Huffman codes

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 Nlines, 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

这个程序我花了不少时间，改改，找错误，放弃，重写。只能 说细节很多，感觉每个程序 都不是那么简单，需要自己 默默地付出许多。

 #include<iostream>
 #include<vector>
 using namespace std;
 #define maxsize 64
 struct node{
 int weight=-;
 node* l=NULL;
 node* r=NULL;
 };
 using haffmantree=node;
 vector<node> Minheap;
 vector<int> no;
 int size,flag=;
 void Createheap(int N){
 Minheap.resize(N+);
 node n; Minheap[]=n;
 size=;
 }
 void Insert(node n){
   int i=++size;
   for(;Minheap[i/].weight>n.weight;i/=)
   Minheap[i]=Minheap[i/];
   Minheap[i]=n;
 }
 void ReadData(int N){
 for(int i=;i<=N;i++){
 string str; int num;
 cin>>str>>num;
 no.push_back(num);
 node n;
 n.weight=num;
 Insert(n);
 }
 }
 node* Delete(){
 node* n=new node();
 n->l=Minheap[].l;
 n->r=Minheap[].r;
 n->weight=Minheap[].weight;
 node temp=Minheap[size--];
 int parent,child;
 for(parent=;parent*<=size;parent=child){
 child=*parent;
 if(child!=size&&Minheap[child+].weight<Minheap[child].weight)
 ++child;
 if(temp.weight<=Minheap[child].weight) break;
 else
 Minheap[parent]=Minheap[child];
 }
 Minheap[parent]=temp;
 return n;
 }
 haffmantree huffman(int N){
     node T;
 for(int i=;i<N;i++){
 node n;
 n.l=Delete();
 n.r=Delete();
 n.weight=n.l->weight+n.r->weight;
 Insert(n);
 }
 T=*Delete();
 return T;
 }
 int WPL(haffmantree T,int depth)
 {
 if(T.l==NULL&&T.r==NULL) return depth*(T.weight);
 else return WPL(*(T.l),depth+)+WPL(*(T.r),depth+);
 }
 void judge(haffmantree* h,string code){
 for(int i=;i<code.length();i++){
 if(code[i]==''){
 if(h->l==NULL){
 node* nod=new node();
 h->l=nod;
 }
 else if(h->l->weight>)
      flag=;
 h=h->l;
 }
 else if(code[i]==''){
 if(h->r==NULL){
 node* nod=new node();
 h->r=nod;
 }else if(h->r->weight>)
      flag=;
     h=h->r;
 }
 }
 if(h->r==NULL&&h->l==NULL)
 h->weight=;
 else flag=;
 }
 int main(){
 int N; cin>>N;
 Createheap(N);
 ReadData(N);
 haffmantree T=huffman(N);
 int wpl=WPL(T,);
 int M; cin>>M;
 for(int i=;i<=M;i++){
 int len=; haffmantree* h=new node();
 for(int j=;j<N;j++){
 string str,code;
 cin>>str>>code;
 judge(h,code);
 len+=no[j]*code.length();
 }
 if(len!=wpl) flag=;
 if(flag==) cout<<"Yes"<<endl;
 else cout<<"No"<<endl;
 flag=;
 }
     return ;
 }