The Blocks Problem UVA - 101

The Blocks Problem UVA - 101

题目描述

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 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

bi adjacent to block bi+1 for all 0 ≤ i < n − 1 as shown in the diagram below



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.

• 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.

• 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.

• 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.

• 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 il
4000
legal commands should be ignored and should have no affect on the configuration of

blocks.

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.

Output

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.

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:

大致题意

给你编号从1到n的n个木块，有四种操作。

1.move a onto b,把a和b上面的木块全都放回原来的位置，然后把a放到b上面。

2.move a over b,把a上面的木块放回原来的位置，然后把a放到b所在的木块堆的最上面

3.pile a onto b,把b上面的木块放回原来的位置然后把a以及a上面的木块全部放到b上

4.pile a over b，把a以及a上面的木块全部放到b所在的木块堆上.

给你一系列的操作后要求你输出最后的结果。

思路

模拟，通过分析，如果是move就先把a上面的木块放回原来的位置，如果是noto就是先把b上面的木块放回原来的位置。然后移动一堆到另一堆上。

下面是代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <fstream>
#include <math.h>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include<sstream>
#include<ctime>
using namespace std;
const int maxn=30;
int n;
vector<int> pile[maxn];
void find_block(int a,int &p,int &h)   //找a木块所在的pile和高度，以引用的形式返回调用者
{
for(p=0;p<n;p++)
for(h=0;h<pile[p].size();h++)
if(pile[p][h]==a)
return;
}
void clear_above(int p,int h)   //把第p堆高度为h的木块上方的所有木块移回原位
{
for(int i=h+1;i<pile[p].size();i++)
{
int b=pile[p][i];
pile[b].push_back(b);
}
pile[p].resize(h+1);
}
void pile_onto(int p,int h,int p2)
{
for(int i=h;i<pile[p].size();i++)
pile[p2].push_back(pile[p][i]);

pile[p].resize(h);
}

void print()
{
for(int i=0;i<n;i++)
{
printf("%d:",i);
for(int j=0;j<pile[i].size();j++)
printf(" %d",pile[i][j]);
cout<<endl;
}
}

int main()
{
int a,b;
cin>>n;
string s1,s2;
for(int i=0;i<n;i++)
pile[i].push_back(i);

while(1)
{
cin>>s1;
if(s1=="quit") break;
cin>>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") clear_above(pb,hb);
if(s1=="move") clear_above(pa,ha);
pile_onto(pa,ha,pb);
}
print();
return 0;
}