题目链接：https://vjudge.net/contest/361484#problem/J
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翻译：
给定一个字符串，只由大写字母和下划线组成。一位字符由8位ASCII编码，也可以通过哈夫曼编码编码
输出8位ASCII编码的位长度 哈夫曼编码的位长度 压缩比
哈夫曼编码：
最优前缀编码。即对于n个字符，分别以它们的使用频度作为叶子权值来构造哈夫曼树。
哈夫曼树：
由n个带权叶子结点构成的所有二叉树中带权路径长度最短的二叉树。
树的带权路径长度：
从根到所有叶子结点的各个带权路径长度之和。
带权路径长度：
从树根到某一结点的路径长度与该结点的权的乘积。
路径长度：
根结点到该结点所经过的分支数目
用静态三叉链表存储哈夫曼树：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 60
#define M 2*N-1
int len;
typedef struct
{
int weight;/*结点的权值*/
int parent;/*双亲的下标*/
int LChild;/*左孩子结点的下标*/
int RChild;/*右孩子结点的下标*/
} HTNode,*HuffmanTree;
void select(HuffmanTree *ht,int n,int *s1,int *s2)
{
int mi;
for(int i=1; i<=n; i++)
{
if((*ht)[i].parent==0)/*找到其中任意一个可能的结果,没有用过的结点parent=0*/
{
mi=i;
i=n+1;
}
}
for(int i=1; i<=n; i++)/*找到其中的一个最小值*/
{
if((*ht)[i].parent==0)
{
if((*ht)[i].weight<(*ht)[mi].weight)
mi=i;
}
}
*s1=mi;
for(int i=1; i<=n; i++)
{
if((*ht)[i].parent==0&&i!=(*s1))
{
mi=i;
i=n+1;
}
}
for(int i=1; i<=n; i++)
{
if((*ht)[i].parent==0&&i!=(*s1))
{
if((*ht)[i].weight<(*ht)[mi].weight)
mi=i;
}
}
*s2=mi;
}
void CrtHuffmanTree(HuffmanTree *ht,int *w,int n)
{
*ht=(HuffmanTree)malloc((M+1)*sizeof(HTNode)); /*0号单元未使用*/
int s1,s2;
for(int i=1; i<=n; i++)
{
(*ht)[i].weight = w[i-1];
(*ht)[i].LChild = 0;
(*ht)[i].parent = 0;
(*ht)[i].RChild = 0;
}
int m=2*n-1;/*n个结点的哈夫曼树共有2*n-1个结点*/
for(int i=n+1; i<=m; i++)
{
(*ht)[i].weight=0;
(*ht)[i].LChild=0;
(*ht)[i].parent=0;
(*ht)[i].RChild=0;
}
for(int i=n+1; i<=m; i++) /*创建非叶子结点,建哈夫曼树*/
{
/*在(*ht)[1]~(*ht)[i-1]的范围内选择两个parent为0且weight最小的结点，其序号分别赋值给s1、s2返回*/
select(ht,i-1,&s1,&s2);
(*ht)[s1].parent=i;
(*ht)[s2].parent=i;
(*ht)[i].LChild=s1;
(*ht)[i].RChild=s2;
(*ht)[i].weight=(*ht)[s1].weight+(*ht)[s2].weight;
}
}
void print(HuffmanTree *ht,int n)
{
int sum=0,cnt,k;
for(int i=1; i<=n; i++)/*n个叶子结点*/
{
k=i;
cnt=0;
while((*ht)[k].parent!=0)
{
k=(*ht)[k].parent;
cnt++;/*分支数目*/
}
sum+=cnt*(*ht)[i].weight;
}
float rat=(len*8.0)/sum;
printf("%d %d %.1f\n",len*8,sum,rat);
}
int main()
{
HuffmanTree HT;
char ch[100000];
int book[N];
int j;
while(gets(ch))
{
if(strcmp(ch,"END")==0)
break;
int *w;
memset(book,0,sizeof(book));
w=(int *)malloc((N+1)*sizeof(int));/*开辟一个w数组*/
int t=0;
len=strlen(ch);
for(int i=0; ch[i]!='\0'; i++)
{
if(ch[i]=='_')
t++;
else
book[ch[i]-'A']++;/*每个字母出现的频度*/
}
book[26]=t;/*0~25是26个英文字母*/
int j=0;/*j个叶子权值*/
for(int i=0; i<26; i++)
{
if(book[i])
w[j++]=book[i];
}
if(t)/*下划线存在*/
w[j++]=t;/*j是叶子结点的个数*/
if(j==1)
printf("%d %d %.1f\n",len*8,w[0],len*8*1.0/w[0]);
else
{
CrtHuffmanTree(&HT,w,j);
print(&HT,j);
}
}
return 0;
}
简化指针：
#include<cstdio>
#include<cstring>
#include<cstdlib>
typedef struct Node
{
int weight;
int parent,left,right;
} HTNode,*HuffmanTree;
typedef char *HuffmanCode;
int len;
void HuffmanCoding(int *w,int n)
{
int *q=w;
HuffmanTree HT;
HuffmanTree p;
int m,i,j,s1,s2,min1,min2,c,f,sum=0;
if(n<1)
return;
m=2*n-1;
HT=new HTNode[m+1]; //0号单元没用
for(p=HT+1,i=1; i<=n; ++i,++p,++w) //初始化
{
p->weight=*w;
p->parent=0;
p->left=0;
p->right=0;
}
for(; i<=m; ++i,++p)
{
p->weight=0;
p->parent=0;
p->left=0;
p->right=0;
}
for(i=n+1; i<=m; ++i)
{
min1=min2=0x7fffffff;
//在HT[1,2，.....i-1]中选择parent为0且weight值最小的两个结点，其序号分别为s1、s2
for(j=1; j<i; ++j)
{
if(HT[j].parent==0&&min1>HT[j].weight)
{
min1=HT[j].weight;
s1=j;
}
}
HT[s1].parent=i;
for(j=1; j<i; ++j)
{
if(HT[j].parent==0&&min2>HT[j].weight)
{
min2=HT[j].weight;
s2=j;
}
}
HT[s2].parent=i;
HT[i].left=s1;
HT[i].right=s2;
HT[i].weight=HT[s1].weight+HT[s2].weight;
}
//从叶子节点到根逆向求
for(i=1; i<=n; ++i) //逐个字符求其编码长度
{
int temp=0;
for(c=i,f=HT[i].parent; f; c=f,f=HT[f].parent) //从叶子到根逆向求编码
++temp;
sum=sum+temp*q[i-1]; //编码长度要乘以权值（即该字符的个数）才是总长度
}
int bit=len*8;
printf("%d %d %.1f\n",len*8,sum,bit*1.0/sum);
}
int main()
{
char s[1000];
int i,j,t;
int weight[30];
while(gets(s))
{
t=0;
if(strcmp(s,"END")==0)
break;
memset(weight,0,sizeof(weight));
len=strlen(s);
for(i=0; i<len; ++i)
{
if(s[i]=='_')
++t;
else
++weight[s[i]-'A'];
}
weight[26]=t;
for(i=0,j=0; i<26; ++i) //去掉没有出现过的字母，将权值都压缩存储到原来的空间中
if(weight[i])
weight[j++]=weight[i];
if(t) //如果有下划线的话
weight[j++]=t;
if(j==1) //就一种字符出现的话直接输出---因为赫夫曼编码要找到最小的两个合并，如果只有一个的话会出错
printf("%d %d %.1f\n",len*8,weight[0],len*8*1.0/weight[0]);
else
HuffmanCoding(weight,j);
}
return 0;
}