数据结构 05-树9 Huffman Codes (30 分)

Posted 2021-05-20 keiiha

 

 

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\', \'_\'}, 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 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
No
No

 

 

参考文章 https://zhuanlan.zhihu.com/p/121684742

借助小顶堆建立huffmanTree ,计算WPL

然后用学生输入的huffmanCode 建立huffmanTree,  

　　　　建树过程中,累加 (对应字符的频率*哈夫曼码长度) 得到学生的树的带权路径长度wplOfTest 

　　　　如果结点存放的位置 非叶结点 或 已经存放了其他结点信息 记为错误huffmanCode, 跳过其他字符的哈夫曼码的处理

　　　　如果学生的huffmanCode可以生成huffmanTree, 则计算学生的带权路径长度 即wplOfTest与WPL是否相等, 不相等也是错误的huffmanCode

 

 

//小顶堆的建立 查找
#include <iostream>
#include <vector>
using namespace std;
class tnode{
public:
    string c{""};
    int f{0};
    tnode* left{nullptr};
    tnode* right{nullptr};
    tnode()=default;
    tnode(string c_,int f_):c{c_},f{f_}{};
};
class minHeap{//小根堆
public:
    vector<tnode*> heap;
    minHeap(){
    }
    int getSize(){
        return heap.size();
    }
    void build(vector<tnode*> list,int n){
        for(int i=0;i<n;i++){
            insertNode(list[i]);
        }
    }
    void insertNode(tnode* newnode){
        heap.push_back(newnode);
        adjustFromBack();
    }
    void insertNode(string c,int f){
        tnode* newnode =new tnode{c,f};
        heap.push_back(newnode);
        adjustFromBack();
    }
    tnode* popMinNode(){
        tnode* temp{nullptr};
        if(heap.size()){
            temp = heap.front();
            swap(heap.front(), heap.back());
            heap.pop_back();
            adjustFromFront();
        }
        return temp;
    }
    void adjustFromBack(){
        for(int i=getSize()-1;i>=0&&getSize()>1;i--){
            if(heap[i]->f<heap[(i-1)/2]->f){
                swap(heap[i], heap[(i-1)/2]);
            }
        }
    }
    void adjustFromFront(){
        for(int i=1;i<getSize()&&getSize()>1;i++){
            if(heap[i]->f<heap[(i-1)/2]->f){
                swap(heap[i], heap[(i-1)/2]);
            }
        }
    }
};

class huffmanTree{
public:
    huffmanTree()=default;
    tnode* root;
    void create(minHeap &minheap){
        tnode* n1,*n2,*newnode;
        int size=minheap.getSize();
        for(int i=0;i<size-1;i++){
            n1 = minheap.popMinNode();
            n2 = minheap.popMinNode();
            newnode = new tnode{"",n1->f+n2->f};
            newnode->left = n1;
            newnode->right = n2;
            minheap.insertNode(newnode);
        }
        newnode = minheap.heap.front();
        root=newnode;
    }
    int WPL(tnode* p,int depth){
        if(!p->left&&!p->right) return p->f*depth;
        return WPL(p->left, depth+1) + WPL(p->right, depth+1);
    }
    int getWPL(){
        return WPL(root, 0);
    }
};
bool judge(int wpl,vector<tnode*>list){
    bool flag=true;
    int wplOfTest{0};
    string c,code;
    tnode* head=new tnode;
    tnode* tail=head;
    tnode* temp;
    for(int i=0;i<list.size();i++){
        cin >> c >> code;
        if(!flag)continue;;
        wplOfTest+=(list[i]->f*code.size());
        for(auto i=code.begin();i!=code.end();i++){
            if(*i==\'0\'){
                if(!tail->left){
                    temp = new tnode;
                    tail->left=temp;
                }
                tail=tail->left;
            }else if(*i==\'1\'){
                if(!tail->right){
                    temp=new tnode;
                    tail->right=temp;
                }
                tail=tail->right;
            }
            if(tail->f){
                flag=false;
                break;
            }
        }
        if(tail->left||tail->right){
            flag=false;
            continue;
        }
        tail->f=list[i]->f;
        tail->c=list[i]->c;
        tail=head;
    }
    if(wplOfTest!=wpl)return false;
    else return flag;
}

int main(){
    int n,m,WPL;
    string c;
    int f;
    cin >> n;
    vector<tnode*> alpha;
    for(int i=0;i<n;i++){
        cin >> c >> f;
        tnode *newnode=new tnode{c,f};
        alpha.push_back(newnode);
    }
    minHeap minheap;
    minheap.build(alpha,n);
    huffmanTree ht;
    ht.create(minheap);
    WPL=ht.getWPL();
    cin >> m;
    for(int i=0;i<m;i++){
        if(judge(WPL,alpha)){
            cout << "Yes"<<endl;
        }else{
            cout << "No"<<endl;
        }
    }
    
    return 0;
}