The Blocks Problem (积木问题) -Uva101

Background
背景

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.
在计算机科学中的很多地方都会使用简单，抽象的方法来做分析和实验验究。比如在早期的规划学和机器人学的人工智能研究就利用一个积木世界，让机械臂执行操作积木的任务。

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.
在这个问题中，你将在确定的规则和约束条件下构建一个简单的积木世界。这不是让你来研究怎样达到某种状态，而是编写一个“机械臂程序”来响应有限的命令集。

 

The Problem
问题

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 0 ≤ i < n - 1 as shown in the diagram below:
问题就是分析一系列的命令，告诉机械臂如何操纵放在一个平台上的积木。最初平台上有n个积木（编号由0到n - 1），对于任意的0 ≤ i < n - 1，积木b_i都与b_{i + 1}相临

Figure: Initial Blocks World
图：积木世界的初始状态

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。

#include<cstdio>
#include<string>
#include<vector>
#include<iostream>
using namespace std;
const int maxn = 30;
int n;
vector<int> pile[maxn];
//查找木块a所在的pile和height
void find_block(int a,int& p,int & h)
{
	for (p = 0; p < n; p++)
	{
		for ( h = 0; 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放回原位
void clearAbove(int p,int h)
{
	for (size_t i = h+1; 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 + 1);//只保留第0-h元素
}
//把第p堆高度为h及以上的移动到p2顶部
void pileonto(int p,int h,int p2)
{
	for (size_t i = h; i < pile[p].size(); i++)
	{
		pile[p2].push_back(pile[p][i]);
	}
	pile[p].resize(h);
}
//输出
void print()
{
	for (size_t i = 0; i < n; i++)
	{
		printf_s("%d", i);
		for (size_t j = 0; j < pile[i].size(); j++)
		{
			printf_s(" %d", pile[i][j]);
		}
		printf_s("\n");

	}
}

int main()
{
	int a, b;
	cin >> n;
	string s1, s2;
	for (size_t i = 0; 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;
		if (s2 == "onto") clearAbove(pb, hb);
		if (s1 == "move") clearAbove(pa, ha);
		pileonto(pa, ha, pb);
	}
	print();
	system("pause");//在平台提交时务必拿掉此句
	return 0;
}