UVa 101 - The Blocks Problem

The Blocks Problem

 

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 n blocks on the table (numbered from 0 to n-1) with block b_i adjacent to block b_i+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 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 aretain 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:

 1 // UVa101 The Blocks Problem

 2 // Rujia Liu

 3 // 代码由刘汝佳提供，我是在我的水平下，加以注释，学习大神的思想

 4 #include <cstdio>

 5 #include <string>

 6 #include <vector>//用vector

 7 #include <iostream>

 8 using namespace std;

 9 

10 const int maxn = 30;

11 int n;

12 vector<int> pile[maxn]; // 每个pile[i]是一个vector

13 

14 // 找木块a所在的pile和height，以引用的形式返回调用者

15 void find_block(int a, int& p, int& h) {

16   for(p = 0; p < n; p++)

17     for(h = 0; h < pile[p].size(); h++)

18       if(pile[p][h] == a) return;

19 }

20 

21 // 把第p堆高度为h的木块上方的所有木块移回原位

22 void clear_above(int p, int h) {

23   for(int i = h+1; i < pile[p].size(); i++) {

24     int b = pile[p][i];

25     pile[b].push_back(b); // 把木块b放回原位

26   }

27   pile[p].resize(h+1); // pile只应保留下标0~h的元素

28 }

29 

30 // 把第p堆高度h及其上方的木块整体移动到p2堆的顶部

31 void pile_onto(int p, int h, int p2) {

32   for(int i = h; i < pile[p].size(); i++)

33     pile[p2].push_back(pile[p][i]);

34   pile[p].resize(h);

35 }

36 

37 void print() {

38   for(int i = 0; i < n; i++) {

39     printf("%d:", i);

40     for(int j = 0; j < pile[i].size(); j++) printf(" %d", pile[i][j]);

41     printf("\n");

42   }  

43 }

44 

45 int main() {

46   int a, b;

47   cin >> n;

48   string s1, s2;

49   for(int i = 0; i < n; i++) pile[i].push_back(i);//初始化所有木块

50   while(cin >> s1 >> a >> s2 >> b) {

51     int pa, pb, ha, hb;//p代表pile，h代表height

52     find_block(a, pa, ha);

53     find_block(b, pb, hb);

54     if(pa == pb) continue; // 非法指令

55     if(s2 == "onto") clear_above(pb, hb);

56     if(s1 == "move") clear_above(pa, ha);//这两行好厉害，反正我想不出。。

57     pile_onto(pa, ha, pb);

58   }

59   print();//我觉得没必要，可能函数好看吧

60   return 0;

61 }

这里主要是用了vector（我暂时把它理解成不定长数组）

若a是一个vector，

//声明方法vector<T> a (要用头文件<vector>)

a.size() 读取a的大小

a.resize() 改变大小//a.resize(x+1)就是只保留0~x的元素

a.push_back()向尾部添加元素

a.pop_back() 删除最后一个元素 //这两条有点像栈

a.clear() 清空

a.empty() 判断是否为空//如果当前vector没有容纳任何元素,则empty()函数返回true,否则返回false

 

这个代码还有个值得学习的地方，就是题目要求4种指令，如果每个独立编写容易出错且不是很好写。

而代码中提取了各个指令的共同点，用了两个函数就写出来了，好厉害。。。