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

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

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。

  1 #include<iostream>
  2 #include<vector>
  3 using namespace std;
  4 #define maxsize 64
  5 struct node{
  6 int weight=-1;
  7 node* l=NULL;
  8 node* r=NULL;
  9 };
 10 using haffmantree=node;
 11 vector<node> Minheap;
 12 vector<int> no;
 13 int size,flag=1;
 14 void Createheap(int N){
 15 Minheap.resize(N+1);
 16 node n; Minheap[0]=n;
 17 size=0;
 18 }
 19 void Insert(node n){
 20   int i=++size;
 21   for(;Minheap[i/2].weight>n.weight;i/=2)
 22   Minheap[i]=Minheap[i/2];
 23   Minheap[i]=n;
 24 }
 25 void ReadData(int N){
 26 for(int i=1;i<=N;i++){
 27 string str; int num;
 28 cin>>str>>num;
 29 no.push_back(num);
 30 node n;
 31 n.weight=num;
 32 Insert(n);
 33 }
 34 }
 35 node* Delete(){
 36 node* n=new node();
 37 n->l=Minheap[1].l;
 38 n->r=Minheap[1].r;
 39 n->weight=Minheap[1].weight;
 40 node temp=Minheap[size--];
 41 int parent,child;
 42 for(parent=1;parent*2<=size;parent=child){
 43 child=2*parent;
 44 if(child!=size&&Minheap[child+1].weight<Minheap[child].weight)
 45 ++child;
 46 if(temp.weight<=Minheap[child].weight) break;
 47 else 
 48 Minheap[parent]=Minheap[child];
 49 }
 50 Minheap[parent]=temp;
 51 return n;
 52 }
 53 haffmantree huffman(int N){
 54     node T;
 55 for(int i=1;i<N;i++){
 56 node n;
 57 n.l=Delete();
 58 n.r=Delete();
 59 n.weight=n.l->weight+n.r->weight;
 60 Insert(n); 
 61 } 
 62 T=*Delete();
 63 return T;
 64 }
 65 int WPL(haffmantree T,int depth)
 66 { 
 67 if(T.l==NULL&&T.r==NULL) return depth*(T.weight);
 68 else return WPL(*(T.l),depth+1)+WPL(*(T.r),depth+1);
 69 }
 70 void judge(haffmantree* h,string code){
 71 for(int i=0;i<code.length();i++){
 72 if(code[i]=='0'){
 73 if(h->l==NULL){
 74 node* nod=new node();
 75 h->l=nod;
 76 }
 77 else if(h->l->weight>0)
 78      flag=0;
 79 h=h->l;
 80 }
 81 else if(code[i]=='1'){
 82 if(h->r==NULL){
 83 node* nod=new node();
 84 h->r=nod;
 85 }else if(h->r->weight>0)
 86      flag=0;
 87     h=h->r;
 88 }
 89 }
 90 if(h->r==NULL&&h->l==NULL)
 91 h->weight=1;
 92 else flag=0;
 93 }
 94 int main(){
 95 int N; cin>>N;
 96 Createheap(N);
 97 ReadData(N);
 98 haffmantree T=huffman(N);
 99 int wpl=WPL(T,0);
100 int M; cin>>M;
101 for(int i=1;i<=M;i++){
102 int len=0; haffmantree* h=new node();
103 for(int j=0;j<N;j++){
104 string str,code;
105 cin>>str>>code;
106 judge(h,code);
107 len+=no[j]*code.length();
108 }
109 if(len!=wpl) flag=0;
110 if(flag==1) cout<<"Yes"<<endl;
111 else cout<<"No"<<endl;
112 flag=1;
113 }
114     return 0; 
115 }

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