首页 > 基础算法 > 模拟法 > hdu 1612 The Blocks Problem-模拟-[解题报告]
2013
12-16

hdu 1612 The Blocks Problem-模拟-[解题报告]

The Blocks Problem

问题描述 :

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 illegal commands should be ignored and should have no affect on the configuration of blocks.

输入:

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.

输出:

The output should consist of the final state of the blocks world. Each original block position numbered i (0≤i<n-1 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).

样例输入:

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

样例输出:

 0: 0
 1: 1 9 2 4
 2:
 3: 3
 4:
 5: 5 8 7 6
 6:
 7:
 8:
 9:

终于AC了,这道题目去年寒假卡得我要死,最后一气之下就不做了。。。想想居然一年之久了,我本来都快忘了这道题了,最近发现白书的奥秘,觉得刘汝佳的题目真的相当练思维以及对代码的操作,决定又刷起题目来,这时候才想起这道题。

用栈进行模拟堆砖块,用个rec[]数组记录其现在所在的栈号,比较麻烦的是pile 操作,为了把a以及a以上的所有砖块都以原秩序放置于b砖块顶端,我用了个临时的栈进行存贮,然后再一个一个放到b栈上面。。这样就不会破坏秩序。。但是感觉这样做挺耗时的,原以为通不过,结果还是通过了。。。22ms,也不算太高吧。。不知道还有没有更好的pile方法

这个题目去年我都没想清楚题意,题目里面有个关键词 initial,意味着所有操作要还原的砖块都应该还原到它原本的位置,即 1还原到1号栈 2还原到2号栈,依次类推,因为根据题目的意思以及几大操作分析,一个栈要么就没元素,要么栈底元素就是栈号对应的元素,一旦移走了,栈必为空,一旦要还原,必定就把还原成最原始的样子

。一年了,觉得自己思维进步了一些,这是好事,继续加油!

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#define N 35
using namespace std;
stack <int> arr[N];
int rec[N];
int n,a,b;
char ch1[5],ch2[5];
void solve()
{
    int temp=arr[rec[a]].top();;
    int t2=arr[rec[b]].top();
    if (temp==t2) return;
    if (ch1[0]=='m' && ch2[1]=='n')
    {

        while (temp!=a)
        {
            arr[temp].push(temp);
            rec[temp]=temp;
            arr[rec[a]].pop();
            temp=arr[rec[a]].top();
        }

        while (t2!=b)
        {
            arr[t2].push(t2);
            rec[t2]=t2;
            arr[rec[b]].pop();
            t2=arr[rec[b]].top();
        }
        arr[rec[b]].push(a);
        arr[rec[a]].pop();
        rec[a]=rec[b];
        return;
    }
    if (ch1[0]=='m' && ch2[1]=='v')
    {

        while (temp!=a)
        {
            arr[temp].push(temp);
            rec[temp]=temp;
            arr[rec[a]].pop();
            temp=arr[rec[a]].top();
        }
        arr[rec[b]].push(a);
        arr[rec[a]].pop();
        rec[a]=rec[b];
        return;
    }
    if (ch1[0]=='p' && ch2[1]=='n')
    {

        while (t2!=b)
        {
            arr[t2].push(t2);
            rec[t2]=t2;
            arr[rec[b]].pop();
            t2=arr[rec[b]].top();
        }
        stack <int> q;
        while (temp!=a)
        {
            q.push(temp);
            arr[rec[a]].pop();
            temp=arr[rec[a]].top();
        }
        arr[rec[b]].push(temp);
        arr[rec[a]].pop();
        rec[a]=rec[b];
        while (!q.empty())
        {
            int tt=q.top();
            q.pop();
            rec[tt]=rec[b];
            arr[rec[b]].push(tt);
        }
        return;
    }
    if (ch1[0]=='p' && ch2[1]=='v')
    {
        stack <int> q;
        while (temp!=a)
        {
            q.push(temp);
            arr[rec[a]].pop();
            temp=arr[rec[a]].top();
        }
        arr[rec[b]].push(temp);
        arr[rec[a]].pop();
        rec[a]=rec[b];
        while (!q.empty())
        {
            int tt=q.top();
            q.pop();
            rec[tt]=rec[b];
            arr[rec[b]].push(tt);
        }
    }
}
void print()
{
    for (int i=0;i<n;i++)
    {
       printf("%d:",i);
       stack<int> q;
       while (!arr[i].empty())
       {
           int temp=arr[i].top();
           q.push(temp);
           arr[i].pop();
       }

       while (!q.empty())
       {
           printf(" %d",q.top());
           q.pop();
       }

       putchar('\n');
    }
}
int main()
{
    scanf("%d",&n);
    int i,j;
    for (i=0;i<n;i++){
        arr[i].push(i);
        rec[i]=i;
    }
    getchar();
    while (scanf("%s",ch1))
    {
        if (ch1[0]=='q')
            break;
        scanf("%d%s%d",&a,ch2,&b);
        getchar();
        solve();
    }
    print();
}

 

解题转自:http://www.cnblogs.com/kkrisen/archive/2013/12/05/3460445.html


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