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。
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 }
