POJ 1208 不定长数组vector

The Blocks Problem

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5404		Accepted: 2314

Description

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 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:

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

const int  maxn = 30;
int n;
vector<int> pile[maxn]; //感觉有点像string pile[maxn]类型

//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);
    }
    pile[p].resize(h+1);//pile只应保留下标0~h的元素，节省内存
}

//把第p堆高度为h及其上方的木块整体移动到p2堆得顶部
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); //重定义大小，节省内存
}

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]);
        printf("\n");
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    int a,b;
    cin>>n;
    string s1,s2;
    for(int i=0;i<n;i++)
        pile[i].push_back(i); //把木块归位
    while(cin>>s1>>a>>s2>>b)
    {    
        
        int pa,pb,ha,hb;
        if(s1=="quit")
            break;
        
        find_block(a,pa,ha);//在pa堆找到a的位置，记录a的高度
        find_block(b,pb,hb);//在pb堆找到b的位置，记录b的高度
        if(pa==pb)
            continue;
        if(s2=="onto")
            clear_above(pb,hb);//把b上方的木块全部归位
         if(s1=="move")
            clear_above(pa,ha);//把a上方的木块全部归位


        pile_onto(pa,ha,pb);
    }
    print();
    return 0;
}

很好的利用了指令的共同点，大大简化了代码量！