UVAOJ101 | 哈喽哈咯

Written by razrlele 09:32 December 14, 2014

101 – The Blocks Problem

Time limit: 3.000 seconds

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 $0 \leq i < n-1$ 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 bwhere 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 bwhere 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 bwhere 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 bwhere 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.
quitterminates 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 ( $0 \leq i < n$ 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).

Sample Input

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

3: 3

5: 5 8 7 6

Miguel Revilla
2000-04-06

最关键的还是读懂题意.

用函数式的方法来说明比较清楚:

首先定义几个函数

putback(i): 把编号为i的砖块所在的堆中i以上的(不包括i)所有砖块都放回其初始所在的堆.

push(a, b): 把a砖块放到b砖块所在的堆的上面.

pushpile(a, b): 把a砖块所在的堆中a以上(包括a)的所有砖块顺序不变地搬到b砖块所在的堆上面.

然后就好说明了:

move a onto b:

先putback(a), putback(b), 再push(a, b).

move a over b:

先putback(a), 再push(a, b).

pile a onto b:

先putvack(b), 再pushpile(a, b).

pile a over b:

直接pushpile(a, b).

一开始想向上道题一样自己手动实现堆栈, 后来发现搞来搞去总爱昏了头, 测试数据各种飙segment fault, 最后还是用STL简化了一下代码, 果然就强多了.

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* File Name: uvaoj101.cpp

* Author: razrLeLe

* Mail: razrlele@outlook.com

* Homepage: https://yueyu.io

* Created Time: 2014年12月13日星期六 18时07分24秒

#define LOCAL

#include "vector"

#include "set"

#include "deque"

#include "queue"

#include "algorithm"

#include "functional"

#include "iostream"

#include "cstdio"

#include "cmath"

#include "cstdlib"

#include "string"

#include "cstring"

#include "string.h"

#include "map"

#include "cctype"

#include "list"

using namespace std;

#define INF 0x3f3f3f3f

const int maxn = 30;

string order;

string order2;

int targeta, targetb;

vector< list<int> > blockposition(maxn);

int block[maxn];

int position_num;

void putback(int a)

{

int tmp = block[a];

list<int>::iterator i = blockposition[tmp].begin();

while(*i != a) i++;

i++;

list<int>::iterator tempi = i;

for(; i != blockposition[tmp].end(); i++)

{

blockposition[*i].push_back(*i);

block[*i] = *i;

}

blockposition[tmp].erase(tempi, blockposition[tmp].end());

return ;

}

void move_onto(int a, int b)

{

putback(a);

putback(b);

blockposition[block[b]].push_back(a);

list<int>::iterator i = blockposition[block[a]].begin();

while(*i != a) i++;

blockposition[block[a]].erase(i);

block[a] = block[b];

return ;

}

void move_over(int a, int b)

{

putback(a);

blockposition[block[b]].push_back(a);

list<int>::iterator i = blockposition[block[a]].begin();

while(*i != a) i++;

blockposition[block[a]].erase(i);

block[a] = block[b];

return ;

}

void pushpile(int a, int b)

{

int tmp = block[a];

list<int>::iterator i = blockposition[tmp].begin();

int count = 0;

while(*i != a) i++;

list<int>::iterator tempi = i;

for(; i != blockposition[tmp].end(); i++)

{

blockposition[b].push_back(*i);

block[*i] = b;

}

blockposition[tmp].erase(tempi, blockposition[tmp].end());

return ;

}

void pile_onto(int a, int b)

{

putback(b);

pushpile(a, block[b]);

return ;

}

void pile_over(int a, int b)

{

pushpile(a, block[b]);

return ;

}

int main()

{

#ifdef LOCAL

freopen("/home/razrlele/build/data.txt", "r", stdin);

freopen("/home/razrlele/build/out.txt", "w", stdout);

#endif

while(~scanf("%d", &position_num))

{

for(int i = 0; i < position_num; i++)

{

blockposition[i].clear();

blockposition[i].push_back(i);

block[i] = i;

}

while(cin >> order)

{

if(order == "quit")

break;

cin >> targeta >> order2 >> targetb;

if(block[targeta] == block[targetb])

continue;

if(order == "move")

{

if(order2 == "onto")

move_onto(targeta, targetb);

else

move_over(targeta, targetb);

}

else

{

if(order2 == "onto")

pile_onto(targeta, targetb);

else

pile_over(targeta, targetb);

}

/*cout << order <<" " << targeta << " " << order2 << " " << targetb << endl;

for(int i = 0; i < position_num; i++)

{

printf("%d:", i);

for(list<int>::iterator j = blockposition[i].begin(); j != blockposition[i].end(); j++)

printf(" %d", *j);

printf("\n");

}

cout << endl;*/

}

for(int i = 0; i < position_num; i++)

{

printf("%d:", i);

for(list<int>::iterator j = blockposition[i].begin(); j != blockposition[i].end(); j++)

printf(" %d", *j);

printf("\n");

}

return 0;

}