[POJ3460] Booksort 题解
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
------------------------------
题目大意:给定n个乱序的1~n的数,每次可以将任意两段不重复的区间交换顺序,问至少交换几次能实现1~n的顺序(若大于等于5次则直接输出5 or more).
思路:看到5 or more我们可以很容易的想到迭代搜索.当某次操作后该序列为递增序列时就找到了最小操作次数.我们很容易想到在判断是否顺序时能够找到'当前可能的最小操作次数'.即:每次操作都可以更改掉3个非顺序的数字(由于每次更改只和前后有关故单次操作最多更改数为3)时还需要的操作次数.我们可以使用这个值进行一个对当前操作的估计:若当前已使用的操作数加上还需最小操作数大于了当前所限制迭代深度,我们就可以省去当前分支.
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <bitset>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define rg register
#define LL long long
#define __space putchar(' ')
#define __endl putchar('\n')
template <typename qwq> inline void read(qwq & x)
{
x = 0;
rg int f = 1;
rg char c = getchar();
while (c < '0' || c > '9')
{
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
{
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
x *= f;
}
template <typename qaq> inline void print(qaq x)
{
if (x < 0)
{
putchar('-');
x = -x;
}
if (x > 9) print(x / 10);
putchar(x % 10 + '0');
}
int t,n,id;
int a[233];
inline void move(int x,int y,int z)
{
//交换[x,y] [y + 1,z]的位置;
int temp[233],cnt = 0;
for (rg int i = y + 1;i <= z;++i) temp[++cnt] = a[i];
for (rg int i = x;i <= y;++i) temp[++cnt] = a[i];
cnt = 0;
for (rg int i = x;i <= z;++i) a[i] = temp[++cnt];
}
inline bool IDA_star(int d)
{
rg int tot = 0;
for (rg int i = 1;i < n;++i)//枚举搬书的起点
{
for (rg int j = i;j < n;++j)//枚举搬书的终点,j - i + 1即为搬的本数
{
for (rg int k = j + 1;k <= n;++k)//枚举和哪些书换位置
{
move(i,j,k);//将[i,j]与到[j + 1,k]换位
tot = 0;
//=====================================
for (rg int l = 1;l < n;++l)
{
if (a[l + 1] != a[l] + 1) ++tot;
}
if (a[n] != n) ++tot;
if (!tot) return true;
//=====================================
// rg int f = ceil(tot * 1.0 / 3);
// if (d + f > id) return false;
if (3 * d + tot <= 3 * id)
{
if (IDA_star(d + 1)) return true;
}
move(i,i - j + k - 1,k);//回溯
}
}
}
return false;
}
int main()
{
read(t);
while (t--)
{
read(n);
id = 0;
for (rg int i = 1;i <= n;++i) read(a[i]);
for (rg int i = 1;i < n;++i)
{
if (a[i + 1] != a[i] + 1) ++id;
}
if (a[n] != n) ++id;
if (!id)
{
print(0),__endl;
continue;
}
id = ceil(id * 1.0 / 3);
while (id <= 4)
{
if (IDA_star(1))
{
print(id),__endl;
break;
}
++id;
}
if (id > 4) puts("5 or more");
}
return 0;
}
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