2013
12-30

# UVa OJ 101 – The Blocks Problem (积木问题)

Time limit: 3.000 second

## Background背景

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问题

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:

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 block b after returning any blocks that are stacked on top of blocks a and b to their initial positions.
• 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
• 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.
• 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
• 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.
• 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
• 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.
• 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。
• quit
• terminates manipulations in the block world.
• 结束积木世界的操纵。

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.

## The Input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.

## The Output 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).

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

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

## Solution 解答

#include <iostream>
#include <string>
using namespace std;

int aStks[25][25];
int aPos[25]; //表示每一块积木的位置，所在栈的编号*所在高度
int aStkLen[25];
int nStkCnt;

struct COMMAND{
int nCom;
int nA;
int nB;
};
bool ParseCommand(const char *pComStr, COMMAND &cmd) {
int nPos = 0, i, j;
static char *pCom[] = {"move ", "pile ", "quit", "dump"};
for (i = 0; i < sizeof(pCom) / sizeof(pCom[0]); ++i) {
for(j = 0; pComStr[j] == pCom[i][j] && pCom[i][j] != 0; ++j);
if (pCom[i][j] == 0) break;
}
if (i >= sizeof(pCom) / sizeof(pCom[0])) {
return false;
}
cmd.nCom = i * 2;
if (cmd.nCom >= 4) {
return true;
}
if (pComStr[j] < '1' || pComStr[j] > '9') {
return false;
}
for (cmd.nA = 0; pComStr[j] >= '0' && pComStr[j] <= '9'; ++j) {
cmd.nA = cmd.nA * 10 + pComStr[j] - '0';
}
static char *pDir[2] = {" onto ", " over "};
for (i = 0, pComStr += j; i < 2; ++i) {
for(j = 0; pComStr[j] == pDir[i][j] && pDir[i][j] != 0; ++j);
if (pDir[i][j] == 0) {
break;
}
}
if (i >= 2) {
return false;
}
cmd.nCom += i;
if (pComStr[j] < '1' || pComStr[j] > '9') {
return false;
}
for (cmd.nB = 0; pComStr[j] >= '0' && pComStr[j] <= '9'; ++j) {
cmd.nB = cmd.nB * 10 + pComStr[j] - '0';
}
if (pComStr[j] != 0 || cmd.nA == cmd.nB) {
return false;
}
return true;
}

void ReturnUpperBlock(int nBaseBlk) {
//计算出积木所在的栈和高度。nStk为栈的编号，nHei为栈的高度
int nStk = aPos[nBaseBlk] / nStkCnt, nHei = aPos[nBaseBlk] % nStkCnt + 1;
//循环将指定积木以上的所有积木归位
for (int &nStkLen = aStkLen[nStk]; nStkLen > nHei; --nStkLen) {
int nUpperBlk = aStks[nStk][nStkLen - 1]; //最顶的积木编号
aStks[nUpperBlk][0] = nUpperBlk; //将其归原位
aPos[nUpperBlk] = nUpperBlk * nStkCnt; //重新计算合成该积木位置
aStkLen[nUpperBlk] = 1; //归位的目标栈的高度置1(只有归位的积木)
}
}

void Dump(void) {
for (int i = 0; i < nStkCnt; ++i, cout << endl) {
cout << i << ':';
for (int j = 0; j < aStkLen[i]; ++j) {
cout << ' ' << aStks[i][j];
}
}
}
int main(void) {
COMMAND cmd;
cin >> nStkCnt;
for (int i = 0; i < nStkCnt; ++i) {
aStks[i][0] = i; aPos[i] = i * nStkCnt; aStkLen[i] = 1;
}
for (string strLine; getline(cin, strLine);) {
if (!ParseCommand(strLine.c_str(), cmd)) {
continue;
}
if (cmd.nCom == 4) {
break; //quit
}
if (cmd.nCom == 6) {
Dump();
continue;
}
int nStkA = aPos[cmd.nA] / nStkCnt, nStkB = aPos[cmd.nB] / nStkCnt;
if (nStkA == nStkB) {
continue;
}
if (cmd.nCom % 2 == 0) {
ReturnUpperBlock(cmd.nB); //onto
}
if (cmd.nCom / 2 == 0) {
ReturnUpperBlock(cmd.nA); //move
}
int nHeiA = aPos[cmd.nA] % nStkCnt;
int nStkLenA = aStkLen[nStkA], &nStkLenB = aStkLen[nStkB];
for (aStkLen[nStkA] = nHeiA; nHeiA < nStkLenA; ++nHeiA) {
int nBlk = aStks[nStkA][nHeiA];
aPos[nBlk] = nStkB * nStkCnt + nStkLenB;
aStks[nStkB][nStkLenB++] = nBlk;
}
}
Dump();
return 0;
}

1. 有一点问题。。后面动态规划的程序中
int dp[n+1][W+1];
会报错 提示表达式必须含有常量值。该怎么修改呢。。