*HDU1053 哈夫曼编码

Entropy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5972    Accepted Submission(s): 2507

Problem Description

An
entropy encoder is a data encoding method that achieves lossless data
compression by encoding a message with “wasted” or “extra” information
removed. In other words, entropy encoding removes information that was
not necessary in the first place to accurately encode the message. A
high degree of entropy implies a message with a great deal of wasted
information; english text encoded in ASCII is an example of a message
type that has very high entropy. Already compressed messages, such as
JPEG graphics or ZIP archives, have very little entropy and do not
benefit from further attempts at entropy encoding.

English text
encoded in ASCII has a high degree of entropy because all characters are
encoded using the same number of bits, eight. It is a known fact that
the letters E, L, N, R, S and T occur at a considerably higher frequency
than do most other letters in english text. If a way could be found to
encode just these letters with four bits, then the new encoding would be
smaller, would contain all the original information, and would have
less entropy. ASCII uses a fixed number of bits for a reason, however:
it’s easy, since one is always dealing with a fixed number of bits to
represent each possible glyph or character. How would an encoding scheme
that used four bits for the above letters be able to distinguish
between the four-bit codes and eight-bit codes? This seemingly difficult
problem is solved using what is known as a “prefix-free
variable-length” encoding.

In such an encoding, any number of
bits can be used to represent any glyph, and glyphs not present in the
message are simply not encoded. However, in order to be able to recover
the information, no bit pattern that encodes a glyph is allowed to be
the prefix of any other encoding bit pattern. This allows the encoded
bitstream to be read bit by bit, and whenever a set of bits is
encountered that represents a glyph, that glyph can be decoded. If the
prefix-free constraint was not enforced, then such a decoding would be
impossible.

Consider the text “AAAAABCD”. Using ASCII, encoding
this would require 64 bits. If, instead, we encode “A” with the bit
pattern “00”, “B” with “01”, “C” with “10”, and “D” with “11” then we
can encode this text in only 16 bits; the resulting bit pattern would be
“0000000000011011”. This is still a fixed-length encoding, however;
we’re using two bits per glyph instead of eight. Since the glyph “A”
occurs with greater frequency, could we do better by encoding it with
fewer bits? In fact we can, but in order to maintain a prefix-free
encoding, some of the other bit patterns will become longer than two
bits. An optimal encoding is to encode “A” with “0”, “B” with “10”, “C”
with “110”, and “D” with “111”. (This is clearly not the only optimal
encoding, as it is obvious that the encodings for B, C and D could be
interchanged freely for any given encoding without increasing the size
of the final encoded message.) Using this encoding, the message encodes
in only 13 bits to “0000010110111”, a compression ratio of 4.9 to 1
(that is, each bit in the final encoded message represents as much
information as did 4.9 bits in the original encoding). Read through this
bit pattern from left to right and you’ll see that the prefix-free
encoding makes it simple to decode this into the original text even
though the codes have varying bit lengths.

As a second example,
consider the text “THE CAT IN THE HAT”. In this text, the letter “T” and
the space character both occur with the highest frequency, so they will
clearly have the shortest encoding bit patterns in an optimal encoding.
The letters “C”, “I’ and “N” only occur once, however, so they will
have the longest codes.

There are many possible sets of
prefix-free variable-length bit patterns that would yield the optimal
encoding, that is, that would allow the text to be encoded in the fewest
number of bits. One such optimal encoding is to encode spaces with
“00”, “A” with “100”, “C” with “1110”, “E” with “1111”, “H” with “110”,
“I” with “1010”, “N” with “1011” and “T” with “01”. The optimal encoding
therefore requires only 51 bits compared to the 144 that would be
necessary to encode the message with 8-bit ASCII encoding, a compression
ratio of 2.8 to 1.

 

Input

The
input file will contain a list of text strings, one per line. The text
strings will consist only of uppercase alphanumeric characters and
underscores (which are used in place of spaces). The end of the input
will be signalled by a line containing only the word “END” as the text
string. This line should not be processed.

 

Output

For
each text string in the input, output the length in bits of the 8-bit
ASCII encoding, the length in bits of an optimal prefix-free
variable-length encoding, and the compression ratio accurate to one
decimal point.

 

Sample Input

AAAAABCD

THE_CAT_IN_THE_HAT
END

 

Sample Output

64 13 4.9

144 51 2.8

 

Source

Greater New York 2000

 

题意：

只有大写字母和下划线的字符串，求哈夫曼编码长度和压缩比例。

代码：

 //搞不懂。。。算出每个字符出现的次数用优先队列从小到大存节点，每次取队列中两个最小的加起来再存入队列至队列中只有一个节点。
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<queue>
 #include<functional>
 #include<vector>
 using namespace std;
 int a[];
 char s[];
 int ans;
 int main()
 {
     while(scanf("%s",s))
     {
         if(!strcmp(s,"END"))
         break;
         priority_queue<int,vector<int>,greater<int> >q;
         int len=strlen(s);
         memset(a,,sizeof(a));
         for(int i=;i<len;i++)
         {
             if(s[i]=='_')
             a[]++;
             else a[s[i]-'A'+]++;
         }
         for(int i=;i<=;i++)
         if(a[i]!=)
         q.push(a[i]);
         if(q.size()==)
         ans=len;
         else
         {
             ans=;
             while(q.size()!=)
             {
                 int x=q.top();
                 q.pop();
                 int y=q.top();
                 q.pop();
                 ans=ans+x+y;
                 q.push(x+y);
             }
         }
         printf("%d %d %.1lf\n",*len,ans,(double)*len/(double)ans);
     }
     return ;
 }