UVA101 The Blocks Problem

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The Blocks Problem

Background Many areas of Computer Science use simple, abstract domains for both analytical and empirical studies. For example, an early AIstudy of planning and robotics (STRIPS) used a block world in which arobot arm performed tasks involving the manipulation of blocks.

In thisproblem you will model a simple block world under certain rules andconstraints. 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 armin how to manipulate blocks that lie on a flat table. Initially thereare n blocks on the table (numbered from 0 to n-1)with block b i adjacent to block b i+1for 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 blockb afterreturning any blocks that are stacked on top of blocks a andb totheir initial positions.

• move a over b

  where a and b are block numbers, puts block a onto the top of thestack containing blockb, after returning any blocks that are stackedon top of block a to their initial positions.

• pile a onto b

  where a and b are block numbers, moves the pile of blocks consistingof blocka, and any blocks that are stacked above block a, ontoblock b. All blocks on top of block b are moved to their initialpositions prior to the pile taking place. The blocks stacked above blocka retain their order when moved.

• pile a over b

  where a and b are block numbers, puts the pile of blocks consistingof blocka, and any blocks that are stacked above block a, ontothe top of the stack containing blockb. The blocks stacked above blocka retain their original order when moved.

• quit

  terminates manipulations in the block world.

Any command in which a = b or in which a and bare in the same stack of blocks is an illegal command. All illegalcommands should be ignored and should have noaffect on the configuration of blocks.

The Input The input begins with an integer n on a line by itself representingthe 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, onecommand per line. Yourprogram should process all commands until thequit command isencountered.

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. Eachoriginal block position numberedi (wheren is the number of blocks) should appearfollowed immediately by a colon.If there is at least a blockon 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 wheren is the integer on the first line of input).

Sample Input 

10move 9 onto 1move 8 over 1move 7 over 1move 6 over 1pile 8 over 6pile 8 over 5move 2 over 1move 4 over 9quit

Sample Output 

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

Miguel Revilla
2000-04-06


題目的意思是：　有一排過去ｎ個木塊，按指令移動木塊；

move a over b:
在木塊ａ所在的那一堆木塊，把ａ上面的的木塊全部放回原地，然後把ａ放到ｂ所在那堆木塊上面。。

move a onto b:
把ａ所在的那堆木塊，ａ上面的全部放回原處，ｂ所在的那堆也是一樣，把ｂ上面的木塊都放回原處，然後把ａ放到ｂ上面；

pile a over b:
把ａ所在的那一堆木塊，ａ上面的全部木塊包括ａ　，都移到ｂ所在的那堆木塊上面。

pile  a onto b:
把ｂ所在那堆木塊，在ｂ上面的木塊全都放回原處，並把ａ所在的那堆木塊ａ上面的全部移到ｂ上面。。

總的來說就是碰到ｍｏｖｅ就要把ａ上面的全部放回原處。如果碰到　ｏｎｔｏ　就要把ｂ上面的全部放到原處。
因為ｍｏｖｅ是只移動ａ一個，所以ａ上面的要歸位，而ｐｉｌｅ是移一堆，所以不用。
ｏｎｔｏ是要和ｂ貼在一起，所以ｂ上面的要歸位，而ｏｖｅｒ是上方，不需要直接接觸，所以不用。。

思路就是用棧來類比，一開始就是ｎ個棧。每個棧裡都是一個元素，然後按照指令移，在這個棧裡ｐｏｐ（）掉它，在另一個棧裡ｐｕｓｈ（）進去。。
分四種情況來做移動，每種情況處理方式不一樣。要注意如果是一堆移過去，因為還是要按照這個順序，多以要先把這一堆放到另一個數組，再按順序ｐｕｓｈj進去。
類比完輸出即可。。

ＡＣ代碼：

#include<iostream>#include<stdio.h>#include<string>#include<stack>using namespace std;int main () {stack <int> sta[100];int t;int num[100];int res[100];string m1,m2;int p1,p2;cin >> t;getchar();for( int i = 0; i < t ;i++) {sta[i].push(i);num[i] = i;}while(1) {cin >> m1;if(m1 == "quit")break;cin >>p1 >>m2 >>p2;if (num[p1] == num[p2])continue;if (m1 == "move" && m2 == "over") {for (;sta[num[p1]].top() != p1; ) {sta[ sta[num[p1]].top() ].push(sta[num[p1]].top());num[sta[num[p1]].top()] = sta[num[p1]].top();sta[num[p1]].pop();}sta[num[p2]].push(p1);sta[num[p1]].pop();num[p1] = num[p2];}if (m1 == "pile" && m2 == "over") {int k = 0;int temp[200];for (;sta[num[p1]].top() != p1; ) {temp[k++] = sta[num[p1]].top();num[sta[num[p1]].top()] = num[p2];sta[num[p1]].pop();}sta[num[p1]].pop();temp[k] = p1;num[p1] = num[p2];for(int w = k ;w >= 0; w--)sta[num[p2]].push(temp[w]);}if (m1 == "move" && m2 == "onto") {for (;sta[num[p1]].top() != p1; ) {sta[ sta[num[p1]].top() ].push(sta[num[p1]].top());num[sta[num[p1]].top()] = sta[num[p1]].top();sta[num[p1]].pop();}for (;sta[num[p2]].top() != p2; ) {sta[ sta[num[p2]].top() ].push(sta[num[p2]].top());num[sta[num[p2]].top()] = sta[num[p2]].top();sta[num[p2]].pop();}sta[num[p2]].push(sta[num[p1]].top());sta[num[p1]].pop();num[p1] = num[p2];}if (m1 == "pile" && m2 == "onto") {int k = 0;int temp[200];for (;sta[num[p1]].top() != p1; ) {temp[k++] = sta[num[p1]].top();num[sta[num[p1]].top()] = num[p2];sta[num[p1]].pop();}sta[num[p1]].pop();temp[k] = p1;num[p1] =num[p2];for (;sta[num[p2]].top() != p2; ) {sta[ sta[num[p2]].top() ].push(sta[num[p2]].top());num[sta[num[p2]].top()] = sta[num[p2]].top();sta[num[p2]].pop();}for(int w = k ;w >= 0; w--)sta[num[p2]].push(temp[w]);}}int j;for(int i = 0;i < t;i++) {cout << i <<":";for( j = 0 ;!sta[i].empty();j++) {res[j] =  sta[i].top();sta[i].pop();}for (j = j -1; j >= 0;j--)cout<<" "<<res[j];cout << endl;}return 0;}

UVA101 The Blocks Problem