UVa 101 The Blocks Problem-优快云博客

UVa 101 The Blocks Problem

1 题目
2 思路
3 代码
4 参考

1题目

===========

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:

===========

2思路

又是一道线性数据结构的模拟题，又写了一大段冗长的代码，写完的时候我一直在想一个问题。到底如何设计出好的数据结构，并且使用这种数据结构写出代码量短小，速度又快，内存又少的程序呢？如何分析问题，练习并习得这种技能呢？

3代码

下面是两个版本的代码。

这一第一版AC的代码，~~还没有第二版，不过我确信会有的。~~

/*
 * Problem: UVa 101 The Blocks Problem
 * Lang: ANSI C
 * Time: 0s
 * Author: minix
 */

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define N 25
#define M 20

typedef struct _block {
  int seq;
  struct _block *top;
}Block;

typedef struct _node {
  Block *top;
  Block *ori;
  Block *pre;
  Block *roof;
  int cur_stack;
}Node;

void move_onto (Node nodes[N], int a, int b);
void move_over (Node nodes[N], int a, int b);
void pile_onto (Node nodes[N], int a, int b);
void pile_over (Node nodes[N], int a, int b);
int is_quit (char cmd[M]);
void implement_cmd (Node nodes[N], char cmd[M]);
void output (Node nodes[N], int n);
void init (Node nodes[N], int n);
Block * new_block (int seq);

void return_up_stacks (Node nodes[N], Block *block) {
  Block *cur = block->top;
  Block *next = NULL;
  int cur_seq;

  while (cur != NULL) {
    next = cur->top;
    cur_seq = cur->seq;
    cur->top = nodes[cur_seq].top;
    nodes[cur_seq].top = cur;
    nodes[cur_seq].cur_stack = cur_seq;
    nodes[cur_seq].pre = NULL;
    nodes[cur_seq].roof = cur;
    cur = next;
  }

  block->top = NULL;
  nodes[nodes[block->seq].cur_stack].roof = block;
}

void a2b (Node nodes[N], int a, int b) {
  Block *block_a = nodes[a].ori;
  Block *block_b = nodes[b].ori;

  if (nodes[a].pre != NULL) {
    nodes[a].pre->top = NULL;
  } else {
    nodes[a].top = NULL;
  }

  nodes[nodes[b].cur_stack].roof = nodes[nodes[a].cur_stack].roof;
  nodes[nodes[a].cur_stack].roof = nodes[a].pre;

  block_b->top = block_a;
  nodes[a].pre = block_b;
  nodes[a].cur_stack = nodes[b].cur_stack;
}

void move_onto (Node nodes[N], int a, int b) {
  Block *block_a = nodes[a].ori;
  Block *block_b = nodes[b].ori;

  return_up_stacks (nodes, block_a);
  return_up_stacks (nodes, block_b);

  a2b (nodes, a, b);
}

void move_over (Node nodes[N], int a, int b) {
  Block *block_a = nodes[a].ori;
  int cur_stack_b = nodes[b].cur_stack;

  return_up_stacks (nodes, block_a);

  a2b (nodes, a, nodes[cur_stack_b].roof->seq);
}

void a_block2b (Node nodes[N], int a, int b) {
  Block *block_a = nodes[a].ori;
  Block *next;

  a2b (nodes, a, b);
  next = block_a->top;
  while (next != NULL) {
    nodes[next->seq].cur_stack = nodes[b].cur_stack;
    next = next->top;
  }
}

void pile_onto (Node nodes[N], int a, int b) {
  Block *block_b = nodes[b].ori;

  return_up_stacks (nodes, block_b);
  a_block2b (nodes, a, b);
}

void pile_over (Node nodes[N], int a, int b) {
  int cur_stack_b = nodes[b].cur_stack;
  a_block2b (nodes, a, nodes[cur_stack_b].roof->seq);
}


void implement_cmd (Node nodes[N], char cmd[M]) {
  char action[M], type[M];
  int a, b;

  sscanf (cmd, "%s %d %s %d\n", action, &a, type, &b);
  if (nodes[a].cur_stack == nodes[b].cur_stack)
    return;

  if (!strcmp (action, "move") && !strcmp (type, "onto"))
    move_onto (nodes, a, b);
  else if (!strcmp (action, "move") && !strcmp (type, "over"))
    move_over (nodes, a, b);
  else if (!strcmp (action, "pile") && !strcmp (type, "onto"))
    pile_onto (nodes, a, b);
  else if (!strcmp (action, "pile") && !strcmp (type, "over"))
    pile_over (nodes, a, b); 
  else
    return;
}


int is_quit (char cmd[M]) {
  if (!strcmp (cmd, "quit\n"))
    return 1;
  else
    return 0;
}

void output (Node nodes[N], int n) {
  int i;
  Block *block;
  for (i=0; i<n; i++) {
    printf ("%d:", i);
    if (nodes[i].top != NULL) {
      block = nodes[i].top;
      while (block != NULL) {
        printf (" %d", block->seq);
        block = block->top;
      }
    }
    printf ("\n");
  }
}

Block * new_block (int seq) {
  Block *block = (Block *)malloc (sizeof(Block));
  block->seq = seq;
  block->top = NULL;
  return block;
}

void init (Node nodes[N], int n) {
  int i;
  Block *block;
  for (i=0; i<n; i++) {
    block = new_block (i);
    nodes[i].top = nodes[i].ori = nodes[i].roof = block;
    nodes[i].cur_stack = i;
    nodes[i].pre = NULL;
  }
}

int main() {
  int n;
  char cmd[M];
  Node nodes[N];
  int i;

  scanf ("%d\n", &n);
  init (nodes, n);

  while (fgets (cmd, M, stdin)) {
    if (is_quit (cmd))
      break;
    implement_cmd (nodes, cmd); 
  } 

   output (nodes, n); 

  return 0;
}

这是第二个版本的代码，使用了栈作为基本的数据结构，代码清晰简洁了不少。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

#define N 30
#define M 30

typedef struct _stack {
  int info[N];
  int top;
}Stack;

void push (Stack *stack, int n) {
  stack->info [++(stack->top)] = n;
}

int pop (Stack *stack) {
  return stack->info [(stack->top)--];
}

int get_top (Stack *stack) {
  return stack->info [stack->top];
}

int is_quit (char cmd[M]) {
  if (!strcmp (cmd, "quit\n"))
    return 1;
  else
    return 0;
}

void pop_to_init (Stack stacks[N], int locations[N], int a) {
  int pos = locations[a];
  int cur = get_top (&stacks[pos]);
  while (cur != a) {
    cur = pop (&stacks[pos]);
    push (&stacks[ cur ], cur);
    locations [cur] = cur;
    cur = get_top (&stacks[pos]);
  }
}

void stack2stack (Stack stacks[N], int locations[N], int src, int des, int top) {
  int cur_top;
  Stack tmp_stack;
  tmp_stack.top = 0;

  do {
    cur_top = pop (&stacks[src]);
    push (&tmp_stack, cur_top);
  }while (cur_top != top);

  while (tmp_stack.top != 0) {
    cur_top = pop (&tmp_stack);
    push (&stacks[des], cur_top);
    locations[cur_top] = des;
  }
}

void move_onto (Stack stacks[N], int locations[N], int a, int b) {
  pop_to_init (stacks, locations, a);
  pop_to_init (stacks, locations, b);
  pop (&stacks [ locations[a] ]);
  push (&stacks[ locations[b] ], a);
  locations[a] = locations[b];
}

void move_over (Stack stacks[N], int locations[N], int a, int b) {
  pop_to_init (stacks, locations, a);
  pop (&stacks [ locations[a] ]);
  push (&stacks [ locations[b] ], a);
  locations[a] = locations[b];
}


void pile_onto (Stack stacks[N], int locations[N], int a, int b) {
  pop_to_init (stacks, locations, b); 
  stack2stack (stacks, locations, locations[a], locations[b], a); 
  locations[a] = locations[b]; 
}

void pile_over (Stack stacks[N], int locations[N], int a, int b) {
  stack2stack (stacks, locations, locations[a], locations[b], a);
  locations[a] = locations[b];
}

void run_cmd(Stack stacks[N], int locations[N], char cmd[M]) {
  char action[M], type[M];
  int a, b;

  sscanf (cmd, "%s %d %s %d\n", action, &a, type, &b);

  if (a == b || locations[a] == locations[b])
    return;

  if (!strcmp (action, "move") && !strcmp (type, "onto"))
    move_onto (stacks, locations, a, b);
  else if (!strcmp (action, "move") && !strcmp (type, "over"))
    move_over (stacks, locations, a, b);
  else if (!strcmp (action, "pile") && !strcmp (type, "onto"))
    pile_onto (stacks, locations, a, b);
  else if (!strcmp (action, "pile") && !strcmp (type, "over"))
    pile_over (stacks, locations, a, b); 
  else
    return;
}

void output (Stack stacks[N], int n) {
  int i, j;
  for (i=0; i<n; i++) {
    printf ("%d:", i);
    if (stacks[i].top != 0) {
      for (j=1; j<=stacks[i].top; j++)
        printf (" %d", stacks[i].info[j]);
    }
    printf ("\n");
  }
}

int main() {
  int n, i;
  int locations[N];
  char cmd[M];
  Stack stacks[N];

  scanf ("%d\n", &n);
  for (i=0; i<n; i++) {
    stacks[i].top = 0;
    push (&stacks[i], i);
    locations[i] = i;
  }

  while (fgets (cmd, M, stdin)) {
    if (is_quit (cmd))
      break;
    run_cmd (stacks, locations, cmd);
  }

  output (stacks, n);
  return 0;
}