浙大数据结构 05-树9 Huffman Codes (30 分)

博主： lnnn
发布时间：2022 年 04 月 03 日
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分类：算法 C++ 学习笔记

` 考察对Huffman编码的理解，程序可能略繁，量力而为。`
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 **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:

```in
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:

```out
Yes
Yes
No
No
```

## 题解

1. 利用堆实现哈夫曼树，计算WPL与测试数据是否相同，若相同且2
2. 判断测试数据是否符合前缀编码（**前缀编码** 是指对字符集进行**编码** 时，要求字符集中任一字符的**编码** 都不是其它字符的**编码** 的**前缀**），如果符合，则Yes，否则No

WPL的计算：
（以下三者等价）

1. 树的带权路径长度等于叶子节点的带权路径长度之和
2. 非叶子结点的权值之和
3. 除根节点以外所有节点的权值之和

```
#include <iostream>
#include <cstring>

using namespace std;

const int N = 70;
int n, m;
int q[N];
char s[N][N];

class MinHeap{
public:
    MinHeap(int capac){
        pRoot = new int[capac + 1];
        size = 0;
        capacity = capac;
        pRoot[0] = -1;
    }
    
    bool isEmpty(){
        return size == 0 ? true : false;
    }
    
    bool isFull(){
        return size >= capacity ? true : false;
    }
    
    void insert(int ele){
        if(isFull())    return;
        
        int i = ++size;
        
        while(ele < pRoot[i/2]){
            pRoot[i] = pRoot[i/2];
            i /= 2;
        }
        
        pRoot[i] = ele;
    }
    
    int pop(){
        if(isEmpty())
            return -1;

int res = pRoot[1];
        int lastEle = pRoot[size--];
        
        int pParent = 1;
        int pChild = pParent*2;
        
        while(pChild <= size){
            if(pChild != size && pRoot[pChild] > pRoot[pChild + 1]) 
                pChild++;
            
            if(lastEle > pRoot[pChild]){
                pRoot[pParent] = pRoot[pChild];
            }else{
                break;
            }
            
            pParent = pChild;
            pChild = pParent*2;
        }
        
        pRoot[pParent] = lastEle;
        
        return res;
    }
    
    void buildHeap(){
        for(int i = size/2; i > 0; i--){
            bmh(i);
        }
    }
    
    void add(int& ele){
        if(isFull())  return;
        
        pRoot[++size] = ele;
    }

void Print(){
        for (int i = 1; i <= size; i++){
            cout << pRoot[i] << " ";
        }
        cout << endl;
    }

int HuffMan(){

int wpl = 0;
        int allSize = size;

for(int i = 0; i < allSize - 1; i++){
            int tmp1 = pop();
            int tmp2 = pop();

wpl += tmp1 + tmp2;

insert(tmp1 + tmp2);
        }
        return wpl;
    }

private:
    void bmh(int pTmpRoot){
        int pParent = pTmpRoot;
        int pChild = pParent*2;
        
        int tmp = pRoot[pParent];
        
        while(pChild <= size){
            if(pChild != size && pRoot[pChild] > pRoot[pChild + 1])
                pChild++;
            
            if(tmp <= pRoot[pChild])    break;
            else    pRoot[pParent] = pRoot[pChild];
            
            pParent = pChild;
            pChild = pParent*2;
        }
        
        pRoot[pParent] = tmp;
    }
    
private:
    int size;
    int capacity;
    int *pRoot;
};

//判断前缀
bool isPre(char* s1, char* s2){
    while(s1 && s2 && *s1 == *s2){
        s1++;
        s2++;
    }
    
    if(!*s1 || !*s2)  return true;
    else    return false;
}

bool isPreCode(char s[][N], int n){
    for(int i = 0; i < n; i++){
        for(int j = i + 1; j < n; j++){
            if(isPre(s[i], s[j])){
                return true;
            }
        }
    }
    return false;
}
int main(){
    char cTmp;
    int inTmp;
    int wpl;
    int num;
    
    cin >> n;
    
    MinHeap* pHeap = new MinHeap(N);
    
    for(int i = 0; i < n; i++){
        cin >> cTmp;
        cin >> inTmp;
        pHeap->add(inTmp);
        q[i] = inTmp;
    }
    
    pHeap->buildHeap();
    wpl = pHeap->HuffMan();
    //cout << wpl << endl;
    
    cin >> m;
    //cout << "m is " << m << endl;
    for(int i = 0; i < m; i++){
        num = 0;
        for(int j = 0; j < n; j++){
            cin >> cTmp;
            cin >> s[j];
            num += strlen(s[j])*q[j];
        }
        //cout << "num is " << num << endl;
        
        if(num == wpl && !isPreCode(s, n)){
            cout << "Yes" << endl;
        }else{
            cout << "No" << endl;
        }
        //cout << num << endl;
    }

return 0;
}
```

最后修改：2022 年 04 月 05 日

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浙大数据结构 05-树9 Huffman Codes (30 分)

lnnn • 2022 年 04 月 03 日

### Input Specification:

```
c[1] f[1] c[2] f[2] ... c[N] f[N]
```

```
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:

### Sample Output:

```out
Yes
Yes
No
No
```

## 题解

WPL的计算：
（以下三者等价）

1. 树的带权路径长度等于叶子节点的带权路径长度之和
2. 非叶子结点的权值之和
3. 除根节点以外所有节点的权值之和

```
#include <iostream>
#include <cstring>

using namespace std;

const int N = 70;
int n, m;
int q[N];
char s[N][N];

void Print(){
        for (int i = 1; i <= size; i++){
            cout << pRoot[i] << " ";
        }
        cout << endl;
    }

int HuffMan(){

int wpl = 0;
        int allSize = size;

for(int i = 0; i < allSize - 1; i++){
            int tmp1 = pop();
            int tmp2 = pop();

wpl += tmp1 + tmp2;

insert(tmp1 + tmp2);
        }
        return wpl;
    }

//判断前缀
bool isPre(char* s1, char* s2){
    while(s1 && s2 && *s1 == *s2){
        s1++;
        s2++;
    }
    
    if(!*s1 || !*s2)  return true;
    else    return false;
}

return 0;
}
```

浙大数据结构 05-树9 Huffman Codes (30 分)

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浙大数据结构 05-树9 Huffman Codes (30 分)

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浙大数据结构 05-树9 Huffman Codes (30 分)

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