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 ;
}