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 Ndistinct 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', '_'}, andf[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 Mstudent 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 NoNo
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<string>
using namespace std;
int TreeHeight;
int cnt=0;
int total=0;
map<char,int>m;
typedef struct TreeNode * Tree;
struct TreeNode{
int data;
char C;
int height;
Tree Left,Right;
};
struct cmp{
bool operator()(Tree T1,Tree T2){
return T1->data>T2->data;
}
};
void Init(priority_queue<Tree,vector<Tree>,cmp>&p,int size){
for(int i=0;i<size;i++){
Tree T=new TreeNode;
T->height=1;
T->Left=T->Right=NULL;
cin>>T->C>>T->data;
m[T->C]=T->data;
p.push(T);
}
while(p.size()>1){
Tree T1=p.top();
p.pop();
Tree T2=p.top();
p.pop();
Tree T3=new TreeNode;
T3->data=T1->data+T2->data;
T3->C=' ';//新树的字符域为空格
T3->Left=T1;
T3->Right=T2;
p.push(T3);
}
return ;
}
void PreOrder(Tree T){
if(!T) return;
else{
if(T->Left==NULL&&T->Right==NULL){
cnt=cnt+T->height*T->data;
total++;
}else{
T->Left->height=T->height+1;
T->Right->height=T->height+1;
PreOrder(T->Left);
PreOrder(T->Right);
}
return;
}
}
void Compare(int size){
int cmpcnt=0;
char A;
string B;
vector<string>v;
bool flag=false;
for(int i=0;i<size;i++){
cin>>A>>B;
int lenv=v.size();
for(int i=0;i<lenv;i++){
if(v[i].find(B)==0) {flag=true;break;}
}
v.push_back(B);
cmpcnt=cmpcnt+m[A]*B.length();
}
if(cmpcnt==cnt){
if(flag==false)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
return ;
}else{
cout<<"No"<<endl;
return ;
}
}
int main(){
priority_queue<Tree,vector<Tree>,cmp>p;
int size;
cin>>size;
Init(p,size);
Tree HuffmanTree=p.top();
p.pop();
HuffmanTree->height=0;
PreOrder(HuffmanTree);
int t;
cin>>t;
while(t--){
Compare(size);
}
return 0;
}