VHUFFM  Variable Radix Huffman Encoding
Huffman encoding is a method of developing an optimal encoding of the symbols in a source alphabet using symbols from a target alphabet when the frequencies of each of the symbols in the source alphabet are known. Optimal means the average length of an encoded message will be minimized. In this problem you are to determine an encoding of the first N uppercase letters (the source alphabet, through , with frequencies through ) into the first R decimal digits (the target alphabet, through ).
Consider determining the encoding when R=2. Encoding proceeds in several passes. In each pass the two source symbols with the lowest frequencies, say and , are grouped to form a new ``combination letter" whose frequency is the sum of and . If there is a tie for the lowest or second lowest frequency, the letter occurring earlier in the alphabet is selected. After some number of passes only two letters remain to be combined. The letters combined in each pass are assigned one of the symbols from the target alphabet.
The letter with the lower frequency is assigned the code 0, and the other letter is assigned the code 1. (If each letter in a combined group has the same frequency, then 0 is assigned to the one earliest in the alphabet. For the purpose of comparisons, the value of a ``combination letter" is the value of the earliest letter in the combination.) The final code sequence for a source symbol is formed by concatenating the target alphabet symbols assigned as each combination letter using the source symbol is formed.
The target symbols are concatenated in the reverse order that they are assigned so that the first symbol in the final code sequence is the last target symbol assigned to a combination letter.
The two illustrations below demonstrate the process for R=2.
When R is larger than 2, R symbols are grouped in each pass. Since each pass effectively replaces R letters or combination letters by 1 combination letter, and the last pass must combine R letters or combination letters, the source alphabet must contain k*(R1)+R letters, for some integer k.
Since N may not be this large, an appropriate number of fictitious letters with zero frequencies must be included. These fictitious letters are not to be included in the output. In making comparisons, the fictitious letters are later than any of the letters in the alphabet.
Now the basic process of determining the Huffman encoding is the same as for the R=2 case. In each pass, the R letters with the lowest frequencies are grouped, forming a new combination letter with a frequency equal to the sum of the letters included in the group. The letters that were grouped are assigned the target alphabet symbols 0 through R1. 0 is assigned to the letter in the combination with the lowest frequency, 1 to the next lowest frequency, and so forth. If several of the letters in the group have the same frequency, the one earliest in the alphabet is assigned the smaller target symbol, and so forth.
The illustration below demonstrates the process for R=3.
Input
The input will contain one or more data sets, one per line. Each data set consists of an integer value for R (between 2 and 10), an integer value for N (between 2 and 26), and the integer frequencies through , each of which is between 1 and 999.
The end of data for the entire input is the number 0 for R; it is not considered to be a separate data set.
Output
For each data set, display its number (numbering is sequential starting with
1) and the average target symbol length (rounded to two decimal places) on one
line. Then display the N letters of the source alphabet and the corresponding
Huffman codes, one letter and code per line.
Print a blank line after each test case.
The examples below illustrate the
required output format.
Example
Input: 2 5 5 10 20 25 40 2 5 4 2 2 1 1 3 7 20 5 8 5 12 6 9 4 6 10 23 18 25 9 12 0 Output: Set 1; average length 2.10 A: 1100 B: 1101 C: 111 D: 10 E: 0 Set 2; average length 2.20 A: 11 B: 00 C: 01 D: 100 E: 101 Set 3; average length 1.69 A: 1 B: 00 C: 20 D: 01 E: 22 F: 02 G: 21 Set 4; average length 1.32 A: 32 B: 1 C: 0 D: 2 E: 31 F: 33
hide comments
chrisea:
20171113 03:15:39
Be careful with the "fictitious" letters! 

yyyuan:
20171112 16:59:48
:) Last edit: 20171112 17:00:19 

cjycjy:
20161119 12:32:25
Cost me a day. Last edit: 20171111 03:27:47 

Amit Jain:
20130228 16:53:48
I implemented this problem using minheap.


changiz:
20101125 15:50:49
dont forget to use fictitious letters! 

Adrian Kuegel:
20090731 06:39:25
My own program used to generate the judge output uses 4 spaces, I guess \t does also work because SPOJ does allow difference in whitespace characters. 

:D:
20090717 18:56:03
Use single \t instead of 4 spaces before every coded letter in the output. Info given by carlosralv on the forum. 
Added by:  Adrian Kuegel 
Date:  20050727 
Time limit:  1s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All 
Resource:  ACM ICPC World Finals 1995 