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 (≤), 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

#include<iostream>
#include<cstring>
using namespace std;
const int maxn = ;

typedef struct TreeNode* Tree;
struct TreeNode
{
    Tree left,right;
    int weight;
};

typedef struct HeapNode* Heap;
struct HeapNode
{
    TreeNode Data[maxn];
    int size;
};

int n,m;
int w[maxn];
char ch[maxn];
int codelen;
int cnt2,cnt0;

Tree creatTree();
Heap creatHeap();
void Insert(Heap H, TreeNode T);
Tree Huffman(Heap H);
Tree Delete(Heap H);
int WPL(Tree T, int depth);
bool Judge();
void JudgeTree(Tree T);

int main()
{
    cin >> n;
    Tree T = creatTree();
    Heap H = creatHeap();

    for (int i = ; i < n; i++)
    {
        getchar();
        cin >> ch[i] >> w[i];
        H->Data[H->size].left = H->Data[H->size].right = NULL;
        T->weight = w[i];
        Insert(H,*T);
    }

    T = Huffman(H);
    codelen = WPL(T,);
    //printf("%d\n",codelen);

    cin >> m;
    while (m--)
    {
        if (Judge())
        {
            printf("Yes\n");
        }
        else
        {
            printf("No\n");
        }
    }

    return ;
}

Tree creatTree()
{
    Tree T = new TreeNode;
    T->left = T->right = NULL;
    T->weight = ;
    return T;
}

Heap creatHeap()
{
    Heap H = new HeapNode;
    H->Data[].weight = -;
    H->size = ;
    return H;
}

void Insert(Heap H, TreeNode T)
{
    int i = ++H->size;
    for (; H->Data[i/].weight > T.weight; i /= )
    {
        H->Data[i] = H->Data[i/];
    }
    H->Data[i] = T;
}

Tree Huffman(Heap H)
{
    Tree T = creatTree();
    while (H->size > )
    {
        T->left = Delete(H);
        T->right = Delete(H);
        T->weight = T->left->weight + T->right->weight;
        Insert(H,*T);
    }
    T = Delete(H);
    return T;
}

Tree Delete(Heap H)
{
    int parent,child;
    TreeNode Tmp = H->Data[H->size--];
    Tree T = creatTree();
    *T = H->Data[];
    for (parent = ; *parent <= H->size; parent = child)
    {
        child = *parent;
        if (child < H->size &&
            H->Data[child+].weight < H->Data[child].weight)
        {
            child++;
        }

        if (H->Data[child].weight > Tmp.weight)
        {
            break;
        }
        H->Data[parent] = H->Data[child];
    }
    H->Data[parent] = Tmp;
    return T;
}

int WPL(Tree T, int depth)
{
    if (!T->left && !T->right)
    {
        return depth * (T->weight);
    }
    else
    {
        return WPL(T->left,depth+) + WPL(T->right,depth+);
    }
}

bool Judge()
{
    char s1[maxn],s2[maxn];
    bool flag = true;
    Tree T = creatTree();
    Tree pt = NULL;
    int wgh;

    for (int i = ;  i < n; i++)
    {
        cin >> s1 >> s2;

        if (strlen(s2) > n)
        {
            return ;
        }

        int j;
        for (j = ; ch[j] != s1[]; j++)
        {
            ;
        }
        wgh = w[j];
        pt = T;
        for (j = ; s2[j]; j++)
        {
            if (s2[j] == '')
            {
                if (!pt->left)
                {
                    pt->left = creatTree();
                }
                pt = pt->left;
            }
            if (s2[j] == '')
            {
                if (!pt->right)
                {
                    pt->right = creatTree();
                }
                pt = pt->right;
            }

            if (pt->weight)
            {
                flag = false;
            }
            if (!s2[j+])
            {
                if (pt->left || pt->right)
                {
                    flag = false;
                }
                else
                {
                    pt->weight = wgh;
                }
            }
        }
    }

    if (!flag)
    {
        return ;
    }
    cnt0 = cnt2 = ;
    JudgeTree(T);

    if (cnt2 != cnt0-)
    {
        return ;
    }
    if (codelen == WPL(T,))
    {
        return ;
    }
    else
    {
        return ;
    }
}

void JudgeTree(Tree T)
{
    if (T)
    {
        if (!T->left && !T->right)
        {
            cnt0++;
        }
        else if(T->left && T->right)
        {
            cnt2++;
        }
        else
        {
            cnt0 = ;
        }

        JudgeTree(T->left);
        JudgeTree(T->right);
    }
}