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Fri, 21 Jan 2011 09:11:00 GMT

1. Compression algorithm (deflate)

The deflation algorithm used by gzip (also zip and zlib) is a variation of
LZ77 (Lempel-Ziv 1977, see reference below). It finds duplicated strings in
the input data. The second occurrence of a string is replaced by a
pointer to the previous string, in the form of a pair (distance,
length). Distances are limited to 32K bytes, and lengths are limited
to 258 bytes. When a string does not occur anywhere in the previous
32K bytes, it is emitted as a sequence of literal bytes. (In this
description, `string' must be taken as an arbitrary sequence of bytes,
and is not restricted to printable characters.)

Literals or match lengths are compressed with one Huffman tree, and
match distances are compressed with another tree. The trees are stored
in a compact form at the start of each block. The blocks can have any
size (except that the compressed data for one block must fit in
available memory). A block is terminated when deflate() determines that
it would be useful to start another block with fresh trees. (This is
somewhat similar to the behavior of LZW-based _compress_.)

Duplicated strings are found using a hash table. All input strings of
length 3 are inserted in the hash table. A hash index is computed for
the next 3 bytes. If the hash chain for this index is not empty, all
strings in the chain are compared with the current input string, and
the longest match is selected.

The hash chains are searched starting with the most recent strings, to
favor small distances and thus take advantage of the Huffman encoding.
The hash chains are singly linked. There are no deletions from the
hash chains, the algorithm simply discards matches that are too old.

To avoid a worst-case situation, very long hash chains are arbitrarily
truncated at a certain length, determined by a runtime option (level
parameter of deflateInit). So deflate() does not always find the longest
possible match but generally finds a match which is long enough.

deflate() also defers the selection of matches with a lazy evaluation
mechanism. After a match of length N has been found, deflate() searches for
a longer match at the next input byte. If a longer match is found, the
previous match is truncated to a length of one (thus producing a single
literal byte) and the process of lazy evaluation begins again. Otherwise,
the original match is kept, and the next match search is attempted only N
steps later.

The lazy match evaluation is also subject to a runtime parameter. If
the current match is long enough, deflate() reduces the search for a longer
match, thus speeding up the whole process. If compression ratio is more
important than speed, deflate() attempts a complete second search even if
the first match is already long enough.

The lazy match evaluation is not performed for the fastest compression
modes (level parameter 1 to 3). For these fast modes, new strings
are inserted in the hash table only when no match was found, or
when the match is not too long. This degrades the compression ratio
but saves time since there are both fewer insertions and fewer searches.

2. Decompression algorithm (inflate)

2.1 Introduction

The key question is how to represent a Huffman code (or any prefix code) so
that you can decode fast. The most important characteristic is that shorter
codes are much more common than longer codes, so pay attention to decoding the
short codes fast, and let the long codes take longer to decode.

inflate() sets up a first level table that covers some number of bits of
input less than the length of longest code. It gets that many bits from the
stream, and looks it up in the table. The table will tell if the next
code is that many bits or less and how many, and if it is, it will tell
the value, else it will point to the next level table for which inflate()
grabs more bits and tries to decode a longer code.

How many bits to make the first lookup is a tradeoff between the time it
takes to decode and the time it takes to build the table. If building the
table took no time (and if you had infinite memory), then there would only
be a first level table to cover all the way to the longest code. However,
building the table ends up taking a lot longer for more bits since short
codes are replicated many times in such a table. What inflate() does is
simply to make the number of bits in the first table a variable, and then
to set that variable for the maximum speed.

For inflate, which has 286 possible codes for the literal/length tree, the size
of the first table is nine bits. Also the distance trees have 30 possible
values, and the size of the first table is six bits. Note that for each of
those cases, the table ended up one bit longer than the ``average'' code
length, i.e. the code length of an approximately flat code which would be a
little more than eight bits for 286 symbols and a little less than five bits
for 30 symbols.

2.2 More details on the inflate table lookup

Ok, you want to know what this cleverly obfuscated inflate tree actually
looks like. You are correct that it's not a Huffman tree. It is simply a
lookup table for the first, let's say, nine bits of a Huffman symbol. The
symbol could be as short as one bit or as long as 15 bits. If a particular
symbol is shorter than nine bits, then that symbol's translation is duplicated
in all those entries that start with that symbol's bits. For example, if the
symbol is four bits, then it's duplicated 32 times in a nine-bit table. If a
symbol is nine bits long, it appears in the table once.

If the symbol is longer than nine bits, then that entry in the table points
to another similar table for the remaining bits. Again, there are duplicated
entries as needed. The idea is that most of the time the symbol will be short
and there will only be one table look up. (That's whole idea behind data
compression in the first place.) For the less frequent long symbols, there
will be two lookups. If you had a compression method with really long
symbols, you could have as many levels of lookups as is efficient. For
inflate, two is enough.

So a table entry either points to another table (in which case nine bits in
the above example are gobbled), or it contains the translation for the symbol
and the number of bits to gobble. Then you start again with the next
ungobbled bit.

You may wonder: why not just have one lookup table for how ever many bits the
longest symbol is? The reason is that if you do that, you end up spending
more time filling in duplicate symbol entries than you do actually decoding.
At least for deflate's output that generates new trees every several 10's of
kbytes. You can imagine that filling in a 2^15 entry table for a 15-bit code
would take too long if you're only decoding several thousand symbols. At the
other extreme, you could make a new table for every bit in the code. In fact,
that's essentially a Huffman tree. But then you spend two much time
traversing the tree while decoding, even for short symbols.

So the number of bits for the first lookup table is a trade of the time to
fill out the table vs. the time spent looking at the second level and above of
the table.

Here is an example, scaled down:

The code being decoded, with 10 symbols, from 1 to 6 bits long:

A: 0
B: 10
C: 1100
D: 11010
E: 11011
F: 11100
G: 11101
H: 11110
I: 111110
J: 111111

Let's make the first table three bits long (eight entries):

000: A,1
001: A,1
010: A,1
011: A,1
100: B,2
101: B,2
110: -> table X (gobble 3 bits)
111: -> table Y (gobble 3 bits)

Each entry is what the bits decode as and how many bits that is, i.e. how
many bits to gobble. Or the entry points to another table, with the number of
bits to gobble implicit in the size of the table.

Table X is two bits long since the longest code starting with 110 is five bits
long:

00: C,1
01: C,1
10: D,2
11: E,2

Table Y is three bits long since the longest code starting with 111 is six
bits long:

000: F,2
001: F,2
010: G,2
011: G,2
100: H,2
101: H,2
110: I,3
111: J,3

So what we have here are three tables with a total of 20 entries that had to
be constructed. That's compared to 64 entries for a single table. Or
compared to 16 entries for a Huffman tree (six two entry tables and one four
entry table). Assuming that the code ideally represents the probability of
the symbols, it takes on the average 1.25 lookups per symbol. That's compared
to one lookup for the single table, or 1.66 lookups per symbol for the
Huffman tree.

There, I think that gives you a picture of what's going on. For inflate, the
meaning of a particular symbol is often more than just a letter. It can be a
byte (a "literal"), or it can be either a length or a distance which
indicates a base value and a number of bits to fetch after the code that is
added to the base value. Or it might be the special end-of-block code. The
data structures created in inftrees.c try to encode all that information
compactly in the tables.

Jean-loup Gailly Mark Adler
jloup@gzip.org madler@alumni.caltech.edu

References:

[LZ77] Ziv J., Lempel A., ``A Universal Algorithm for Sequential Data
Compression,'' IEEE Transactions on Information Theory, Vol. 23, No. 3,
pp. 337-343.

``DEFLATE Compressed Data Format Specification'' available in
http://www.ietf.org/rfc/rfc1951.txt

蔡东�?/a> 2011-01-21 17:11 发表评论

Thu, 17 Jun 2010 00:50:00 GMT

想了几个月�?br />捣鼓了十多天实验�Q�和分析搞出来的。BP��经�|�络的异常点��应用可行性研�I?br />�q�就是我杯具的毕业设计�?br />
在这些日子，�U�压�?�q�有一切的一切，我失��M��你�?br />

http://www.shnenglu.com/Files/cdy20/2.4.rar

�?/span> �?/span>

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关键�?/span>�Q�异常，BP�Q�异常点��，��经�|�络

引用论文注明出处。妈的国内还没有一个�h做，国外tmd都是聚类�Q�神�l�网�l�只是辅�?

add ps�Q�你们反应也太激烈了吧！吓死我。祝大家工作生活��利�Q�~

蔡东�?/a> 2010-06-17 08:50 发表评论

Thu, 26 Nov 2009 04:40:00 GMT

O(n)旉��O(1)辅助�I�间�Q��@环移�?br>

�q�一道，google的笔试题�Q�网易也用过�Q�学校的数据�l�构题目�pȝ��也有�Q�之前我都是卡着机器出的

现在整理一�?br>
一 3�ơ翻转做�?br>/*about 循环�U�M��
实例�Q�abcdefgh 向左循环�U�M��3�?br>     �l�果 defghabc

      1.abc做翻�?nbsp; cba defgh

      2 defgh做翻�?nbsp; cba hgfed

      3.�W�二�l�果全部在做��{    成�ؓ defghabc
*/

template<class T>
void reverse(T a[],int i,int j)
{
     while(i < j)
     {
             swap(a[i],a[j]);
             ++i;
             --j;
     }
}

template<class T>
void exch1(T a[],int n,int k)
{
     reverse(a,0,k-1);
     reverse(a,k,n-1);
     reverse(a,0,n-1);
}

     reverse(a,0,k-1); // k/2
     reverse(a,k,n-1); // (n-k)/2
     reverse(a,0,n-1); // n/2
�?k/2+(n-k)/2+k/2=n/2 + n/2 = n

�?...

蔡东�?/a> 2009-11-26 12:40 发表评论

Tue, 22 Sep 2009 02:38:00 GMT

摘要: 阅读全文

蔡东�?/a> 2009-09-22 10:38 发表评论

hdu Northcott Game ��姆博奕 (Nimm Game)

Thu, 23 Apr 2009 02:56:00 GMT

昨晚热��n�Q�a了个背包�Q�tle了搜索（好久没练的）�Q�剩下的队友出，早上��Z��博弈�?br>

Northcott Game

Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 11 Accepted Submission(s) : 3

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

　　Tom和Jerry正在玩一�U�Northcott游戏�Q�可是Tom老是输，因此他怀疑这个游戏是不是有某�U�必胜策略，郁闷的Tom现在向你求救了，你能帮帮他么�Q?br>游戏规则是这��L��Q?br>　　如图所�C�，游戏在一个n行m列（1 ≤ n ≤ 1000�? ≤ m ≤ 100�Q�的��盘上进行，每行有一个黑子（黑方�Q�和一个白子（白方�Q�。执黑的一方先行，每次玩家可以�U�d��己方的�Q何一枚棋子到同一行的��M��一个空��g��Q�当然这�q�程中不许越�q�该行的敌方��子。双方轮��移动，直到某一�Ҏ��法行动�ؓ止，�U�d��最后一步的玩家莯��。Tom��L��先下�Q�黑方）。图1是某个初始局面，图二是Tom�U�d��一个棋子后的局面（�W�一行的黑子左移两步�Q��?br>

�?

�?

Input

　　输入数据有多�l�。每�l�数据第一行�ؓ两个整数n和m�Q�由�I�格分开。接下来有n行，每行两个数Ti�Q�Ji (1 ≤ Ti, Ji ≤ m)分别表示Tom和Jerry在该行棋子所处的列数�?br>　　注意�Q�各�l�测试数据之间有不定数量的空行。你必须处理到文件末�?/strong>

Output

对于每组��试数据输出一行你的结果。如果当前局面下Tom有必胜策略则输出“I WIN!”�Q�否则输�?#8220;BAD LUCK!”�?

Sample Input

3 6 4 5 1 2 1 2 3 6 4 5 1 3 1 2

Sample Output

BAD LUCK! I WIN!

/**//*
  ��姆博奕(Nimm Game):
有三堆各若干个物�?两个��从某一堆取��L��多的物品,规定每次臛_��取一�?多者不�?最后取光者得�?
�q�种情况最有意�?它与二进制有密切关系,我们�?a,b,c)表示某种局�?首先(0,0,0)昄��是奇异局�?无论谁面对奇异局�?都必然失�?�W�二�U�奇异局势是(0,n,n),只要与对手拿��C��样多的物�?最后都��导�?0,0,0).仔细分析一�?(1,2,3)也是奇异局�?无论�Ҏ��如何�?接下来都可以变�ؓ(0,n,n)的情�?
计算机算法里面有一�U�叫做按位模2�?也叫做异或的�q�算,我们用符�?+)表示�q�种�q�算.�q�种�q�算和一般加法不同的一�Ҏ��1+1=0.先看(1,2,3)的按位模2加的�l�果�Q?br>1 =二进�?1
2 =二进�?0
3 =二进�?1 (+)
-------
0 =二进�?0 (注意不进�?
对于奇异局�?0,n,n)也一�?�l�果也是0.
��M��奇异局�?a,b,c)都有a(+)b(+)c =0.
如果我们面对的是一个非奇异局�?a,b,c),要如何变为奇异局势呢�Q�假�?nbsp;a < b< c,我们只要��?nbsp;c 变�ؓ a(+)b,卛_��,因�ؓ有如下的�q�算�l�果: a(+)b(+)(a(+)b)=(a(+)a)(+)(b(+)b)=0(+)0=0.要将c 变�ؓa(+)b,只要�?nbsp;c中减�?nbsp;c-(a(+)b)卛_��.

*/
#include<iostream>
using namespace std;
#define MAXN 10001
int an[MAXN];

int Nimm_Game(int n)
{
      int ans=0;

      for(int i=0;i<n;++i){
         ans^=an[i];
      }
      return ans;
}
int main()
{
    int n,m,a,b;
    while(scanf("%d %d",&n,&m) !=EOF)
    {
        for( int i=0;i<n;++i){
             scanf("%d%d",&a,&b);
             an[i]=abs(a-b)-1;
         }

        if(!Nimm_Game(n)) printf("BAD LUCK!\n");

        else printf("I WIN!\n");
    }
    return 0;
}

蔡东�?/a> 2009-04-23 10:56 发表评论

威佐夫博�?Wythoff Game)

Sat, 18 Apr 2009 01:57:00 GMT
     摘要:   阅读全文

蔡东�?/a> 2009-04-18 09:57 发表评论

Sun, 26 Oct 2008 15:33:00 GMT

* 求有向图的强�q�通分�?(Strongerst Connected Component)(cut)

o Kosaraju��法

o Gabow��法

o Tarjan��法

* 求最��生成树 (Minimal Spanning Trees) (cut)

o Kruskal��法(cut�Ҏ��?

o Prim��法(cut�Ҏ��?

* 最短�\径问�?cut)

o SSSP(Single-source Shortest Paths)

* Dijkstra��法 (cut)

* Bellman-Ford��法(SPFA��法)(cut)

o APSP(All-pairs Shortest Paths)

* Floyd-Warshall��法(cut)

* Johnson��法

* �|�络��问�?

o 最大网�l�流

* 增广路算�?/div>
* Ford-Fulkerson��法

* Edmonds-Karp��法

* Dinic

* 预流推进��法

o 最��费用流

* 囑֌�配问�?nbsp; �Q�部分cut�Q?br>

o 匈牙利算�?cut)

o Kuhn-Munkres��法

o Edmonds' blossom-contraction ��法

�ơ小生成�?K��生成树)
最��树形图
最��K限制度生成树
最优比率生成树(0-1分数规划)
�W�K最短�\
LP问题以及Primal-Dual(单纯型法)
最大流(最短增�q��\、最高标号预��推�q?
最��费用流(最��费用�\、Primal-Dual��法)
二分图最优匹�?原始-对偶KM��法)

acm.pku.edu.cn 的：
最��生成树
1251(cut)

1258(cut)

1789(cut)

2485(cut)

最短�\

1062(cut 建模的时侯要注意 �Q�不断徏�W�合�{��差的�?做最短�\�?

1125�Q�cut 做全源最短�\
   再对以每各点为根的树�Q�找最长的�?
再对每棵树的最长边找最短的那一�?nbsp;
int ans=maxint;
    for(i=1;i<=n;i++){
      tmp=-1;
       for(j=1;j<=n;j++){
          tmp=max(tmp,a[i][j]);
       }
       if(tmp < ans){ans=tmp;val=i;}
    }
�Q?br>
1797(cut
    ��L��到n�?路径�?nbsp; 所能承受的最大重量的�?br>
    dist[k]   源点到k 路径中最��的那个�Ҏ��?nbsp; mat[k][i]边k-i权�?nbsp;
   取�\�?dist[k] 和mat【k】【i】边最大那�?nbsp; 更新 dist[i] )

2253(cut 要求的与 1797相反)

Johnson��法适用于求All Pairs Shortest Path. Johnson��法应用了重标号技术，先进行一�ơBellman-Ford��法�Q�然后对原图�q�行重标��P��w'(i,j)=h[i]-h[j]+w(i,j)。然后对每个点进行一�ơDijkstra�Q�每�ơDijkstra的复杂度为O(nlogn+m)�Q�于是算法复杂度为O(n^2logn+m)�?/p>

蔡东�?/a> 2008-10-26 23:33 发表评论

麻豆亚洲AV永久无码精品久久,四虎久久影院,99久久婷婷国产综合亚洲

hdu Northcott Game ���姆博奕 (Nimm Game)

Northcott Game

Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 11 Accepted Submission(s) : 3

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

Input

Output

Sample Input

Sample Output

威佐夫博�?Wythoff Game)

hdu Northcott Game ��姆博奕 (Nimm Game)