The Blocks Problem UVA - 101

ManyareasofComputerScienceusesimple,abstractdomainsforbothanalyticalandempiricalstudies. 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. Theproblemistoparseaseriesofcommandsthatinstructarobotarminhowtomanipulateblocks 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:
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 initialpositions priorto thepile takingplace. Theblocksstackedaboveblock a retaintheir 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. Youmayassumethatallcommandswillbeoftheformspecifiedabove. Therewillbenosyntactically 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 isatleastablockonit, thecolonmustbefollowedbyonespace, followedbyalistofblocksthatappear stackedin that position with eachblocknumber separated from other blocknumbers bya 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:

题意：从左到右有n个木板（0~n-1）要求模拟一下四种操作。

　　　　　　move a onto b:  把a和b上方的木块全部归位，然后把a摞在b上面。

　　　　　　move a over b:  把a上方的木块全部归位，然后把a放在b所在木块锥的顶部。

　　　　　　pile    a onto b:  把b上方的木块全部归位，然后把a及上面的整体摞在b上面。

　　　　　　pile    a over b:  把a及上面的木块整体摞在b所在木块锥的顶部。

书上的写法相当棒，从四种操作中提取出了共同点；

对数据结构中的vector要熟练才行啊。

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<string>
 5 #include<cstdio>
 6 #include<vector> 
 7 using namespace std;
 8 
 9 int n;
10 vector<int> pile[30];
11 
12 void Find(int a,int& p,int& h){
13     for(p=0;p<n;p++){
14         for(h=0;h<pile[p].size();h++)
15             if(pile[p][h]==a) return;
16     }
17 }
18 
19 void Clear(int p,int h){
20     for(int i=h+1;i<pile[p].size();i++){
21         int tem=pile[p][i];
22         pile[tem].push_back(tem);
23     }
24     pile[p].resize(h+1);
25 }
26 
27 void Pile(int p,int h,int p2){
28     for(int i=h;i<pile[p].size();i++){
29         int tem=pile[p][i];
30         pile[p2].push_back(tem);
31     }
32     pile[p].resize(h);
33 }
34 
35 void print(){
36     for(int i=0;i<n;i++){
37         printf("%d:",i);
38         for(int j=0;j<pile[i].size();j++)
39             printf(" %d",pile[i][j]);
40         printf("\n");
41     }
42 }
43     
44 int main()
45 {   cin>>n;
46     for(int i=0;i<n;i++) pile[i].push_back(i);
47     int x,y;
48     char a[4],b[4];
49     while(true){
50         scanf("%s",a);
51         if(a[0]=='q') break;
52         scanf("%d%s%d",&x,b,&y);
53         int p1,p2,h1,h2;
54         Find(x,p1,h1);
55         Find(y,p2,h2);
56         if(p1==p2) continue;
57         if(b[1]=='n') Clear(p2,h2);
58         if(a[0]=='m') Clear(p1,h1);
59         Pile(p1,h1,p2);
60     }
61     print();
62     return 0;
63 }

 

转载于:https://www.cnblogs.com/zgglj-com/p/7237027.html