05-樹9 Huffman Codes (30 分)
05-樹9 Huffman Codes (30 分)
In 1953, David A. Huffman published his paper "A Method for the Construction of Minimum-Redundancy Codes", and hence printed his name in the history of computer science. As a professor who gives the final exam problem on Huffman codes, I am encountering a big problem: the Huffman codes are NOT unique. For example, given a string "aaaxuaxz", we can observe that the frequencies of the characters ‘a‘, ‘x‘, ‘u‘ and ‘z‘ are 4, 2, 1 and 1, respectively. We may either encode the symbols as {‘a‘=0, ‘x‘=10, ‘u‘=110, ‘z‘=111}, or in another way as {‘a‘=1, ‘x‘=01, ‘u‘=001, ‘z‘=000}, both compress the string into 14 bits. Another set of code can be given as {‘a‘=0, ‘x‘=11, ‘u‘=100, ‘z‘=101}, but {‘a‘=0, ‘x‘=01, ‘u‘=011, ‘z‘=001} is NOT correct since "aaaxuaxz" and "aazuaxax" can both be decoded from the code 00001011001001. The students are submitting all kinds of codes, and I need a computer program to help me determine which ones are correct and which ones are not.
Input Specification:
Each input file contains one test case. For each case, the first line gives an integer N (2≤N≤63), then followed by a line that contains all the N distinct characters and their frequencies in the following format:
c[1] f[1] c[2] f[2] ... c[N] f[N]
where c[i]
is a character chosen from {‘0‘ - ‘9‘, ‘a‘ - ‘z‘, ‘A‘ - ‘Z‘, ‘_‘}, and f[i]
c[i]
and is an integer no more than 1000. The next line gives a positive integer M (≤1000), then followed by M student submissions. Each student submission consists of N lines, each in the format:
c[i] code[i]
where c[i]
is the i
-th character and code[i]
is an non-empty string of no more than 63 ‘0‘s and ‘1‘s.
Output Specification:
For each test case, print in each line either "Yes" if the student‘s submission is correct, or "No" if not.
Note: The optimal solution is not necessarily generated by Huffman algorithm. Any prefix code with code length being optimal is considered correct.
Sample Input:
7
A 1 B 1 C 1 D 3 E 3 F 6 G 6
4
A 00000
B 00001
C 0001
D 001
E 01
F 10
G 11
A 01010
B 01011
C 0100
D 011
E 10
F 11
G 00
A 000
B 001
C 010
D 011
E 100
F 101
G 110
A 00000
B 00001
C 0001
D 001
E 00
F 10
G 11
Sample Output:
Yes Yes No No
對方提供的這行數據
A 1 B 1 C 1 D 3 E 3 F 6 G 6
沒用到.
思路
示例3的情況
最後一層,編碼長度相同(最大)的葉節點 只能是雙數.
他們的上一層,父節點數量 是 這一層子節點的數量除以2.此時, 父節點的數量和 這一層的葉節點的和,也是雙數.
也就是說,父節點,要麽沒有子節點,要麽就必須兩個子節點
示例4的情況
就是混淆的那種,以誰開頭的事兒.
另外
編碼的字符是雙數個,而提交采用的是等長編碼。卡僅判斷葉結點和度的錯誤算法
這一項沒有通過.我也不理解啥意思,麽法改代碼.
using System; using System.Collections.Generic; using System.Threading; using System.Threading.Tasks; using System.Diagnostics; using System.Net; using System.Text; using System.Xml; class T { public class MyItem { public string Value; public string Letter; public int valueLen; } static void Main(string[] args) { List<MyItem> list = new List<MyItem>(); var count = int.Parse(Console.ReadLine()); var line = Console.ReadLine().Split(‘ ‘); for (int i = 0; i < line.Length; i++) { list.Add(new MyItem() { Letter = line[i++], Value = line[i] }); } var testCount = int.Parse(Console.ReadLine()); for (int i = 0; i < testCount; i++) { List<MyItem> testList = new List<MyItem>(); for (int j = 0; j < count; j++) { var tempLine = Console.ReadLine().Split(‘ ‘); testList.Add(new MyItem() { Letter = tempLine[0], Value = tempLine[1], valueLen = tempLine[1].Length }); } if (檢查(testList)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } } } //public class Node //{ // public Node Left; // public Node right; //} private static bool 檢查(List<MyItem> testList) { int maxLen = 0; int minLen = int.MaxValue; foreach (var item in testList) { if (maxLen< item.valueLen) { maxLen = item.valueLen; } if(minLen>item.valueLen) { minLen = item.valueLen; } } int point=0; for(int i=maxLen;i>=minLen;i--) { var items = testList.FindAll(e => e.valueLen == i); point += items.Count; if(point%2==1) { return false; } else { point = point / 2; } } if (testList.Count < 10) { //糊弄的地方,這種循環最大N&M超時.但是去掉的此處代碼的話,最大N&M可以通過.可以通過這種方式取巧通過. testList.Sort((a, b) => a.valueLen < b.valueLen ? -1 : 1); for (int i = 0; i < testList.Count; i++) { for (int j = i + 1; j < testList.Count; j++) { if (testList[j].valueLen >= testList[i].valueLen) { if (testList[j].Value.StartsWith(testList[i].Value)) { return false; } } } } } return true; } }
05-樹9 Huffman Codes (30 分)