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 $N$ ($2\le N\le 63$), 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 $M$ ($\le 1000$), 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

主要思路：

1、通过小根堆构建Huffman树

2、通过Huffman树计算最小编码长度

3、判断输入的编码是否符合要求（1、编码长度与Huffman编码长度相同 2、前缀编码）

#include
#include 
using namespace std;
#define MaxNum 64

struct TreeNode//树的结点
{
    int Weight=0;
    TreeNode *Left=nullptr;
    TreeNode *Right=nullptr;
};

struct HeapNode//堆
{
    TreeNode Data[MaxNum];
    int Size=0;
};

HeapNode *CreateHeap(int N)//创建一个新的小根堆
{
    HeapNode *H=new(HeapNode);
    H->Data[0].Weight=-1;
    return H;
}

TreeNode *DeleteMin(HeapNode *H)//从堆中删除一个结点
{
    int Parent=0,Child=0;
    TreeNode temp;
    TreeNode *MinItem=new(TreeNode);
    *MinItem=H->Data[1];
    
    temp=(H->Data[(H->Size)--]);
    
    for (Parent = 1; Parent*2 <= H->Size ; Parent=Child)//寻找删除结点前堆中最后一个结点在新堆中的插入位置
    {
        Child=Parent*2;
        if ((Child!=H->Size)&&((H->Data[Child].Weight)>(H->Data[Child+1].Weight)))
        {
            Child++;
        }
        if ((temp.Weight)<=(H->Data[Child].Weight))
        {
            break;
        }else
        {
            H->Data[Parent]=H->Data[Child];
        }
        
    }
    H->Data[Parent]=temp;
    return MinItem;
    
}

void Insert(HeapNode *H,TreeNode *item)//插入新结点到堆中
{
    int i=0;
    i=++(H->Size);
    for (;H->Data[i/2].Weight>item->Weight; i/=2)
    {
        H->Data[i]=H->Data[i/2];
        
    }
    H->Data[i]=*item;
    
}

HeapNode *ReadData(int N,HeapNode *H,int A[])//读取各个节点的权值输入数据
{
    char s='\0';
    int value=0;
    for (int i=0; i>s;
        cin>>value;
        A[i]=value;
        TreeNode *T=new(TreeNode);
        T->Weight=value;
        Insert(H, T);
    }
    return H;
}

TreeNode *Huffman(HeapNode *H)//构建Huffman树
{
    TreeNode *T=nullptr;
    int num=H->Size;
    for (int i=0; iLeft=DeleteMin(H);
        
        T->Right=DeleteMin(H);
        
        T->Weight=T->Left->Weight+T->Right->Weight;
        Insert(H, T);
        
    }
    T=DeleteMin(H);
    return T;
}

int WPL(TreeNode *T,int Depth)//计算Huffman树的编码长度
{
    if ((T->Left==nullptr)&&(T->Right==nullptr))
    {
        return Depth*(T->Weight);
    }else
    {
        return (WPL(T->Left,Depth+1)+WPL(T->Right,Depth+1));
    }
}
struct JNode
{
    int Flag=0;
    JNode *Left=nullptr;
    JNode *Right=nullptr;
    
};
bool Judge(string S,JNode *J)//判断该次编码能否符合前缀编码的要求
{
    int i=0;
    for (; iLeft==nullptr)
            {
                JNode *J_1=new(JNode);
                J->Left=J_1;
            }else
            {
                if (J->Left->Flag==1)
                {
                    return false;
                }
            }
            J=J->Left;
        }else
        {
            if (J->Right==nullptr)
            {
                JNode *J_1=new(JNode);
                J->Right=J_1;
            }else
            {
                if (J->Right->Flag==1)
                {
                    return false;
                }
            }
            J=J->Right;
        }
    }
    J->Flag=1;
    if (J->Left==nullptr&&J->Right==nullptr)
    {
        return true;
    }else
    {
        return false;
    }
}

int main(int argc, char const *argv[])
{
    int N=0,n=0;
    cin>>N;
    HeapNode *H=CreateHeap(N);
    int Value[MaxNum]={};
    H=ReadData(N,H,Value);
    
    TreeNode *T=Huffman(H);
    
    int CodeLen=WPL(T,0);
    
    cin>>n;
    string temp="\0";
    char c='\0';
    bool result=false;
    
    
    for (int i=0; i>c>>temp;
            count+=temp.length()*Value[k];
            if (!flag)
            {
                result=Judge(temp,J);
                if (!result)
                {
                    flag=1;
                }
            }
            
            
        }
        delete J;
        if (result&&(count==CodeLen))//前缀编码且编码长度之和与Huffman编码相同
        {
            cout<<"Yes"<