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

#include 
    
     #include 
     
      using namespace std;const int minData=-1;struct HuffmanTree{    int f;    HuffmanTree *left=NULL;    HuffmanTree *right=NULL;};struct minHeap{    HuffmanTree *data;    int Size=0;};minHeap* Create(int maxSize){    minHeap* H=new minHeap;    H->data=new HuffmanTree[maxSize+1];    H->data[0].f=minData;    return H;}void Insert(HuffmanTree* x,minHeap* H){    int i=++H->Size;    for(;H->data[i/2].f>x->f;i/=2){        H->data[i]=H->data[i/2];    }    H->data[i]=*x;}HuffmanTree* Delete(minHeap *H){    int parent,child;    HuffmanTree temp=H->data[H->Size--];    HuffmanTree *minItem=new HuffmanTree;    *minItem=H->data[1];    for(parent=1;parent*2<=H->Size;parent=child){        child=parent*2;        if(child+1<=H->Size           && H->data[child].f>H->data[child+1].f){            child++;        }        if(temp.f<=H->data[child].f) break;        else H->data[parent]=H->data[child];    }    H->data[parent]=temp;    return minItem;}HuffmanTree* BuildHuffmanTree(int m,minHeap* H){    HuffmanTree* T;    for(int i=0;i
      
       left=Delete(H);        T->right=Delete(H);        T->f=T->left->f+T->right->f;        Insert(T,H);    }    return Delete(H);}int wpl(HuffmanTree* T,int h){    if(!T->left && !T->right) return h*(T->f);    else return wpl(T->left,h+1) + wpl(T->right,h+1);}void check(int n,int freq[],HuffmanTree* T){    int i,j,is=1,sum=0;    char c;    string s[n];    for(i=0;i
       
        >c>>s[i];        if(s[i].size()>n-1) break;        for(j=0;j
        
         >c>>s[i];    if(is) cout<<"Yes"<
         
          >m; int freq[m]; minHeap* H=Create(m); HuffmanTree* T; for(int i=0;i
          
           >c>>freq[i]; HuffmanTree* node=new HuffmanTree; node->f=freq[i]; Insert(node,H); } T=BuildHuffmanTree(m,H); cin>>n; for(int i=0;i