UVA-101 The Blocks Problem 栈

Description

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 nblocks
on the table (numbered from 0 to n-1) with block bi adjacent
to block bi+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.

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 aretain 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 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:

题目：给你n个方块，有四种操作：
1.move a onto b，把a和b上面的方块都放回原来位置，然后把a放到b上面；
2.move a over b，把a上面的放回原处，然后把a放在b所在的方块堆的上面；
3.pile a onto b，把b上面的放回原来位置，然后把a和a上面的方块整体放到b上面；
4.pile a over b，把a和a上面的方块整体放到b所在堆的上面。
定义两个基本操作：
1.将a上面的还原init_place(a)；
2.将a和上面的（可以没有上面的）放到b上面pile_a_to_b(a,b)。
那么上述的四组操作就变成下面了：
1.move a onto b，init_place(a)；init_place(b)；pile_a_to_b(a,b)；
2.move a over b，init_place(a)；pile_a_to_b(a,b)；
3.pile a onto b，init_place(b)；pile_a_to_b(a,b)；
4.pile a over b，pile_a_to_b(a,b)。

#include<iostream>
#include<vector>
#include<cstdio>
#include<stack>
#include<cstring>
using namespace std;
stack<int> vv[300];
int place[25],top[25],s[30][30];
void init_place(int a)//将a还原
{
int block=0,id=place[a];
if(vv[id].empty())
return ;

//cout<<block<<endl;
while(vv[id].top()!=a)
{ //cout<<vv[id].top()<<endl;
block=vv[id].top();
place[block]=block;
vv[block].push(block);//
vv[id].pop();
}
//cout<<vv[a].top()<<endl;
}
int temp[30];
void pile_a_to_b(int a,int b)
{
int topt=-1,id=place[a],ID=place[b];
while(vv[id].top()!=a)
temp[++topt]=vv[id].top(),vv[id].pop();
int x=vv[id].top();
//	printf(" %d\n",x);
if(vv[id].top()==a)
vv[id].pop();
place[a]=ID;
vv[ID].push(a);
while(topt>=0)
{
place[temp[topt]]=ID;
vv[ID].push(temp[topt--]);
}
}
int main()
{
int n,a,b;
char op[6],type[6],x[30];
while ( ~scanf("%d",&n) ) {
for ( int i = 0 ; i < n ; ++ i ) {
vv[i].push(i) ;
place[i] = i;
//top[i] = 0;
}
while(~scanf("%s",op)&&op[0]!='q')
{
scanf("%d%s%d",&a,type,&b);
if(place[a]==place[b])
continue;
if(op[0]=='m')
init_place(a);
if(type[1]=='n')
init_place(b);
pile_a_to_b(a,b);
}
memset(x,0,sizeof x);
for(int i=0;i<n;i++)
{
printf("%d:",i);
int now=0,j=0;
while(!vv[i].empty())
{

x[j++]=vv[i].top();

vv[i].pop();
}
for(int i=j-1;i>=0;i--)
printf(" %d",x[i]);
puts("");
}

}
return 0;
}