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:
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 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 illegal 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:
模拟4种操作：
move a onto b:把a和b上方的木块全部归位，然后把a摞在b上面；
move a over b：把a上方的木块全部归位，然后把a放在b所在木块堆的顶部；
pile a onto b ： 把b上方的木块全部归位，然后把a及a上方的木块全部摞在b上面；
pile a over b：把a及a上方的木块全部摞在b所在木块堆的顶部；
quit是终止操作；
输入一共有四种指令，观察指令的不同，编写函数以减少重复代码；
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <string>
#include <queue>
#include <stack>
using namespace std;
typedef long long ll;
vector <int> pile[30];//每个pile[i]是一个vector
int n;
//找木块a所在的pile和height,以引用的形式返回调用者
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移回原位
void clear_above(int p,int h)
{
for(int 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);//pile只应保留下标0-h的元素
}
//把第p堆高度为h及其上方的木块整体移动到p2堆的顶部
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++){
cout << i <<':';
for(int j = 0;j < pile[i].size();j++)
cout <<' ' << pile[i][j];
cout << endl;
}
}
int main()
{
int i,a,b;
cin >>n;
for(i = 0;i < n;i++)
pile[i].push_back(i);
string s1,s2;
while(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;//非法指令
if(s2 == "onto") clear_above(pb,hb);
if(s1 == "move") clear_above(pa,ha);
pile_onto(pa,ha,pb);
}
print();
return 0;
}