The Blocks Problem
Time Limit: 1000 MS Memory Limit: 10000 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
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:
Source
Duke Internet Programming Contest 1990,uva 101
题意：
从左到右有n个木块，编号为0~n-1，要求模拟以下四种操作（下面的a和b都是木块编号）：
<1> move a onto b : 把a和b上方的木块全部归位，然后把a放在b上面。
<2> 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。
<3> pile a onto b : 把b上方的木块全部归位，然后把a及上面的木块整体放在b上面。
<4> pile a over b : 把a及上面的木块整体放在b所在木块堆的顶部。
遇到quit时终止数据。a和b在同一堆的指令是非法指令，忽略。
所有操作结束后，输出每个位置的木块列表，按照从底部到顶部的顺序排列。
分析：
每个木块堆的高度不确定，所以用vector来保存很合适：而木块堆的个数不超过n,所以用一个数组来保存即可。
代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <vector>
4 #include <string>
5 using namespace std;
6 int n;
7 vector<int>pile[30];
8 //找到木块所在的堆pile和高度height,以引用的形式返回调用者
9 void find_block(int a, int& p,int& h){ //引用是传递地址...当在函数里修改其值,变量的值也会随着改变
10 for(p=0;p<n;p++)
11 for(h=0;h<pile[p].size();h++)
12 if(pile[p][h]==a) return ;//void中的return:退出函数体，如果在函数体结尾处不加也可自动退出，
13 //如果在中途需要退出的话就用return
14 }
15 //把p堆高度为h的木块上方所有的木块移回原位置
16 void clear_above(int p,int h){
17 for(int i=h+1;i<pile[p].size();i++){
18 int t=pile[p][i];
19 pile[t].push_back(t); //把标号为t的木块放回原位，即t堆
20 }
21 pile[p].resize(h+1); //重新定义p堆的长度
22 }
23 //把p堆高度为h及其上方的木块整体移到p2堆得尾部
24 void pile_onto(int p,int h,int p2){
25 for(int i=h;i<pile[p].size();i++) pile[p2].push_back(pile[p][i]);
26 pile[p].resize(h);
27 }
28 //输出木块最后的状态
29 void print(){
30 for(int i=0;i<n;i++){
31 printf("%d:",i);
32 for(int j=0;j<pile[i].size();j++){
33 printf(" %d",pile[i][j]);
34 }
35 printf("\n");
36 }
37 }
38 int main(){
39 int a,b;
40 string s1,s2;
41 scanf("%d",&n);
42 for(int i=0;i<n;i++) pile[i].push_back(i);
43 while(cin>>s1){
44 if(s1=="quit") break;
45 cin>>a>>s2>>b;
46 int pa,pb,ha,hb;
47 find_block(a,pa,ha);
48 find_block(b,pb,hb);
49 if(pa==pb) continue;
50 if(s2=="onto") clear_above(pb,hb);
51 if(s1=="move") clear_above(pa,ha);
52 pile_onto(pa,ha,pb);
53 }
54 print();
55 return 0;
56 }
题后总结：
数据结构的核心是vector<int>pile[maxn]，所有操作都是围绕它进行的。vector就像一个二维数组，只是第一维大小固定（最大maxn），二维大小不固定。上述代码还有一个值得学习的技巧就是：输入一共有4种指令，提取指令之间的共同点，编写函数以减少重复代码。