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$N$ ($2\le N\le 63$), 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', '_'}, andf[i] is the frequency of c[i] and is an integer no more than 1000. The next line gives a positive integer$M$ ($\le 1000$), 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


/*
体会：
这道题写了有8个小时了。
这题主要考察:
1.哈夫曼树，建立依靠最小堆，在堆中存储树的节点 。
2.最小wpl的判断。只要符合长度的都是最小wpl，哈夫曼树不唯一，但最小值是唯一的，所以只要判断是否为前缀码，再比较wpl值即可。
*/ 

#include 
#include 
#include 
#define Maxn 64

int N, w[Maxn];
char ch[Maxn];
int codelen, cnt1, cnt2;

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 = (Tree)malloc(sizeof(struct TreeNode));
    T->Left = T->Right = NULL;
    T->Weight = 0;
    return T;
}

Heap CreatHeap() {
    Heap H;
    H = (Heap)malloc(sizeof(struct HeapNode));
    H->Size = 0;
    H->Data[0].Weight = -1;
    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 parent, child;
    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(Temp.Weight < H->Data[child].Weight) break;
        H->Data[parent] = H->Data[child];
    }
    H->Data[parent] = Temp;
    return T;
}

Tree Huffman(Heap H) {
    Tree T = CreatTree(); //分配空间
    while(H->Size != 1) {
        T->Left = Delete(H);
        T->Right = Delete(H);
        T->Weight = T->Left->Weight + T->Right->Weight;
        //printf("l = %d, r = %d, t = %d\n", T->Left->Weight, T->Right->Weight, T->Weight);
        Insert(H, *T);
    }
    T = Delete(H);
    return T;
}

void PreTravel(Tree T) {
    if(T) {
        printf("%d ", T->Weight);
        PreTravel(T->Left);
        PreTravel(T->Right);
    }
}

int WPL(Tree T, int Depth) {
    if(!T->Left && !T->Right) { /*printf("d = %d w = %d\n", Depth, T->Weight);*/ return Depth*T->Weight; }
    else return WPL(T->Left, Depth+1) + WPL(T->Right, Depth+1);
}

void JudgeTree(Tree T) {
    if(T) {
        if(T->Left && T->Right) 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++) {
        scanf("%s%s", s1, s2);
        if(strlen(s2) > N) return 0;
        for(j = 0; s1[0] != ch[j]; j++); wgh = w[j];
        pt = T;//每次建树前先将指针移动到根节点上；
        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;//判断是否为前缀
                pt->Weight = wgh;
            }
            //printf("w = %d\n", pt->Weight);
        }
    }
    if(!flag) return 0;
    cnt1 = cnt2 = 0;
    JudgeTree(T);//判断是否不存在度数1的节点
    if(cnt1 != cnt2 + 1) return 0;
    //printf("wpl = %d\n", WPL(T, 0));
    if(codelen == WPL(T, 0)) return 1;
    else return 0;
}

int main() {
    int i, n;
    Heap H;
    Tree T;
    H = CreatHeap();
    T = CreatTree();
    scanf("%d", &N);
    for(i = 0; i < N; i++) {
        getchar();
        scanf("%c %d", &ch[i], &w[i]);
        H->Data[H->Size].Left = H->Data[H->Size].Right = NULL;
        T->Weight = w[i];
        Insert(H, *T);
    }
    //for(i = 1; i <= H->Size; i++) printf("%d ", H->Data[i].Weight);
    T = Huffman(H);     //PreTravel(T);
    codelen = WPL(T, 0);
    scanf("%d", &n);
    while(n--) {
        if(Judge()) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}