表块号，值代表存储位置

void init_blocks(int size)
{
	blocks.reserve(size);
    for(int i=0;i<size;++i){
		blocks.push_back(stack<int>());
	    blocks[i].push(i);
	}
}

void init_save_pos(int size)
{
	save_pos.reserve(size);
	for(int i=0;i<size;++i){
		save_pos.push_back(i);
	}
}

void back_block(size_t block)
{//把pos这个块之上的块还原到初始位置
	stack<int> &s=blocks[save_pos[block]];
	while(!s.empty() && s.top()!=block){
		blocks[s.top()].push(s.top());//把块s.top()放回初始位置
		save_pos[s.top()]=s.top();//更新块s.top()的存储位置
		s.pop();//从当前容器中删除块s.top()
	}
}

void move_block(size_t a,size_t b)
{//把块a加入到块b所在的那叠块中
	int a_pos=save_pos[a];
	blocks[a_pos].pop();//把块a从块a所在的那叠积木中挪开
	int b_pos=save_pos[b];
	blocks[b_pos].push(a);//把块a加入到块b所在的那叠块中
	save_pos[a]=b_pos;//更新a的存储位置(即a在第几个容器中)
}

void move_pile_block(size_t a,size_t b)
{//把块a及之上的块全部挪到块b之上
	stack<int> &s=blocks[save_pos[a]];
	stack<int> temp;

	while(!s.empty() && s.top()!=a){
		temp.push(s.top());//把块s.top()放回初始位置
		s.pop();//从当前容器中删除块s.top()
	}
	if(!s.empty()){
		temp.push(s.top());
		s.pop();
	}

	int _pos=save_pos[b];
	while(!temp.empty()){
		blocks[_pos].push(temp.top());
		save_pos[temp.top()]=_pos;//更新块temp.top()的存储位置
		temp.pop();
	}
}

bool is_legal_command(int a,int b)
{//判断指令是否合法
	if(a==b) return false;
	int a_pos=save_pos[a];
	int b_pos=save_pos[b];
	if(a_pos==b_pos) return false;
	return true;
}

void print(size_t size)
{
    for(size_t i=0;i<size;++i){
		cout<<i<<":";
		stack<int> temp;
		stack<int> &s=blocks[i];
		while(!s.empty()){
			temp.push(s.top());
			s.pop();
		}
		while(temp.size()){
			cout<<" "<<temp.top();
			temp.pop();
		}
		cout<<endl;
	}
}

int main()
{
	int num;
	string first,second;
	size_t block_num1,block_num2;
	while(cin>>num){
		init_blocks(num);
		init_save_pos(num);
		while(cin>>first){
			if(first=="quit") break;
			cin>>block_num1>>second>>block_num2;
			if(!is_legal_command(block_num1,block_num2)) continue;
			if(first=="move" && second=="onto"){
				 back_block(block_num1);
				 back_block(block_num2);
				 move_block(block_num1,block_num2);
			}
			if(first=="move" && second=="over"){
				back_block(block_num1);
			    move_block(block_num1,block_num2);
			}
			if(first=="pile" && second=="onto"){
			    back_block(block_num2);//把block_num1上的所有积木位置放回初始位置
				move_pile_block(block_num1,block_num2);
			}
			if(first=="pile" && second=="over"){
				move_pile_block(block_num1,block_num2);
			}
		}
		print(num);
		blocks.clear();
		save_pos.clear();
	}
}