UVA101 The Blocks Problem

题目

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.

输入

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.

输出

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).

样例输入

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

样例输出

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

代码

#include<stdio.h>
#include<string.h>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rev(i,a,b) for(int i=a;i>=b;i--)

int block[30][30];
int len[30];

inline void initialize(int n){
   
    rep(i,0,n){
   
        block[i][0]=i;
        len[i]=1;
    }
}

inline int getnum(){
   
    int ans;
    if(scanf("%d",&ans)==EOF) return EOF;
    return ans;
}

inline int getcmd(char cmd1[],int &a,char cmd2[],int &b){
   
    scanf("%s",cmd1);
    if(strcmp(cmd1,"quit")==0) return -1;
    scanf("%d%s%d",&a,cmd2,&b);
    return 0;
}

inline void belong(int num,int &pile,int &height,int n){
   
    rep(i,0,n)
    rep(j,0,len[i]){
   
        if(block[i][j]==num){
   
            pile=i;
            height=j;
            return ;
        }
    }
}

inline void moveup(int p,int l,int length){
   
    rev(i,len[p]-1,l){
   
        block[p][i+length]=block[p][i];
    }
    len[p]+=length;
}

inline void transfer(int pa,int la,int pb,int lb,int length){
   
    rep(i,0,length){
   
        block[pb][lb+i]=block[pa][la+i];
    }
}

inline void movedown(int p,int l,int length){
   
    rep(i,l,len[p]){
   
        block[p][i-length]=block[p][i];
    }
    len[p]-=length;
}

inline void moveonto(int pa,int la,int pb,int lb){
   
    moveup(pb,lb+1,1);
    transfer(pa,la,pb,lb+1,1);
    movedown(pa,la+1,1);
}

inline void moveover(int pa,int la,int pb){
   
    moveup(pb,len[pb],1);
    transfer(pa,la,pb,len[pb]-1,1);
    movedown(pa,la+1,1);
}

inline void pileonto(int pa,int la,int pb,int lb){
   
    int length=len[pa]-la;
    moveup(pb,lb+1,length);
    transfer(pa,la,pb,lb+1,length);
    movedown(pa,len[pa],length);
}

inline void pileover(int pa,int la,int pb){
   
    int length=len[pa]-la;
    moveup(pb,len[pb],length);
    transfer(pa,la,pb,len[pb]-length,length);
    movedown(pa,len[pa],length);
}

inline void print(int n){
   
    rep(i,0,n){
   
        printf("%d:",i);
        rep(j,0,len[i]){
   
            printf(" %d",block[i][j]);
        }
        putchar('\n');
    }
}

int main(){
   
    int n;
    while((n=getnum())!=EOF){
   
        initialize(n);
        while(true){
   
            char cmd1[10],cmd2[10];int a,b;
            if(getcmd(cmd1,a,cmd2,b)==-1) break;
            int pa,la,pb,lb;
            belong(a,pa,la,n);
            belong(b,pb,lb,n);
            if(a==b||pa==pb) continue;
            if(strcmp(cmd1,"move")==0){
   
                if(strcmp(cmd2,"onto")==0){
   
                    moveonto(pa,la,pb,lb);
                }else{
   
                    moveover(pa,la,pb);
                }
            }else{
   
                if(strcmp(cmd2,"onto")==0){
   
                    pileonto(pa,la,pb,lb);
                }else{
   
                    pileover(pa,la,pb);
                }
            }
        }
        print(n);
    }
    return 0;
}

后记

萌新第一次写CSDN博客请多多指教!

全部评论

相关推荐

我见java多妩媚:大外包
点赞 评论 收藏
分享
点赞 收藏 评论
分享
牛客网
牛客企业服务