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), 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;
#define maxn 70
int N,codelen,cnt1,cnt2,w[maxn];
char ch[maxn];
typedef struct TreeNode* Tree;
struct TreeNode{
    int weight;
    Tree Left,Right;
};
typedef struct HeapNode* Heap;
struct HeapNode{
    struct TreeNode Data[maxn];
    int size;
};

Tree creatTree(){
      Tree T;
      T = new struct TreeNode;
      T->weight = 0;
      T->Left = T->Right = NULL;
      return T;
}

Heap creatHeap(){
    Heap H;
    H = new struct HeapNode;
    H->Data[0].weight = -1;
    H->size = 0;
    return H;
}

void Insert(Heap H,struct TreeNode T){
    int i = ++H->size;
    for(; T.weight < H->Data[i/2].weight; i /= 2)
         H->Data[i] = H->Data[i/2];
    H->Data[i] = T;
}

Tree Delete(Heap H){
    int child,parent;
    struct TreeNode Temp = H->Data[H->size--];
    Tree T = creatTree();
    *T = H->Data[1];
    for(parent = 1; 2 * parent <= H->size; parent = child){
        child = 2 * parent;
        if(child < H->size && H->Data[child].weight > H->Data[child+1].weight)
        child++;
        if(H->Data[child].weight > Temp.weight) break;
        H->Data[parent] = H->Data[child];
    }
    H->Data[parent] = Temp;
    return T;
}

Tree Huffman(Heap H){
    Tree T = creatTree();
    while(H->size != 1){
        T->Left = Delete(H);
        T->Right = Delete(H);
        T->weight = T->Right->weight + T->Right->weight;
        Insert(H,*T);
    }
    T = Delete(H);
    return T;
}

int WPL(Tree T,int depth){
    if(!T->Left && !T->Right) return(depth*T->weight);
    else return WPL(T->Left,depth+1)+WPL(T->Right,depth+1);
}

void JudgeTree(Tree T){
    if(T){
        if(T->Right && T->Left) cnt2++;
        else if(!T->Left && !T->Right) cnt1++;
        else cnt1 = 0;
        JudgeTree(T->Left);
        JudgeTree(T->Right);
    }
}

int Judge(){
    int i,j,wgh,flag = 1;;
    char s1[maxn],s2[maxn];
    Tree T = creatTree(), pt = NULL;
    for(i = 0; i < N; i++){
        cin >> s1 >> s2;
        if(strlen(s2) > N) return 0;
        for(j = 0; s1[0] != ch[j]; j++); wgh = w[j];
        pt = T;
        for(j = 0; s2[j] ; j++){
            if(s2[j] == '0'){
                if(!pt->Left) pt->Left = creatTree();
                pt = pt->Left;
            }
            if(s2[j] == '1'){
                if(!pt->Right) pt->Right = creatTree();
                pt = pt->Right;
            }
            if(pt->weight) flag = 0;
            if(!s2[j+1]){
                if(pt->Left || pt->Right) flag = 0;
                else pt->weight = wgh;
            }
        }
    }
    if(flag == 0) return 0;
    cnt1 = cnt2 = 0;
    JudgeTree(T);
    if(cnt1 != cnt2 + 1) return 0;
    if(codelen == WPL(T,0)) return 1;
    else return 0;
}

int main(){
    int i,n;
    Tree T;
    Heap H;
    T = creatTree();
    H = creatHeap();
    cin >> N;
    for(i = 0; 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,0);
    cin >> n;
    while(n--){
        if(Judge()) cout<< "Yes" << endl;
        else cout << "No" << endl;
    }
    return 0;
}