POJ 1208 The Blocks Problem

The Blocks Problem

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5397		Accepted: 2312

Description

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

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:

Source

Duke Internet Programming Contest 1990,uva 101

 

 

 

解析：模拟题，有以下四种操作：

move a onto b：a和b都是积木的编号，先将a和b上面所有的积木都放回原处，再将a放在b上。

move a over b：a和b都是积木的编号，先将a上面所有的积木放回原处，再将a放在b上。（b上原有积木不动）

pile a onto b：a和b都是积木的编号，将a和其上面所有的积极组成的一摞整体移动到b上。在移动前要先将b上面所有的积极都放回原处。

　　　　　　　 移动的一摞积木要保持原来的顺序不变。

pile a over b：a和b都是积木的编号，将a和其上面所有的积极组成的一摞整体移动到b所在一摞积木的最上面一个积木上。移动的一摞积木

　　　　　　　要保持原来的顺序不变。

运用STL可以方便解决。

 

 

 

#include <cstdio>
#include <vector>
using namespace std;

int n, a, b;
char op1[5], op2[5];
vector<int> v[30];

//找到所在积木的位置及高度
void findBlock(int a, int& p, int& h)
{
    for(p = 0; p < n; ++p)
        for(h = 0; h < (int)v[p].size(); ++h)
            if(v[p][h] == a)
                return;
}

//清除位置p上高度h以上的积木
void clearAbove(int p, int h)
{
    for(size_t i = h+1; i < v[p].size(); ++i){
        int tmp = v[p][i];
        v[tmp].push_back(tmp);
    }
    v[p].resize(h+1);
}

//移动位置p上高度h及以上的积木到位置p2上
void moveBlock(int p, int h, int p2)
{
    for(size_t i = h; i < v[p].size(); ++i)
        v[p2].push_back(v[p][i]);
    v[p].resize(h);
}

int main()
{
    scanf("%d", &n);
    for(int i = 0; i < n; ++i)
        v[i].push_back(i);
    while(scanf("%s", op1), op1[0] != 'q'){
        scanf("%d%s%d", &a, op2, &b);
        int pa, pb, ha, hb;
        findBlock(a, pa, ha);
        findBlock(b, pb, hb);
        if(pa == pb)
            continue;
        if(op1[0] == 'm')
            clearAbove(pa, ha);
        if(op2[1] == 'n')
            clearAbove(pb, hb);
        moveBlock(pa, ha, pb);
    }
    for(int i = 0; i < n; ++i){
        printf("%d:", i);
        for(size_t j = 0; j < v[i].size(); ++j)
            printf(" %d", v[i][j]);
        printf("\n");
    }
    return 0;
}