1125 POJ#1208 The Blocks Problem

摘要

-一个早期的移动方块的AI程序的模拟实现

原题目摘要

-The Blocks Problem  http://poj.org/problem?id=1208

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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

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: 01: 1 9 2 42:3: 34:5: 5 8 7 66:7:8:9:

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题目理解

-程序的目的是按照指定的命令移动响应的木块；最后给出结果；实现的时候，以栈为实现的位置，每个栈的存放对应的位置的方块情况；主要实现清除a上的方块，移动a到b，移动到所在b的上面；pile 的时候在借助一个栈来完成；

注意

-识别命令的 我看命令不多 就借用命令的第二个字母来完成判断

日期

-2017 11 25

附加

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代码

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#include <iostream>#include <cstdio>#include <stack>#include <algorithm>#include <cstring>#include <memory>using namespace std;#define MAX 30/*move    onto  : _find ->a  _find->b _put a onto b    over  : _find ->a  _put a onto which contain bpile    onto  :__find top to a, _find b; pile them onto b    over  :_find top to a, _find b; which contain bquit*/int n;stack<int>  blockpos[MAX],tmp;int  ptr[MAX];//the i-th block 's stack_ordervoid init(){    scanf("%d",&n);    for(int i=0;i<n;i++){        blockpos[i].push(i);        ptr[i] = i;    }}void clearTop(int i){    int s_i_top = blockpos[ptr[i]].top();    while(s_i_top!=i){        blockpos[s_i_top].push(s_i_top);        blockpos[ptr[i]].pop();        ptr[s_i_top] = s_i_top;        s_i_top =blockpos[ptr[i]].top();    }}void moveOver(int a,int b){    clearTop(a);    blockpos[ptr[b]].push(blockpos[ptr[a]].top());    blockpos[ptr[a]].pop();    ptr[a] =  ptr[b];}void moveOnto(int a,int b){    clearTop(a);    clearTop(b);    moveOver(a,b);}void pileOver(int a,int b){    while(blockpos[ptr[a]].top()!=a){        tmp.push(blockpos[ptr[a]].top());        blockpos[ptr[a]].pop();    }    tmp.push(blockpos[ptr[a]].top());    blockpos[ptr[a]].pop();    int top_id;    while(!tmp.empty()){        top_id = tmp.top();        ptr[top_id] = ptr[b];        blockpos[ptr[b]].push(top_id);        tmp.pop();    }}void pileOnto(int a,int b){    clearTop(b);    pileOver(a,b);}void solve(){    init();    char com1[5],com2[5];    int a,b;    while(true){        scanf("%s%d%s%d",com1,&a,com2,&b);        switch(com1[1]){            case 'o':{//move                switch(com2[1]){                    case 'n':{//move a onto b                        moveOnto(a,b);                    break;}                    case 'v':{//move a over b                        moveOver(a,b);                    break;}                } break;            }            case 'i':{//pile                switch(com2[1]){                    case 'n':{//pile a onto b                        pileOnto(a,b);                    break;}                    case 'v':{//pile a over b                        pileOver(a,b);                    break;}                } break;            }            case 'u'://q u it            default:                return;        }    }}int main(){    solve();    for(int i=0;i<n;i++){        printf("%d:",i);        while(!blockpos[i].empty()){            tmp.push(blockpos[i].top());            blockpos[i].pop();        }        while(!tmp.empty()){            printf(" %d",tmp.top());            tmp.pop();        }        printf("\n");    }    return 0;}