hdu 1053 Entropy

Entropy

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2962 Accepted Submission(s): 1121

Problem DescriptionAn 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.

InputThe 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.

OutputFor 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

AAAAABCDTHE_CAT_IN_THE_HATEND

Sample Output

64 13 4.9144 51 2.8

SourceGreater New York 2000 

題目大意：求輸入的字串的霍夫曼編碼的編碼效率。

題意分析：此題思路比較簡單，基本原理就是大學資料結構中的霍夫曼編碼的實現：把給每個字元在字串中出現的次數作為該字元的權值，然後建立二叉樹儲存該字元，權值越小，在二叉樹中的深度越深，權值越大，在二叉樹中的深度越淺。方法就是：

1、每次選擇所有字串隊列中權值最小的兩個字元，然後建立一個新節點，將這兩字元節點串連起來，新節點的權值為這兩個字元節點的權值之和。

2、然後將新節點加入原隊列中，再迭代進行上述操作，直到最後建成一棵樹，即只剩一個節點。因為每次從隊列中選擇權值最小的節點，故採用優先隊列比較合適。

3、得到的這棵樹中，所有字元節點都在分葉節點上，用分葉節點的權值乘以該分葉節點的深度（根節點的深度為0），即為該霍夫曼編碼的編碼長度。

#include<stdio.h>#include<string.h>int INF=0x3f3f3f3f;//無窮大typedef struct{    int data;//記錄字母出現次數    int pa;//記錄父親結點} jd;jd d[80];//建立80個結點。應該夠用了char s[20005];//讀取文本void huffman(int n){    int m1,m2,x1,x2,t,i,j;//m1,m2記錄兩個權值較小的結點，x1,x2記錄對應下標    for(i=0;i<n;i++)    {        m1=m2=INF;//初始化為無窮大        for(j=0;j<n+i;j++)//找到兩個較小值的根結點        {            if(m1>d[j].data&&d[j].pa==-1&&d[j].data!=0)            {                m2=m1;                m1=d[j].data;                x2=x1;                x1=j;            }            else if(m2>d[j].data&&d[j].pa==-1&&d[j].data!=0)            {                x2=j;                m2=d[j].data;            }        }        if(m2!=INF)//如果找到兩個較小值        {            t=n+i;            d[t].data=m1+m2;//加入結點            d[x1].pa=d[x2].pa=t;//刪除兩個已最小的結點        }    }}int main(){    int i,l,c,p,ans;    while(scanf("%s",s))    {        if(strcmp(s,"END")==0)//結束標誌          break;        for(i=0;i<80;i++)//初始化        {            d[i].data=0;            d[i].pa=-1;        }        l=strlen(s);//記錄字串長度        for(i=0;i<l;i++)            if(s[i]=='_')                d[26].data++;//建立字元與數字下標的映射            else                d[s[i]-'A'].data++;        huffman(27);//建立哈弗曼樹        for(i=0;i<=26;i++)        {            c=0;//c記錄編碼所需長度            if(d[i].data!=0)            {                p=i;                while(d[p].pa!=-1)                {                    c++;                    p=d[p].pa;                }                if(c==0)//如果為根節點長度為1                    c=1;                d[i].data=c;//將data賦值為所需位元組數。方便下面算總數            }        }        ans=0;//ans記錄所需總位元組數        for(i=0;i<l;i++)        {            if(s[i]=='_')                ans+=d[26].data;            else                ans+=d[s[i]-'A'].data;        }        printf("%d %d %.1f\n",8*l,ans,8.0*l*1.0/ans);//如題意    }    return 0;}