7-1 Huffman Codes

7-1 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.
翻译
1953年，David A.Huffman发表了他的论文“构造最小冗余代码的方法”，从而在计算机科学史上留下了他的名字。作为一名给霍夫曼码做期末考试问题的教授，我遇到了一个大问题：霍夫曼码不是唯一的。例如，给定字符串“a a a x u a x z”，我们可以观察到字符“a”、“x”、“u”和“z”的频率分别为4、2、1和1。我们可以将符号编码为{‘a’=0，‘x’=10，‘u’=110，‘z’=111}，或者以另一种方式编码为{‘a’=1，‘x’=01，‘u’=001，‘z’=000}，两者都将字符串压缩为14位。另一组代码可以给出为{‘a’=0，‘x’=11，‘u’=100，‘z’=101}，但是{‘a’=0，‘x’=01，‘u’=011，‘z’=001}是不正确的因为“a a a x u a x z”和“a a z u a x a x”都可以从代码000011001解码。学生们正在提交各种代码，我需要一个计算机程序来帮助我判断哪些是正确的，哪些是不正确的。

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:
翻译
每个输入文件包含一个测试用例。对于每种情况，第一行给出整数N(2≤N≤63)，然后是包含所有N个不同字符及其频率的行，其格式如下：

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:
翻译
其中c[i]是选自{‘0’-‘9’、‘a’-‘z’、‘A’-‘Z’、’’}的字符，f[i]是c[i]的频率，并且是不超过1000的整数。下一行给出正整数M（小于1000），然后是M学生提交。每个学生提交由n行组成，每种格式：

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.
翻译
其中C[i]是第i个字符，而代码[i]是一个不超过63’0’和‘1’的非空字符串。

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.
翻译
对于每个测试用例，如果学生的提交正确，或者“否”，则在每行中打印“yes”。

注：最优解不一定是由赫夫曼算法生成的。任何代码长度最佳的前缀码都被认为是正确的。

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

思路

一、构建Huffman树

1. 输入n，malloc链表LNode hf，大小为2n；
2. for hf[0]~hf[n-1]，输入n组数初始化，
3. for hf[n]~hf[2*n-2]，构建huffman树（*包含查找权重最小的两个数），计算WPL值

二、判断学生提交结果是否正确

1. 输入m，while（m–）
2. 若学生数据WPL2值（*包含获取权重）与WPL不等，输出No
3. 否则判断是否为前缀码，是则Yes，否则No

代码部分

1.结构体

typedef struct LNode{
    char name;
    int weight;
    int parent;
}LNode;

2.查找权重最小的两个数函数

void Select_2min(LNode *hf,int n,int &min1,int &min2){
    int flag=0;
    for(int i=0;i<n;i++)
        if(!flag && hf[i].parent==-1){min1=i;flag=1;}
        else if(flag && hf[i].parent==-1){min2=i;break;}
    if(hf[min1].weight > hf[min2].weight){
        int t=min1; min1=min2; min2=t;
    }
    for(int i=0; i<n; i++){
        if(hf[i].parent!=-1 || i==min1 || i==min2) continue;
        else if(hf[i].weight < hf[min1].weight){
            min2=min1; min1=i;
        }
        else if(hf[i].weight < hf[min2].weight)
            min2=i;
    }
}

3.获取权重函数

int GetWeight(LNode *hf, char c, int n){
    for(int i=0; i<n; i++)
        if(hf[i].name == c)
            return hf[i].weight;
}

4.main函数

int main()
{
    int n; scanf("%d",&n);
    LNode *hf = (LNode*)malloc(sizeof(LNode)*(2*n));
    for(int i=0; i<n; i++){
        getchar();
        scanf("%c %d",&hf[i].name,&hf[i].weight);
        hf[i].parent=-1;
    }
    int min1, min2, WPL=0;
    for(int i=n; i<2*n-1; i++){
        Select_2min(hf, i, min1, min2);
        hf[i].weight = hf[min1].weight+hf[min2].weight;
        hf[min1].parent = i;
        hf[min2].parent = i;
        WPL += hf[i].weight;
        hf[i].parent = -1;
    }
    int m; scanf("%d",&m);
    while(m--){
        string code[1005];
        int sum=0;
        char c;
        for(int i=0; i<n; i++){
            getchar();
            cin>>c>>code[i];
            sum += GetWeight(hf,c,n)*code[i].length();
        }
        if(sum!=WPL) printf("No\n");
        else{
            int flag=0;
            for(int i=0; i<n; i++){
                int len=code[i].length();
                for(int j=0;j<n;j++){
                    if(i==j) continue;
                        int len_other=code[j].length();
                        if(len<=len_other&&code[i]==code[j].substr(0,len)){
                            flag=1;
                            break;
                        }
                }
                if(flag) break;
            }
            if(flag) printf("No\n");
            else printf("Yes\n");
        }
    }
}

正确代码

#include <stdio.h>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <malloc.h>
#include <iostream>
using namespace std;
typedef struct LNode{
    char name;
    int weight;
    int parent;
}LNode;
void Select_2min(LNode *hf,int i,int &min1,int &min2);
int GetWeight(LNode *hf, char c, int n);
int main()
{
    int n; scanf("%d",&n);
    LNode *hf = (LNode*)malloc(sizeof(LNode)*(2*n));
    for(int i=0; i<n; i++){
        getchar();
        scanf("%c %d",&hf[i].name,&hf[i].weight);
        hf[i].parent=-1;
    }
    int min1, min2, WPL=0;
    for(int i=n; i<2*n-1; i++){
        Select_2min(hf, i, min1, min2);
        hf[i].weight = hf[min1].weight+hf[min2].weight;
        hf[min1].parent = i;
        hf[min2].parent = i;
        WPL += hf[i].weight;
        hf[i].parent = -1;
    }
    int m; scanf("%d",&m);
    while(m--){
        string code[1005];
        int sum=0;
        char c;
        for(int i=0; i<n; i++){
            getchar();
            cin>>c>>code[i];
            sum += GetWeight(hf,c,n)*code[i].length();
        }
        if(sum!=WPL) printf("No\n");
        else{
            int flag=0;
            for(int i=0; i<n; i++){
                int len=code[i].length();
                for(int j=0;j<n;j++){
                    if(i==j) continue;
                        int len_other=code[j].length();
                        if(len<=len_other&&code[i]==code[j].substr(0,len)){
                            flag=1;
                            break;
                        }
                }
                if(flag) break;
            }
            if(flag) printf("No\n");
            else printf("Yes\n");
        }
    }
}
int GetWeight(LNode *hf, char c, int n){
    for(int i=0; i<n; i++)
        if(hf[i].name == c)
            return hf[i].weight;
}
void Select_2min(LNode *hf,int n,int &min1,int &min2){
    int flag=0;
    for(int i=0;i<n;i++)
        if(!flag && hf[i].parent==-1){min1=i;flag=1;}
        else if(flag && hf[i].parent==-1){min2=i;break;}
    if(hf[min1].weight > hf[min2].weight){
        int t=min1; min1=min2; min2=t;
    }
    for(int i=0; i<n; i++){
        if(hf[i].parent!=-1 || i==min1 || i==min2) continue;
        else if(hf[i].weight < hf[min1].weight){
            min2=min1; min1=i;
        }
        else if(hf[i].weight < hf[min2].weight)
            min2=i;
    }
}

参考链接: wuqi5328_CSDN.