05-树9 Huffman Codes （30 分）

05-树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] is the frequency of 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

对方提供的这行数据

A 1 B 1 C 1 D 3 E 3 F 6 G 6没用到.

思路示例3的情况

最后一层,编码长度相同(最大)的叶节点 只能是双数.他们的上一层,父节点数量 是 这一层子节点的数量除以2.此时, 父节点的数量和 这一层的叶节点的和,也是双数.也就是说,父节点,要么没有子节点,要么就必须两个子节点

示例4的情况就是混淆的那种,以谁开头的事儿.

另外 编码的字符是双数个，而提交采用的是等长编码。卡仅判断叶结点和度的错误算法

这一项没有通过.我也不理解啥意思,么法改代码.

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 (检查(testList))
            {
                Console.WriteLine("Yes");
            }
            else
            {
                Console.WriteLine("No");
            }

        }

    }
    //public class Node
    //{
    //    public Node Left;
    //    public Node right;

    //}
    private static bool 检查(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,这种循环最大N&M超时.但是去掉的此处代码的话,最大N&M可以通过.可以通过这种方式取巧通过.
            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;

    }
}

原文地址：https://www.cnblogs.com/interim/p/9746203.html