[Andrew Stankevich's Contest#21] Lempel-Ziv Compression
Problem Description
Most modern archivers, such as WinRAR or WinZIP, use modifications of Lempel-Ziv method as their primary compression algorithm. Although decompression of LZ-compressed archives is usually easy and fast, the compression process itself is often rather complicated and slow. Therefore professional archivers use approximation methods that sometimes do not allow to achieve the best possible compression.
This situation doesn’t satisfy your chief George. He would like to create the best archiver WinGOR. The archiver will use the following modification of LZ77 algorithm.
The text is partitioned to chunks of length not exceeding 4096. Each chunk is compressed independently. We will describe the decompression of one chunk t. Based on this description, you will have to create a compression algorithm that will create the shortest possible compressed chunk x from the given chunk t.
The compressed chunk is written down as the sequence of plain characters and repetition blocks. Plain character is 8 bits long. When decompressing, plain character c is simply copied to output. Repetition block (r; l) consists of two parts: reference r and length l, each 12 bits long. Reference r is an integer number between 1 and 4095. When repetition block (r; l) is obtained after decompressing i − 1 characters of text, characters t[i − r ... i − r + l − 1] are copied to output. Note, that r can be less than l, in this
case recently copied characters are copied to output as well.
To help decompressor distinguish between plain characters and repetition blocks a leading bit is prepended to each element of the compressed text: 0 means plain character follows, 1 — repetition block follows.
For example, “aaabbaaabababababab” can be compressed as “aaabb(5,4)(2,10)”. The compressed variant has 8 + 8 + 8 + 8 + 8 + 24 + 24 + 7 = 95 bits instead of 152 in the original text (additional 7 bits are used to distinguish between plain characters and repetition blocks).
Given a text chunk, find its compressed representation which needs fewest number of bits to encode
Input
Output
Sample Input
aaabbaaabababababab
Sample Output
95
aaabb(5,4)(2,10)
题意概述
分析
这里需要的一点小trick在于如何寻找尽量靠前的一段最长重复子串。代入kmp算法的思路,我们可以对原字符串的每一段后缀预处理出next数组,这样原字符串中每个前缀的最长重复子串就可以在O(n)的时间内得到了。于是这里可以得出一个复杂度O(n^2)的动态规划:
记minbit[i]为压缩前i个字符所需的最小数据量,str[i]为此时的决策,r[i],l[i]为这次转移时压缩成的数对(若不需压缩,str[i]=i, r[i]储存不需压缩的部分的起始点)
那么,每次转移时就可以从0到i-1枚举决策j,选择可以使(minbit[j] + (i-j)*9)或minbit[i - next[j][i-1]] + 25取得最小值的j,并保存决策。最后可以通过一个递归函数推出最终答案。
(这里需要注意一点问题……next数组是二维的,数据范围maxn为2^12,而内存限制为64000KB……如果用next[maxn][maxn]的方式开数组明显会MLE……所以我鼓起勇气采用了动态内存分配……)
AC代码
1 //Verdict: Accepted 2 // Submission Date: -- ::
3 // Time: 8256MS
4 // Memory: 34624KB
5
6 /*=============================================================================================================================*/
7 /*======================================================Code by Asm.Def========================================================*/
8 /*=============================================================================================================================*/
9 #include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <memory.h>
#include <cstring>
#include <cstdlib>
using namespace std;
#define maxn ((int)4.1e3)
/*===========================================================TYPES=============================================================*/
typedef long long LL;
/*======================================================GLOBAL VARIABLES=======================================================*/
char ch[maxn];
int len = , minbit[maxn], *next[maxn];
int l[maxn], r[maxn], str[maxn];
/*==========================================================FUNCTIONS==========================================================*/
inline void getnext(int l){
int i, j, L = len - l;
next[l] = new int[L+];
int *Next = next[l];
Next[] = ;
for(i = ;i < L;++i){
j = Next[i-] - ;
while(ch[l+i] != ch[l+j+] && j >= )
j = Next[j] - ;
if(ch[l+i] == ch[l+j+])
Next[i] = j + ;
else Next[i] = ;
}
}
void printpro(int i){
if(str[i] == i){
if(r[i])printpro(r[i]-);
int j;
for(j = r[i];j <= i;++j)putchar(ch[j]);
return;
}
printpro(str[i]);
printf("(%d,%d)", r[i], l[i]);
}
int main(){
#ifdef DEBUG
assert(freopen("test","r",stdin));
#endif
//--------------------------------------------------variables-----------------------------------------------------------
//-----------------------------------------------------work-------------------------------------------------------------
char c;
while(isalpha(c = getchar()))str[len] = len, ch[len++] = c;
int i, j, Min, t;
for(i = ;i < len - ; ++i)
getnext(i);
minbit[] = ;
for(i = ;i < len; ++i){
Min = 0x7fffffff;
for(j = ;j < i;++j)
if(minbit[j] + (i-j)* < Min){
Min = minbit[j] + (i-j)*;
str[i] = i;
r[i] = j+;
}
for(j = ;j < i; ++j){
t = next[j][i-j];
if(!t)continue;
if(minbit[i-t] + < Min){
Min = minbit[i-t] + ;
str[i] = i-t;
r[i] = i+-t-j;
l[i] = t;
}
}
minbit[i] = Min;
}
printf("%d\n", minbit[len-]);
printpro(len-);
return ;
}
/*=============================================================================================================================*/
[Andrew Stankevich's Contest#21] Lempel-Ziv Compression的更多相关文章
- Andrew Stankevich's Contest (21) J dp+组合数
坑爹的,,组合数模板,,, 6132 njczy2010 1412 Accepted 5572 MS 50620 KB C++ 1844 B 2014-10-02 21:41:15 J - 2-3 T ...
- 【模拟ACM排名】ZOJ-2593 Ranking (Andrew Stankevich’s Contest #5)
真心是道水题,但找bug找的我想剁手了/(ㄒoㄒ)/~~ 注意几个坑点, 1.输入,getline(cin); / gets(); 一行输入,注意前面要加getchar(); 输入运行记录的时候可 ...
- Andrew Stankevich's Contest (1)
Andrew Stankevich's Contest (1) 打一半出门了,回来才补完了...各种大数又不能上java..也是蛋疼无比 A:依据置换循环节非常easy得出要gcd(x, n) = 1 ...
- acdream:Andrew Stankevich Contest 3:Two Cylinders:数值积分
Two Cylinders Special JudgeTime Limit: 10000/5000MS (Java/Others)Memory Limit: 128000/64000KB (Java/ ...
- GYM 100608G 记忆化搜索+概率 2014-2015 Winter Petrozavodsk Camp, Andrew Stankevich Contest 47 (ASC 47)
https://codeforces.com/gym/100608 题意: 两个人玩游戏,每个人有一个长为d的b进制数字,两个人轮流摇一个$[0,b-1]$的骰子,并将选出的数字填入自己的d个空位之中 ...
- LeetCode Weekly Contest 21
1. 530. Minimum Absolute Difference in BST 最小的差一定发生在有序数组的相邻两个数之间,所以对每一个数,找他的前驱和后继,更新结果即可!再仔细一想,bst的中 ...
- NOIP模拟·20141105题解
[A.韩信点兵] 结论题+模板题,用到了中国剩余定理,维基百科上讲的就比较详细,这里就不再赘述了…… 对于这题,我们先利用中国剩余定理($x \equiv \sum{(a_i m_i (m_i^{-1 ...
- ZJU 2605 Under Control
Under Control Time Limit: 2000ms Memory Limit: 65536KB This problem will be judged on ZJU. Original ...
- AC自动机-算法详解
What's Aho-Corasick automaton? 一种多模式串匹配算法,该算法在1975年产生于贝尔实验室,是著名的多模式匹配算法之一. 简单的说,KMP用来在一篇文章中匹配一个模式串:但 ...
随机推荐
- 安全测试===appscan扫描工具介绍
IBM AppScan该产品是一个领先的 Web 应用安全测试工具,曾以 Watchfire AppScan 的名称享誉业界.Rational AppScan 可自动化 Web 应用的安全漏洞评估工作 ...
- 【模板】BZOJ 1692:队列变换—后缀数组 Suffix Array
传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=1692 题意: 给出一个长度为N的字符串,每次可以从串头或串尾取一个字符,添加到新串中,使新串 ...
- https配置注意细节
直接将阿里云https的ca配置配置好之后如果不通的话很有可能是防火墙原因造成的,还有就是nginx要用1.10以上版本的
- windows server 2012 IIS配置之FTP站点
原文地址:[原创]winserver2012IIS配置之FTP站点作者:hkmysterious 一.实验拓扑: 使server2012客户计算机通过ftp方式从FTP服务器上下载已上传并共享的文 ...
- gridcontrol的列头右键菜单问题
Dev控件GridControl设置了一个右键菜单 this.gridControl1.ContextMenu = contextMenu2; 而GridControl在运行排序的时候,即 gridv ...
- AIOps实践三板斧:从可视化、自动化到智能化
http://ai.51cto.com/art/201806/576881.htm?mobile
- 【JBPM4】State 节点
State状态节点 相比 Task 节点的区别为: 主要是没有“操作人员”assignee 流程操作方便基本相同.如下: 部署流程: repositoryService.createDeploymen ...
- 使用python获取网易云音乐无损音频教程
博客园主页:http://www.cnblogs.com/handoing/ github项目:https://github.com/handoing/get-163-music 环境:Python ...
- js对象替换键值名称
js对象替换键值名称 将obj中的id和name字段替换分别替换成为“@id”,“@name” 代码如下: let obj = [{id:,name:,name:"李四"}].ma ...
- python模式匹配,提取指定字段
re匹配时分多行模式(re.M)与单行模式(rs.S),多行模式是每一行单独匹配,单行模式是把所有的行当成一行来匹配. 单行模式下.可以匹配换行符. ^$匹配所有字符 import re s='1_2 ...