Uva 101 - The Blocks Problem

The Blocks Problem

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

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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

题目大意：

输入一系列指令让你移到木块

move a onto b

在將a搬到b上之前，先將a和b上的積木放回原來的位置（例如：1就放回1的最開始位罝）

move a over b

在將a搬到b所在的那堆積木之上之前，先將a上的積木放回原來的位罝（b所在的那堆積木不動）

pile a onto b

將a本身和其上的積木一起放到b上，在搬之前b上方的積木放回原位

pile a over b

將a本身和其上的積木一起搬到到b所在的那堆積木之上

quit

结束

解析：

这题最大的难点就是比较繁琐，但是耐心写还是能完成的。

原来想用字符串来模拟，但是不知道为什么一直RE

后来改用结构体模拟就AC了。

另外附上一组测试样例：

输入：

21

move 2 onto 1

move 3 onto 2

move 4 onto 3

move 5 over 1

pile 1 over 10

move 9 over 8

move 11 over 8

pile 3 over 8

pile 8 over 3

move 20 over 19

pile 19 over 18

pile 18 onto 15

move 15 over 3

pile 20 onto 19

pile 19 onto 18

pile 18 over 17

quit

输出：

0: 0

1:

2:

3:

4:

5:

6: 6

7: 7

8: 8 9 11 3 4 5 15

9:

10: 10 1 2

11:

12: 12

13: 13

14: 14

15:

16: 16

17: 17 18 19 20

18:

19:

20:

AC代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N = 30;
struct Stack {
	int top;
	int num
;
};
struct Block {
	int row;
	int col;
};
Stack stack
;
Block block
;
//初始化
void init(int n) {
	for(int i = 0 ;i < n; i++) {
		stack[i].top = 1;
		stack[i].num[0] = i;
		block[i].row = i;
		block[i].col = 0;
	}
}

void back(int a) {
	int x = block[a].row;
	int y = block[a].col;
	int top = stack[x].top;
	for(int i = top -1 ; i > y; i--) { //将a上面的所有元素都归位
		int topnum = stack[x].num[i]; //栈顶的数字
		stack[topnum].num[stack[topnum].top] = topnum;
		block[topnum].row = topnum;
		block[topnum].col = stack[topnum].top;
		stack[topnum].top++;
		stack[x].top--;
	}
}

void move(int a,int b) {
	int x_a = block[a].row;
	int y_a = block[a].col;
	int x_b = block[b].row;
	int y_b = block[b].col;
	stack[x_b].num[stack[x_b].top] = a;
	block[a].row = x_b;
	block[a].col = stack[x_b].top;
	stack[x_b].top++;
	stack[x_a].top--;
}

void pile(int a,int b) {
	int x_a = block[a].row;
	int y_a = block[a].col;
	int x_b = block[b].row;
	int y_b = block[b].col;
	int top = stack[x_a].top;
	for(int i = y_a; i < top; i++) {
		stack[x_b].num[stack[x_b].top] = stack[x_a].num[i];
		block[stack[x_a].num[i]].row = x_b;
		block[stack[x_a].num[i]].col = stack[x_b].top;
		stack[x_b].top++;
		stack[x_a].top--;
	}
}
int main() {
	int n;
	while(scanf("%d",&n) != EOF) {
		getchar();
		init(n);
		char str1[10],str2[10];
		int a,b;
		while(scanf("%s",str1)) {
			if(!strcmp(str1,"quit")) {
				break;
			}
			scanf("%d%s%d",&a,str2,&b);
			if(block[a].row == block[b].row) {
				continue;
			}
			if(!strcmp(str1,"move")) {
				if(!strcmp(str2,"onto")) {
					back(a);
					back(b);
					move(a,b);
				}
				else {
					back(a);
					move(a,b);
				}
			}
			else {
				if(!strcmp(str2,"onto")) {
					back(b);
					pile(a,b);
				}
				else {
					pile(a,b);
				}
			}
		}
		for(int i = 0; i < n; i++) {
			printf("%d:",i);
			for(int j = 0; j < stack[i].top; j++) {
				printf(" %d",stack[i].num[j]);
			}
			printf("\n");
		}
	}
	return 0;
}