POJ3460 Booksort

飞来山上千寻塔，闻说鸡鸣见日升。

不畏浮云遮望眼，自缘身在最高层。——王安石

题目：Booksort

网址：http://poj.org/problem?id=3460

Description

The Leiden University Library has millions of books. When a student wants to borrow a certain book, he usually submits an online loan form. If the book is available, then the next day the student can go and get it at the loan counter. This is the modern way of borrowing books at the library.

There is one department in the library, full of bookcases, where still the old way of borrowing is in use. Students can simply walk around there, pick out the books they like and, after registration, take them home for at most three weeks.

Quite often, however, it happens that a student takes a book from the shelf, takes a closer look at it, decides that he does not want to read it, and puts it back. Unfortunately, not all students are very careful with this last step. Although each book has a unique identification code, by which the books are sorted in the bookcase, some students put back the books they have considered at the wrong place. They do put it back onto the right shelf. However, not at the right position on the shelf.

Other students use the unique identification code (which they can find in an online catalogue) to find the books they want to borrow. For them, it is important that the books are really sorted on this code. Also for the librarian, it is important that the books are sorted. It makes it much easier to check if perhaps some books are stolen: not borrowed, but yet missing.

Therefore, every week, the librarian makes a round through the department and sorts the books on every shelf. Sorting one shelf is doable, but still quite some work. The librarian has considered several algorithms for it, and decided that the easiest way for him to sort the books on a shelf, is by sorting by transpositions: as long as the books are not sorted,

take out a block of books (a number of books standing next to each other),

shift another block of books from the left or the right of the resulting ‘hole’, into this hole,

and put back the first block of books into the hole left open by the second block.

One such sequence of steps is called a transposition.

The following picture may clarify the steps of the algorithm, where X denotes the first block of books, and Y denotes the second block.

Original situation:

After step 1:

After step 2:

After step 3:

Of course, the librarian wants to minimize the work he has to do. That is, for every bookshelf, he wants to minimize the number of transpositions he must carry out to sort the books. In particular, he wants to know if the books on the shelf can be sorted by at most 4 transpositions. Can you tell him?

Input

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:

One line with one integer n with 1 ≤ n ≤ 15: the number of books on a certain shelf.

One line with the n integers 1, 2, …, n in some order, separated by single spaces: the unique identification codes of the n books in their current order on the shelf.

Output

For every test case in the input file, the output should contain a single line, containing:

if the minimal number of transpositions to sort the books on their unique identification codes (in increasing order) is T ≤ 4, then this minimal number T;

if at least 5 transpositions are needed to sort the books, then the message "5 or more".

Sample Input

3
6
1 3 4 6 2 5
5
5 4 3 2 1
10
6 8 5 3 4 7 2 9 1 10

Sample Output

2
3
5 or more

首先，理解：选取某一段剪切至另一位置的后面，等价于将该段到该位置的区间到该段前面；

若选区间[L, R)将其剪切至pos的后面（pos >= R），等价于将[R, pos ]剪切至L之前。

故只考虑将一区间剪切至后面的位置即可，避免重复计算相同状态了。

看起来整一棵搜索树的规模是无穷无尽的，因此即便限制深度为5，效率并不高；

可以考虑以下做法：

• 双向广搜
  
  不解释；

• IDA*
  
  估价函数为错误后继的个数tot除以三；
  
  若[tot / 3] + now.dep > dep，剪枝；
  
  不解释。

代码如下：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>

using namespace std;

int n, p[16];
bool valid(int *s)
{
	for(int i = 1; i <= n; ++ i) if(s[i - 1] != i) return false;
	return true;
}
void flip(int *s, int cur, int len, int *c, int pos)
{
	vector <int> q;
	q.clear();
	for(int i = 0; i < cur; ++ i) q.push_back(c[i]);

	for(int i = cur + len; i <= pos; ++ i) q.push_back(c[i]);
	for(int i = 0; i < len; ++ i) q.push_back(s[i]);
	for(int i = pos + 1; i < n; ++ i) q.push_back(c[i]);

	for(int i = 0; i < n; ++ i) c[i] = q[i];
	return;
}
bool dfs(int *a, int dep)
{
	if(valid(a)) return true;
	if(dep == 0) return false;
	int num = 0;
	for(int i = 0; i < n - 1; ++ i) if(a[i + 1] != a[i] + 1) ++ num;
	if(num > dep * 3) return false;
	int copy[16], s[15];
	memcpy(copy, a, sizeof(copy));
	for(int len = 1; len < n; ++ len)//限制了长度
	{
		for(int i = 0; i + len < n; ++ i)//定位
		{
			for(int k = i; k < i + len; ++ k) s[k - i] = a[k];
			for(int j = i + len; j < n; ++ j)
			{
				flip(s, i, len, copy, j);//复制粘贴函数
				if(dfs(copy, dep - 1)) return true;
				memcpy(copy, a, sizeof(copy));
			}
		}
	}
	return false;
}
int main()
{
	int T, copy[16];
	int dep;
	scanf("%d", &T);
	bool ok = false;
	while(T --)
	{
		scanf("%d", &n);
		memset(p, 0, sizeof(p));
		for(int i = 0; i < n; ++ i) scanf("%d", &p[i]);
		for(dep = 0; dep < 5; ++ dep)
		{
			memcpy(copy, p, sizeof(copy));
			if(dfs(copy, dep))
			{
				printf("%d\n", dep);
				ok = true;
				break;
			}
		}
		if(ok) ok = false;
		else puts("5 or more");
	}
	return 0;
}