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;
}
POJ3460 Booksort的更多相关文章
- POJ3460 Booksort(IDA*)
POJ3460 Booksort 题意:给定一个长度为n的序列,每次可以取出其中的一段数,插入任意一个位置,问最少需要几次操作才能使整个序列变为1~n 思路:IDA*+迭代加深搜索 小技巧:将一段数插 ...
- A*专题训练
POJ2449 Remmarguts' Date UDF's capital consists of N stations. The hall is numbered S, while the sta ...
- hdu 1685 Booksort (IDA*)
Booksort Time Limit: 10000/10000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others) Tot ...
- POJ 3460 Booksort(算竞进阶习题)
IDA* 这题真不会写..估价函数太巧妙了.. 按照lyd神牛的说法我们把a[i+1]=a[i]+1记为正确后继,反之则记为错误后继 那么考虑最优的一次交换区间,至多能够纠正三个错误后继,所以我们统计 ...
- Booksort POJ - 3460 (IDA*)
Description The Leiden University Library has millions of books. When a student wants to borrow a ce ...
- 【POJ 3460】 Booksort
[题目链接] http://poj.org/problem?id=3460 [算法] IDA* 注意特判答案为0的情况 [代码] #include <algorithm> #include ...
- acm位运算应用 搜索
acm位运算应用 搜索 搜索 此处不讲题目,只讲位运算是怎样在这些题中实现和应用的.由于搜索题往往是基于对状态的操作,位运算往往特别有效,优化之后的效果可以有目共睹. 例1.POJ 132 ...
- Server Tomcat v7.0 Server at localhost failed to start.
1:这里记录一下这个错误,反正百度一大推,但是很长很长,我感觉这个问题肯定是servlet引起的,因为我遇到的总是如此: 2:我的问题如下所示: <servlet> <servlet ...
- scrapy爬去京东书籍信息
# -*- coding: utf-8 -*- import scrapy import urllib import json from copy import deepcopy class JdSp ...
随机推荐
- javascript实现组合列表框中元素移动效果
应用背景:在页面中有两个列表框,需要把其中一个列表框的元素移动到另一个列表框 . 实现的基本思想: (1)编写init方法对两个列表框进行初始化: (2)为body添加onload事件调用init方 ...
- 老技术新谈,Java应用监控利器JMX(2)
各位坐稳扶好,我们要开车了.不过在开车之前,我们还是例行回顾一下上期分享的要点. 上期由于架不住来自于程序员内心的灵魂的拷问,于是我们潜心修炼,与 Java 应用监控利器 JMX 正式打了个照面. J ...
- findbugs过滤R.java文件
在第一次使用findbugs时检查出100多个Bad pratice,仔细一看原来全是R文件里面的类名首字母没有大写导致的. 于是只有自己在findbugs设置界面中添加过滤条件来忽略掉R文件. 在F ...
- 1038 Recover the Smallest Number (30分)(贪心)
Given a collection of number segments, you are supposed to recover the smallest number from them. Fo ...
- 标准与扩展ACL
标准与扩展ACL 案例1:配置标准ACL 案例2:配置扩展ACL 案例3:配置标准命名ACL 配置扩展命名ACL 1 案例1:配置标准ACL 1.1 问题 络调通后,保证网络是通畅的.同时也很可能出现 ...
- fdisk分区规划和添加wap交换空间
分区规划和添加wap交换空间 1 案例1:硬盘分区及格式化 注意:fdisk只能分区小容量的磁盘 1.1 问题 本例要求熟悉硬盘分区结构,使用fdisk分区工具在磁盘 /dev/vdb 上按以下要 ...
- 部署并测试动态WSGI站点
部署并测试动态WSGI站点 5.1问题 本例要求为站点webapp0.example.c ...
- 微信小程序template富文本插件image宽度被js强制设置
这段时间一直做微信小程序,过程中遇到了一个问题,这个问题一直没有得到完美的解决. 问题描述: 在Web编程中经常会引入template插件,这个插件是封装好,我们通常的做法是直接引入,配置简单,好用, ...
- Dos 命令启动网卡
禁用网卡: netsh interface set interface "本地连接" disabled 启用网卡 : netsh interface set interface & ...
- 28.6 Integer 自动装箱和拆箱
public class IntegerDemo2 { public static void main(String[] args) { //自动装箱 // Integer i = new Integ ...