UVa OJ 101 The Blocks Problem (积木问题)

Time limit: 3.000 second
限时：3.000秒

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，图示如下：

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。

The Input
输入

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.
输入由1个整数n开始开始，该整数独占一行，表示积木世界中的积木数量。你可以假定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.
从积木数量值的下一行开始是一系列的命令，每条命令独占一行。你的程序要处理所有的命令直到输入退出命令。

You may assume that all commands will be of the form specified above. There will be no syntactically incorrect commands.
你可以假定所有的命令都按上文所示的格式给出。不会出现语法错误的命令。

The Output
输出

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.
以积木世界的最终状态作为输出。每一个原始积木的位置i（0 ≤ i < n，n为积木数量）后面都要紧跟一个冒号。如果至少有一个积木在该位置上，冒号后面都要紧跟一个空格，然后是该位置上所有积木编号的序列。每2个积木的编号之间以一个空格隔开。行尾不能出现多余的空格。

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).
每个积木位置独占一行（即第一行输入的n，对应输出n行数据）。

Sample Input
输入示例

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
输出示例

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

Analysis
分析

模拟类型的题，编程比较繁琐。无需考虑将积木放回原位时原位被占的情况，因为没有任何一个操作可以把积木放在空的位置上，因此也不可能出现在第i个位置上，第i个积木的下面有其它积木的情况。可将操作分为两类，一类需要清空(return)上面的积木，一类不需要，这样可以统一处理清空的操作。下面的代码中还有一个附加的命令：dump，可以随时查看当前的状态，方便调试。

Solution
解答

#include <iostream>
#include <string>
using namespace std;

int aStks[25][25];
int aPos[25]; //表示每一块积木的位置，所在栈的编号*所在高度
int aStkLen[25];
int nStkCnt;

struct COMMAND{
	int nCom;
	int nA;
	int nB;
};
bool ParseCommand(const char *pComStr, COMMAND &cmd) {
	int nPos = 0, i, j;
	static char *pCom[] = {"move ", "pile ", "quit", "dump"};
	for (i = 0; i < sizeof(pCom) / sizeof(pCom[0]); ++i) {
		for(j = 0; pComStr[j] == pCom[i][j] && pCom[i][j] != 0; ++j);
		if (pCom[i][j] == 0) break;
	}
	if (i >= sizeof(pCom) / sizeof(pCom[0])) {
		return false;
	}
	cmd.nCom = i * 2;
	if (cmd.nCom >= 4) {
		return true;
	}
	if (pComStr[j] < '1' || pComStr[j] > '9') {
		return false;
	}
	for (cmd.nA = 0; pComStr[j] >= '0' && pComStr[j] <= '9'; ++j) {
		cmd.nA = cmd.nA * 10 + pComStr[j] - '0';
	}
	static char *pDir[2] = {" onto ", " over "};
	for (i = 0, pComStr += j; i < 2; ++i) {
		for(j = 0; pComStr[j] == pDir[i][j] && pDir[i][j] != 0; ++j);
		if (pDir[i][j] == 0) {
			break;
		}
	}
	if (i >= 2) {
		return false;
	}
	cmd.nCom += i;
	if (pComStr[j] < '1' || pComStr[j] > '9') {
		return false;
	}
	for (cmd.nB = 0; pComStr[j] >= '0' && pComStr[j] <= '9'; ++j) {
		cmd.nB = cmd.nB * 10 + pComStr[j] - '0';
	}
	if (pComStr[j] != 0 || cmd.nA == cmd.nB) {
		return false;
	}
	return true;
}

void ReturnUpperBlock(int nBaseBlk) {
	//计算出积木所在的栈和高度。nStk为栈的编号，nHei为栈的高度
	int nStk = aPos[nBaseBlk] / nStkCnt, nHei = aPos[nBaseBlk] % nStkCnt + 1;
	//循环将指定积木以上的所有积木归位
	for (int &nStkLen = aStkLen[nStk]; nStkLen > nHei; --nStkLen) {
		int nUpperBlk = aStks[nStk][nStkLen - 1]; //最顶的积木编号
		aStks[nUpperBlk][0] = nUpperBlk; //将其归原位
		aPos[nUpperBlk] = nUpperBlk * nStkCnt; //重新计算合成该积木位置
		aStkLen[nUpperBlk] = 1; //归位的目标栈的高度置1(只有归位的积木)
	}
}

void Dump(void) {
	for (int i = 0; i < nStkCnt; ++i, cout << endl) {
		cout << i << ':';
		for (int j = 0; j < aStkLen[i]; ++j) {
			cout << ' ' << aStks[i][j];
		}
	}
}
int main(void) {
	COMMAND cmd;
	cin >> nStkCnt;
	for (int i = 0; i < nStkCnt; ++i) {
		aStks[i][0] = i; aPos[i] = i * nStkCnt; aStkLen[i] = 1;
	}
	for (string strLine; getline(cin, strLine);) {
		if (!ParseCommand(strLine.c_str(), cmd)) {
			continue;
		}
		if (cmd.nCom == 4) {
			break; //quit
		}
		if (cmd.nCom == 6) {
			Dump();
			continue;
		}
		int nStkA = aPos[cmd.nA] / nStkCnt, nStkB = aPos[cmd.nB] / nStkCnt;
		if (nStkA == nStkB) {
			continue;
		}
		if (cmd.nCom % 2 == 0) {
			ReturnUpperBlock(cmd.nB); //onto
		}
		if (cmd.nCom / 2 == 0) {
			ReturnUpperBlock(cmd.nA); //move
		}
		int nHeiA = aPos[cmd.nA] % nStkCnt;
		int nStkLenA = aStkLen[nStkA], &nStkLenB = aStkLen[nStkB];
		for (aStkLen[nStkA] = nHeiA; nHeiA < nStkLenA; ++nHeiA) {
			int nBlk = aStks[nStkA][nHeiA];
			aPos[nBlk] = nStkB * nStkCnt + nStkLenB;
			aStks[nStkB][nStkLenB++] = nBlk;
		}
	}
	Dump();
	return 0;
}