EOJ 1501/UVa The Blocks Problem
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:

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 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-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).
Examples
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
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>
#define MAXN 25
using namespace std;
int n;
vector<int> blocks[MAXN];
void findb(int x,int& p,int& h){
for(p=;p<n;p++)
for(h=;h<blocks[p].size();h++){
if(x==blocks[p][h]) return;
}
}
void clear_above(int p,int h){
for(int i=h+;i<blocks[p].size();i++){
int tmp=blocks[p][i];
blocks[tmp].push_back(tmp);
}
blocks[p].resize(h+);
}
void pile_over(int pa,int ha,int pb){
for(int i=ha;i<blocks[pa].size();i++){
int tmp=blocks[pa][i];
blocks[pb].push_back(tmp);
}
blocks[pa].resize(ha);
}
int main()
{
cin>>n;
for(int i=;i<n;i++) blocks[i].push_back(i);
int a,b;string s1,s2;
while(cin>>s1&&s1!="quit"){
cin>>a>>s2>>b; int pa,pb,ha,hb;
findb(a,pa,ha);
findb(b,pb,hb);
if(pa==pb) continue;
if(s1=="move") clear_above(pa,ha);
if(s2=="onto") clear_above(pb,hb);
pile_over(pa,ha,pb); }
for(int i=;i<n;i++){
printf("%d:",i);
for(int j=;j<blocks[i].size();j++)
printf(" %d",blocks[i][j]);
printf("\n");
}
return ;
}
翻译一下:用vector来实现我啊(误
紫书STL入门章节经典题,来自UVaOJ。给定木块数量,进行四种操作,结束后输出。
此题数据结构的核心时vector<int> blocks[MAXN],所有的操作围绕其进行。vector就像一个二位数组,,只是第一维的大小是固定的(不超过MAXN),但第二维的大小不固定......输入共有4中指令......更好的方法是提取指令之间的共同点,编写函数以减少重复代码。
EOJ 1501/UVa The Blocks Problem的更多相关文章
- UVa 101 The Blocks Problem Vector基本操作
UVa 101 The Blocks Problem 一道纯模拟题 The Problem The problem is to parse a series of commands that inst ...
- UVa 101 - The Blocks Problem(积木问题,指令操作)
题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&pa ...
- 木块问题(The Blocks Problem,Uva 101)
不定长数组:vector vector就是一个不定长数组.不仅如此,它把一些常用操作“封装”在了vector类型内部. 例如,若a是一个vector,可以用a.size( )读取它的大小,a.resi ...
- 【UVA - 101】The Blocks Problem(vector+模拟)
The Blocks Problem Descriptions:(英语就不说了,直接上翻译吧) 初始时从左到右有n个木块,编号为0~n-1,要求实现下列四种操作: move a onto b: 把a和 ...
- POJ 1208 The Blocks Problem
The Blocks Problem Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5397 Accepted: 231 ...
- uva 11991 - Easy Problem from Rujia Liu?(STL)
option=com_onlinejudge&Itemid=8&page=show_problem&problem=3142" target="_blank ...
- CJOJ 2485 UVa 11991 生日礼物 / UVa 11991 Easy Problem from Rujia Liu?
CJOJ 2485 UVa 11991 生日礼物 / UVa 11991 Easy Problem from Rujia Liu? Description (原题来自刘汝佳<训练指南>Pa ...
- The Blocks Problem(vector)
题目链接:http://poj.org/problem?id=1208 The Blocks Problem Time Limit: 1000MS Memory Limit: 10000K Tot ...
- UVa101 The Blocks Problem(不定长数组vector)
The Blocks Problem 书上的一道例题,代码思路比较清晰,可以看懂. 相关知识: 若a是一个vector,则: a.size():读取它的大小 a.resize():改变大小 a.pus ...
随机推荐
- gitlab 第1次提交代码到1个新仓库
1.如果是本地刚刚搭建好git环境,第一次和gitlab服务器产生连接 参照这个文 https://www.cnblogs.com/kaerxifa/p/10929098.html 2.已经和gitl ...
- http://blog.csdn.net/pizi0475/article/details/48286579 -------------(Collada 快速入门)
http://blog.csdn.net/zhouhangjay/article/details/8469085 说明:Collada的文件格式,中文版的很少,在csdn上看到了一个Sleepy的,感 ...
- 6 个 Linux 运维典型问题,大牛的分析解决思路在这里
作为一名合格的 Linux 运维工程师,一定要有一套清晰.明确的解决故障思路,当问题出现时,才能迅速定位.解决问题,这里给出一个处理问题的一般思路: 重视报错提示信息:每个错误的出现,都是给出错误提示 ...
- 【特 性】Attribute
1 AttributeUsage [AttributeUsageAttribute(AttributeTargets.All, AllowMultiple = true, Inherited = tr ...
- DP背包问题小总结
DP的背包问题可谓是最基础的DP了,分为01背包,完全背包,多重背包 01背包 装与不装是一个问题 01背包基本模型,背包的总体积为v,总共有n件物体,每件物品的体积为v[i],价值为w[i],每件物 ...
- 利用WMITool解决浏览器快捷方式启动参数被篡改以及浏览器主页被劫持的问题
先说说症状 症状①:通过快捷方式启动浏览器,首页跳转到2345以及hao123网址导航页,切系统内安装的多款浏览器(IE.Chrome.Firefox.Opera.Safari.Maxthon)症状相 ...
- PAT 1094. The Largest Generation (层级遍历)
A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level bel ...
- .Net防sql注入的方法总结
#防sql注入的常用方法: 1.服务端对前端传过来的参数值进行类型验证: 2.服务端执行sql,使用参数化传值,而不要使用sql字符串拼接: 3.服务端对前端传过来的数据进行sql关键词过来与检测: ...
- TypeError: CleanWebpackPlugin is not a constructor
在项目中引入clean-webpack-plugin打包后报错 new CleanWebpackPlugin(), ^ TypeError: CleanWebpackPlugin is not a c ...
- Beetl学习总结(2)——基本用法
2.1. 安装 如果使用maven,使用如下坐标 <dependency> <groupId>com.ibeetl</groupId> <artifactId ...