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 armperformed 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 ofcommands.

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 blockson the table (numbered from 0 to n-1) with block bi adjacentto block bi+1 for all  asshown in the diagram below: Figure: Initial Blocks WorldThe valid commands for the robot arm that manipulates blocks are:move a onto b //将堆在a和b上的block全部返回到初始位置，然后在将a移动到b上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 // 将堆在a上的block全部返回到初始位置上，然后在将a移动到含b的堆上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 // 将堆在b上的block全部返回到初始位置上，然后在将a和堆在a上的全部block按原顺序堆到b上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 positionsprior to the pile taking place. The blocks stacked above block a retain their order when moved.pile a over b // 将a和堆在a上的全部block按原顺序堆到含b的堆上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 aretaintheir original order when moved.quitterminates 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 configurationof blocks.

The Input 

The input begins with an integer n on a line by itself representing the number of blocks in theblock 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 isthe 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 otherblock 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:

简单的模拟题，用vector就可以实现了，再加上一个数组记录每个block所处的位置下标。

题目总而言之就是，给你一条命令，遇到move就要将堆在a上面的全部返回到初始位置，然后再移动a。遇到pile就将a和a上面全部的block一起移动

遇到over就直接将要移动的放到含b的堆上，遇到onto就需要直接放在b上面所以要将堆在b上的全部返回到初始位置

#include <iostream>#include <vector>using namespace std;int addr[25]; //记录每一块block在哪个编号vector< vector<int> > vec;void ret( int a)    //将所有压在a上的block返回初始位置{int i = addr[a];vector<int>::iterator it;for( it = vec[i].begin(); it != vec[i].end(); it++)if( *it == a) break;for( vector<int>::iterator it1 = it+1; it1!=vec[i].end(); ){vec[*it1].push_back(*it1);addr[*it1] = *it1;it1 = vec[i].erase( it1);}}void moveOver( int a, int b){ret( a);  //返回a上的blockint i = addr[a];vector<int>::iterator it;for( it = vec[i].begin(); it != vec[i].end(); it++)if( *it == a) break;vec[i].erase( it);vec[addr[b]].push_back(a);  //将a推入b中addr[a] = addr[b];   //更新地址}void moveOnto( int a, int b){ret( b);moveOver( a, b);}void pileOver( int a, int b){int i = addr[a];vector<int>::iterator it;for( it = vec[i].begin(); it != vec[i].end(); it++)if( *it == a) break;for( it; it != vec[i].end(); ){      //将a及以上的block推入b中vec[addr[b]].push_back( *it);addr[*it] = addr[b];it = vec[i].erase( it);}}void pileOnto( int a, int b){ret( b);pileOver( a, b);}int main(){int n, a, b;string str1, str2;cin >> n;for( int i = 0; i < n; i++){vector<int> v;v.push_back(i);vec.push_back(v);addr[i] = i;}while( cin >> str1){if( str1 == "quit") break;cin >> a >> str2 >> b;if( a==b || addr[a] == addr[b]) continue;if( str1 == "move"){if( str2 == "over") moveOver( a, b);else moveOnto( a, b);}else{if( str2 == "over") pileOver( a, b);else pileOnto( a, b);}}for( int i = 0; i < n; i++){cout << i << ":";vector<int>::iterator it;b3a4for( it = vec[i].begin(); it != vec[i].end(); it++)cout << " " << *it;cout << endl;}return 0;}