Huffman Codes

Huffman Codes

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), 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 (≤), 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

 

解题思路

　　要判断编码是否为最优编码，需要对编码进行两个方面的检验：对每一组编码来判断WPL是否为最小的，以及是否为前缀码。

　　下面给出第一种思路，需要构建Huffman Tree的方法，同时也是用堆来实现的。

　　树节点的定义如下：

struct Data {
    char letter;
    int freq;
};

struct TNode {
    Data data;
    TNode *left, *right;
};

　　首先我们要根据给出的字符频率来构造出一颗对应的Huffman Tree，同时计算出WPL。

　　又因为我们是用堆来实现的，所以要构造一颗Huffman Tree我们先要把给定的频率来构建一个最小堆。

　　这里我们需要对堆进行定义，同时还要定义堆的相关操作。这里就直接上代码，不解释了。

struct Heap {
    TNode *H;    // 堆的每一个元素的数据类型是树节点TNode 
    int size;
    int capacity;
};

Heap *createMinHeap(int n) {
    Heap *minHeap = new Heap;
    minHeap->size = 0;
    minHeap->capacity = n;
    minHeap->H = new TNode[n + 1];
    minHeap->H[0].data.freq = -1;
    
    for (int i = 0; i < minHeap->capacity; i++) {
        TNode *tmp = new TNode;
        tmp->left = tmp->right = NULL;
        getchar();
        scanf("%c %d", &tmp->data.letter, &tmp->data.freq);
        insertHeap(minHeap, tmp);
    }
    
    return minHeap;
}

void insertHeap(Heap *minHeap, TNode *treeNode) {
    int pos = ++minHeap->size;
    for ( ; treeNode->data.freq < minHeap->H[pos / 2].data.freq; pos /= 2) {
        minHeap->H[pos] = minHeap->H[pos / 2];
    }
    minHeap->H[pos] = *treeNode;
}

TNode *deleteMin(Heap *minHeap) {
    TNode *minTreeNode = new TNode;
    *minTreeNode = minHeap->H[1];
    TNode tmp = minHeap->H[minHeap->size--];
    
    int parent = 1, child;
    for ( ; parent * 2 <= minHeap->size; parent = child) {
        child = parent * 2;
        if (child != minHeap->size && minHeap->H[child].data.freq > minHeap->H[child + 1].data.freq) child++;
        if (tmp.data.freq < minHeap->H[child].data.freq) break;
        else minHeap->H[parent] = minHeap->H[child];
    }
    minHeap->H[parent] = tmp;
    
    return minTreeNode;
}

　　构建好最小堆后，下一步我们需要通过这个堆来构造出对应的Huffman Tree。

　　就是每次从堆中弹出最小频率的那两个节点，然后把这两个节点分别插在新节点的左右边，作为左右孩子。再把新节点压入堆中。如此循环n-1次后（其中n代表节点的个数），堆中就只剩下一个元素，那个元素就是Huffman Tree的根节点，我们直接返回即可。

　　按照上面构造Huffman Tree的思路，相应的代码如下：

TNode *createHuffmanTree(Heap *minHeap) {
    int n = minHeap->size - 1;
    while (n--) {
        TNode *tmp = new TNode;
        tmp->left = deleteMin(minHeap);
        tmp->right = deleteMin(minHeap);
        tmp->data.freq = tmp->left->data.freq + tmp->right->data.freq;
        
        insertHeap(minHeap, tmp);
    }
    
    return deleteMin(minHeap);
}

　　然后，我们要根据这颗Huffman Tree来计算WPL。我们用递归来实现计算WPL。

　　如果这个节点是叶子节点，那么就用当前深度乘以对应的频率，然后返回。如果不是叶子节点，就递归来计算左右子树的WPL，相加后返回。

int WPL(TNode *T, int depth) {
    if (T->left == NULL && T->right == NULL) return depth * T->data.freq;   // 叶子节点直接返回结果 
    else return WPL(T->left, depth + 1) + WPL(T->right, depth + 1);         // 不是叶子节点，计算左右子树的WPL并返回，同时由于左右子树的深度加深一层，记得depth+1 
}

　　好了，折腾了这么久，终于计算出给定频率的WPL了。

　　接下来我们先对每一组编码来检测其WPL是不是最小的，也就是每组编码的WPL是否与给定频率的WPL相等。

　　计算方法很简单，每一组编码的WPL计算公式为：

　　再判断codeLen是否与上面求出的给定频率的WPL相等，如果不相等，就说明这个编码不是最优编码，就不需要再判断是否为前缀码了。如果相等再去判断是否为前缀码。

　　这里还有个陷阱。首先我们要知道，一个最优编码的长度是不会超过n-1的。所以如果某个编码的长度大于n-1也说明该编码不是最优编码。

　　这里同时给出计算编码长度和判断是否是前缀码的函数：

bool check(TNode *huffmanTree, int n) {
    int wpl = WPL(huffmanTree, 0);  // 计算给定频率构成的Huuffman Tree的WPL 

    std::string code[n];            // 存放每一个字符的编码 
    int codeLen = 0;
    bool ret = true;                // 用来标记该组编码是否为最优编码 
    
    for (int i = 0; i < n; i++) {
        char letter;
        getchar();
        scanf("%c", &letter);
        getchar();
        std::cin >> code[i];
        
        if (ret) {                  // 如果已经知道该组编码不是最优编码就不需要再计算编码长度了，但仍要继续输入 
            if (code[i].size() > n - 1) ret = false;                    // 如果某个字符的编码长度大于n-1，说明该组编码不是最优编码 
            codeLen += code[i].size() * findFreq(huffmanTree, letter);  // 计算编码长度 
        }
    }
    
    if (ret && codeLen == wpl) {        // 如果ret == true并且编码长度与WPL相同，接着判断该组编码是否为前缀码 
        TNode *T = new TNode;           // 为这组编码构造一颗Huffman Tree，初始化Huffman Tree的根节点 
        T->data.freq = 0;
        T->left = T->right = NULL;
            
        for (int i = 0; i < n; i++) {   // 有n个节点，需要判断n次 
            TNode *pre = T;             // 每次判断一个字符都从根节点开始 
            
            for (std::string::iterator it = code[i].begin(); it != code[i].end(); it++) {   // 对该字符的每一个编码进行判断 
                if (*it == '0') {                   // 如果编码是0 
                    if (pre->left == NULL) {        // 如果当前节点的左子树为空 
                        TNode *tmp = new TNode;     // 就为当前节点生成一颗左子树 
                        tmp->data.freq = 0;         // 该节点的频率标记为0，表示该节点还没有字符占用 
                        tmp->left = tmp->right = NULL;
                        pre->left = tmp;
                    }
                    pre = pre->left;                // pre指针指向左子树 
                }
                else {                              // 如果编码是1 
                    if (pre->right == NULL) {       // 如果当前节点的右子树为空
                        TNode *tmp = new TNode;     // 就为当前节点生成一颗右子树
                        tmp->data.freq = 0;         // 该节点的频率标记为0，表示该节点还没有字符占用
                        tmp->left = tmp->right = NULL;
                        pre->right = tmp;
                    }
                    pre = pre->right;                // pre指针指向左子树
                }
            }
            
            // 读完了字符的编码后，pre指针就指向这个字符应该占用的位置
            // 这时需要判断pre指向的这个节点是否为叶子节点，并且该节点有没有被其他字符占用 
            if (pre->left == NULL && pre->right == NULL && pre->data.freq == 0) {
                pre->data.freq = 1;                  //  如果是叶子节点并且没有被占用，该字符就占用了这个节点，并把这个节点的频率标记为1 
            }
            else {                                   // 否则，如果这些条件中有一个不满足 
                ret = false;                         // 就说明该组字符不满足前缀码的要求，ret赋值为false 
                break;                               // 后面的字符不需要判断了，直接退出退出判断前缀码的循环 
            }
        }
    }
    else {          // 如果ret == false并且编码长度不等于WPL，就说明该组编码不是最优编码 
        ret = false;
    }
    
    return ret;
}

　　这里是通过构造一颗Huffman Tree来判断该组编码是否符合前缀码。

　　判断的过程如下：

　　有一个指向Huffman Tree根节点的指针。

• 如果编码是'0'，先判断当前节点的左子树是否存在，如果不存在先生成左子树，再让指针移到左子树的节点。如果存在那么直接让指针移到左子树的节点即可。
• 如果编码是'1'，先判断当前节点的右子树是否存在，如果不存在先生成右子树，再让指针移到右子树的节点。如果存在那么直接让指针移到右子树的节点即可。

　　读完该字符的编码后，那么此时字符应该放入这个指针指向的节点。这个节点要满足两个条件才可以放入：

• 该节点的左右孩子都为空，也就是该节点为叶子节点。
• 该节点每有被标记过，也就是说该节点没有存放其他的字符。

　　如果有一个条件不满足，就说明该组编码不是前缀码。

　　最后，给出这种方法的完整AC代码，代码量有点多。

#include <cstdio>
#include <iostream>
#include <string>

struct Data {
    char letter;
    int freq;
};

struct TNode {
    Data data;
    TNode *left, *right;
};

struct Heap {
    TNode *H;
    int size;
    int capacity;
};

Heap *createMinHeap(int n);
void insertHeap(Heap *minHeap, TNode *treeNode);
TNode *deleteHeap(Heap *minHeap);
TNode *createHuffmanTree(Heap *minHeap);
bool check(TNode *huffmanTree, int n);
int WPL(TNode *T, int depth);
int findFreq(TNode *huffmanTree, char letter);

int main() {
    int n;
    scanf("%d", &n);
    
    Heap *minHeap = createMinHeap(n);
    TNode *huffmanTree = createHuffmanTree(minHeap);
    
    int m;
    scanf("%d", &m);
    for (int i = 0; i < m; i++) {
        bool ret = check(huffmanTree, n);
        printf("%s\n", ret ? "Yes" : "No");
    }
    
    return 0;
}

Heap *createMinHeap(int n) {
    Heap *minHeap = new Heap;
    minHeap->size = 0;
    minHeap->capacity = n;
    minHeap->H = new TNode[n + 1];
    minHeap->H[0].data.freq = -1;
    
    for (int i = 0; i < minHeap->capacity; i++) {
        TNode *tmp = new TNode;
        tmp->left = tmp->right = NULL;
        getchar();
        scanf("%c %d", &tmp->data.letter, &tmp->data.freq);
        insertHeap(minHeap, tmp);
    }
    
    return minHeap;
}

void insertHeap(Heap *minHeap, TNode *treeNode) {
    int pos = ++minHeap->size;
    for ( ; treeNode->data.freq < minHeap->H[pos / 2].data.freq; pos /= 2) {
        minHeap->H[pos] = minHeap->H[pos / 2];
    }
    minHeap->H[pos] = *treeNode;
}

TNode *deleteHeap(Heap *minHeap) {
    TNode *minTreeNode = new TNode;
    *minTreeNode = minHeap->H[1];
    TNode tmp = minHeap->H[minHeap->size--];
    
    int parent = 1, child;
    for ( ; parent * 2 <= minHeap->size; parent = child) {
        child = parent * 2;
        if (child != minHeap->size && minHeap->H[child].data.freq > minHeap->H[child + 1].data.freq) child++;
        if (tmp.data.freq < minHeap->H[child].data.freq) break;
        else minHeap->H[parent] = minHeap->H[child];
    }
    minHeap->H[parent] = tmp;
    
    return minTreeNode;
}

TNode *createHuffmanTree(Heap *minHeap) {
    int n = minHeap->size - 1;
    while (n--) {
        TNode *tmp = new TNode;
        tmp->left = deleteHeap(minHeap);
        tmp->right = deleteHeap(minHeap);
        tmp->data.freq = tmp->left->data.freq + tmp->right->data.freq;
        
        insertHeap(minHeap, tmp);
    }
    
    return deleteHeap(minHeap);
}

bool check(TNode *huffmanTree, int n) {
    int wpl = WPL(huffmanTree, 0);

    std::string code[n];
    int codeLen = 0;
    bool ret = true;
    
    for (int i = 0; i < n; i++) {
        char letter;
        getchar();
        scanf("%c", &letter);
        getchar();
        std::cin >> code[i];
        
        if (ret) {
            if (code[i].size() > n - 1) ret = false;
            codeLen += code[i].size() * findFreq(huffmanTree, letter);
        }
    }
    
    if (ret && codeLen == wpl) {
        TNode *T = new TNode;
        T->data.freq = 0;
        T->left = T->right = NULL;
            
        for (int i = 0; i < n; i++) {
            TNode *pre = T;
            
            for (std::string::iterator it = code[i].begin(); it != code[i].end(); it++) {
                if (*it == '0') {
                    if (pre->left == NULL) {
                        TNode *tmp = new TNode;
                        tmp->data.freq = 0;
                        tmp->left = tmp->right = NULL;
                        pre->left = tmp;
                    }
                    pre = pre->left;
                }
                else {
                    if (pre->right == NULL) {
                        TNode *tmp = new TNode;
                        tmp->data.freq = 0;
                        tmp->left = tmp->right = NULL;
                        pre->right = tmp;
                    }
                    pre = pre->right;
                }
            }
            
            if (pre->left == NULL && pre->right == NULL && pre->data.freq == 0) {
                pre->data.freq = 1;
            }
            else {
                ret = false;
                break;
            }
        }
    }
    else {
        ret = false;
    }
    
    return ret;
}

int WPL(TNode *T, int depth) {
    if (T->left == NULL && T->right == NULL) return depth * T->data.freq;
    else return WPL(T->left, depth + 1) + WPL(T->right, depth + 1);
}

int findFreq(TNode *huffmanTree, char letter) {
    int ret = 0;
    if (huffmanTree) {
        if (huffmanTree->left == NULL && huffmanTree->right == NULL && huffmanTree->data.letter == letter) ret = huffmanTree->data.freq;
        if (ret == 0) ret = findFreq(huffmanTree->left, letter);
        if (ret == 0) ret = findFreq(huffmanTree->right, letter);
    }
    
    return ret;
}

　　然后，我们来对判定前缀码的代码进行改进，下面给出判定前缀码的另外一种思路，这个方法不需要构造Huffman Tree。

　　首先，假设现在有两个编码，如果这两个编码不满足前缀码的话，比如"110"和"1101"，那么其中一个编码会与另外一个编码前的m个位置的相同（其中m是指这两个编码长度中最小的那个长度）。也就是说"110"，与"1101"的前3个位置的"110"相同，就说明"110"和"1101"不满足前缀码。

　　我们需要对同组编码的每两个字符进行比较，需要比较的次数为 C(n, 2) = n * (n - 1) / 2 。

　　check函数改进的代码如下：

bool check(TNode *huffmanTree, int n) {
    int wpl = WPL(huffmanTree, 0);

    std::string code[n];
    int codeLen = 0;
    bool ret = true;
    
    for (int i = 0; i < n; i++) {
        char letter;
        getchar();
        scanf("%c", &letter);
        getchar();
        std::cin >> code[i];
        
        if (ret) {
            if (code[i].size() > n - 1) ret = false;
            codeLen += code[i].size() * findFreq(huffmanTree, letter);
        }
    }
    
    if (ret && codeLen == wpl) {        // 一样的，如果ret == true并且编码长度与WPL相同，才判断该组编码是否为前缀码
        for (int i = 0; i < n; i++) {   // 每个字符都跟它之后的字符进行判断是否满足前缀码的要求 
            for (int j = i + 1; j < n; j++) {
                // 判断某个编码是否与另外一个编码前m个位置的相同，详细请看图片 
                if (code[i].substr(0, code[j].size()) == code[j].substr(0, code[i].size())) {
                    ret = false;        // 只要有一对编码的前缀相同，就说明这组的编码不满足前缀码 
                    break;              // 后面的字符不需要判断了，直接退出退出判断前缀码的循环 
                }
            }
            if (ret == false) break;
        }
    }
    else {
        ret = false;
    }
    
    return ret;
}

 　　code[i].substr(0, code[j].size()) == code[j].substr(0, code[i].size()) ，这么做始终能够保证取到两个编码中，长度最小那个编码的全部，以及另外一个编码的前面同样长度的部分，来进行判断是否满足前缀码。

　　这个方法也可以通过，下面给出完整的AC代码，其中改动的部分就是check部分，其他的不变。

#include <cstdio>
#include <iostream>
#include <string>

struct Data {
    char letter;
    int freq;
};

struct TNode {
    Data data;
    TNode *left, *right;
};

struct Heap {
    TNode *H;
    int size;
    int capacity;
};

Heap *createMinHeap(int n);
void insertHeap(Heap *minHeap, TNode *treeNode);
TNode *deleteHeap(Heap *minHeap);
TNode *createHuffmanTree(Heap *minHeap);
bool check(TNode *huffmanTree, int n);
int WPL(TNode *T, int depth);
int findFreq(TNode *huffmanTree, char letter);

int main() {
    int n;
    scanf("%d", &n);
    
    Heap *minHeap = createMinHeap(n);
    TNode *huffmanTree = createHuffmanTree(minHeap);
    
    int m;
    scanf("%d", &m);
    for (int i = 0; i < m; i++) {
        bool ret = check(huffmanTree, n);
        printf("%s\n", ret ? "Yes" : "No");
    }
    
    return 0;
}

Heap *createMinHeap(int n) {
    Heap *minHeap = new Heap;
    minHeap->size = 0;
    minHeap->capacity = n;
    minHeap->H = new TNode[n + 1];
    minHeap->H[0].data.freq = -1;
    
    for (int i = 0; i < minHeap->capacity; i++) {
        TNode *tmp = new TNode;
        tmp->left = tmp->right = NULL;
        getchar();
        scanf("%c %d", &tmp->data.letter, &tmp->data.freq);
        insertHeap(minHeap, tmp);
    }
    
    return minHeap;
}

void insertHeap(Heap *minHeap, TNode *treeNode) {
    int pos = ++minHeap->size;
    for ( ; treeNode->data.freq < minHeap->H[pos / 2].data.freq; pos /= 2) {
        minHeap->H[pos] = minHeap->H[pos / 2];
    }
    minHeap->H[pos] = *treeNode;
}

TNode *deleteHeap(Heap *minHeap) {
    TNode *minTreeNode = new TNode;
    *minTreeNode = minHeap->H[1];
    TNode tmp = minHeap->H[minHeap->size--];
    
    int parent = 1, child;
    for ( ; parent * 2 <= minHeap->size; parent = child) {
        child = parent * 2;
        if (child != minHeap->size && minHeap->H[child].data.freq > minHeap->H[child + 1].data.freq) child++;
        if (tmp.data.freq < minHeap->H[child].data.freq) break;
        else minHeap->H[parent] = minHeap->H[child];
    }
    minHeap->H[parent] = tmp;
    
    return minTreeNode;
}

TNode *createHuffmanTree(Heap *minHeap) {
    int n = minHeap->size - 1;
    while (n--) {
        TNode *tmp = new TNode;
        tmp->left = deleteHeap(minHeap);
        tmp->right = deleteHeap(minHeap);
        tmp->data.freq = tmp->left->data.freq + tmp->right->data.freq;
        
        insertHeap(minHeap, tmp);
    }
    
    return deleteHeap(minHeap);
}

bool check(TNode *huffmanTree, int n) {
    int wpl = WPL(huffmanTree, 0);

    std::string code[n];
    int codeLen = 0;
    bool ret = true;
    
    for (int i = 0; i < n; i++) {
        char letter;
        getchar();
        scanf("%c", &letter);
        getchar();
        std::cin >> code[i];
        
        if (ret) {
            if (code[i].size() > n - 1) ret = false;
            codeLen += code[i].size() * findFreq(huffmanTree, letter);
        }
    }

    if (ret && codeLen == wpl) {
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (code[i].substr(0, code[j].size()) == code[j].substr(0, code[i].size())) {
                    ret = false;
                    break;
                }
            }
            if (ret == false) break;
        }
    }
    else {
        ret = false;
    }
    
    return ret;
}

int WPL(TNode *T, int depth) {
    if (T->left == NULL && T->right == NULL) return depth * T->data.freq;
    else return WPL(T->left, depth + 1) + WPL(T->right, depth + 1);
}

int findFreq(TNode *huffmanTree, char letter) {
    int ret = 0;
    if (huffmanTree) {
        if (huffmanTree->left == NULL && huffmanTree->right == NULL && huffmanTree->data.letter == letter) ret = huffmanTree->data.freq;
        if (ret == 0) ret = findFreq(huffmanTree->left, letter);
        if (ret == 0) ret = findFreq(huffmanTree->right, letter);
    }
    
    return ret;
}

　　还可以再改进！我们不用堆，而改用优先队列，同时不需要构造任何的Huffman Tree，甚至不需要定义树节点，也可以计算出给定频率的WPL！并且代码长度也缩短许多。

　　这里主要说明如何通过不构造Huffman Tree来计算给定频率的WPL。其实计算WPL不一定要用深度乘以频率再求和来得到。另外一种方法是把Huffman Tree中度为2的节点存放的频率都相加起来，最后得到的结果也是WPL。这是因为叶子节点被重复计算，和用深度乘以频率的原理基本一样。

　　就拿题目给的测试样例来举例：

　　我们往优先队列里面压的就是每个字符对应的频率，而不是树节点。代码实现的过程是：我们要有一个变量来累加如上图度为2节点存放频率。每次从优先队列里弹出两个频率，这两个频率是优先队列中所包含频率里面最小的那两个，然后把这两个频率相加，相加的结果其实就对应上图度为2节点存放的频率，也就是红色的数字。然后把相加的结果累加到那个变量，同时把相加的结果压入优先队列中。其实这个累加的过程就是累加上图红色的那些数字。一直重复，直到优先队列为空，那么那个变量最后累加的结果就是我们要计算的WPL。

　　AC代码如下：

#include <cstdio>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <map>
using namespace std;

void readLetterFreq(map<char, int> &letterFreq, priority_queue< int, vector<int>, greater<int> > &pq, int n);
void checkOptimalCode(map<char, int> &letterFreq, priority_queue< int, vector<int>, greater<int> > &pq, int n);
int getWPL(priority_queue< int, vector<int>, greater<int> > &pq);

int main() {
    map<char, int> letterFreq;        // 用map来存储字符和对应的频率，字符映射为对应的频率 
    priority_queue< int, vector<int>, greater<int> > pq;
    
    int n;
    cin >> n;
    readLetterFreq(letterFreq, pq, n);
    checkOptimalCode(letterFreq, pq, n);
    
    return 0;
}

void readLetterFreq(map<char, int> &letterFreq, priority_queue< int, vector<int>, greater<int> > &pq, int n) {
    for (int i = 0; i < n; i++) {
        char letter;
        getchar();
        cin >> letter;              // 读入字符 
        getchar();
        cin >> letterFreq[letter];  // 读入频率，为字符的映射 
        
        pq.push(letterFreq[letter]);// 把读入的频率压入到优先队列中 
    }
}

void checkOptimalCode(map<char, int> &letterFreq, priority_queue< int, vector<int>, greater<int> > &pq, int n) {
    int wpl = getWPL(pq);           // 用不构造Huffman Tree的方法来计算WPL 
    
    int m;
    cin >> m;
    for (int i = 0; i < m; i++) {
        string code[n];
        int codeLen = 0;
        bool ret = true;
        
        for (int i = 0; i < n; i++) {
            char letter;
            getchar();
            cin >> letter >> code[i];
            
            if (ret) {
                if (code[i].size() > n - 1) ret = false;
                codeLen += code[i].size() * letterFreq[letter];
            } 
        }
        
        if (ret && codeLen == wpl) {
            for (int i = 0; i < n; i++) {
                for (int j = i + 1; j < n; j++) {
                    if (code[i].substr(0, code[j].size()) == code[j].substr(0, code[i].size())) {
                        ret = false;
                        break;
                    }
                }
                if (ret == false) break;
            }
        }
        else {
            ret = false;
        }
        
        cout << (ret ? "Yes\n" : "No\n");
    }
}

int getWPL(priority_queue< int, vector<int>, greater<int> > &pq) {
    int wpl = 0;                // 用来保存累加的结果 
    while (!pq.empty()) {       // 当优先队列不为空 
        int tmp = pq.top();     // 从优先队列弹出一个元素，这个元素就是最小频率 
        pq.pop();
        
        if (pq.empty()) break;  // 如果弹出那个频元素优先队列就为空了，退出循环 
        
        tmp += pq.top();        // 如果优先队列不为空，再弹出一个元素，同时把两个频率进行相加 
        pq.pop();
        pq.push(tmp);           // 把两个频率相加的结果压入优先队列中 
        
        wpl += tmp;             // 同时，把这个相加结果进行累加，对应着累加度为2节点存放的频率 
    }
    
    return wpl;
}

 

参考资料

　　浙江大学——数据结构：https://www.icourse163.org/course/ZJU-93001?tid=1461682474

　　priority_queue的用法：https://www.cnblogs.com/Deribs4/p/5657746.html

　　pta5-9 Huffman Codes (30分)：https://www.cnblogs.com/Deribs4/p/4801656.html