基于哈夫曼树的数据压缩算法

最新推荐文章于 2023-11-19 22:06:49 发布

aaaliaosha

最新推荐文章于 2023-11-19 22:06:49 发布

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本文链接：https://blog.youkuaiyun.com/aaaliaosha/article/details/78463191

本文介绍了一种基于哈夫曼树实现数据压缩与解压缩的算法。通过统计字符串中字符出现频率，构建哈夫曼树及编码表，进而完成数据的压缩与解压缩过程。

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基于哈夫曼树的数据压缩算法

发布时间: 2017年10月30日 19:30 时间限制: 1000ms 内存限制: 128M

描述

输入一串字符串，根据给定的字符串中字符出现的频率建立相应哈夫曼树，构造哈夫曼编码表，在此基础上可以对待压缩文件进行压缩（即编码），同时可以对压缩后的二进制编码文件进行解压（即译码）。

输入

多组数据，每组数据一行，为一个字符串（只考虑26个小写字母即可）。当输入字符串为“0”时，输入结束。

输出

每组数据输出2n+3行（n为输入串中字符类别的个数）。第一行为统计出来的字符出现频率（只输出存在的字符，格式为：字符：频度），每两组字符之间用一个空格分隔，字符按照ASCII码从小到大的顺序排列。第二行至第2n行为哈夫曼树的存储结构的终态（形如教材139页表5.2（b），一行当中的数据用空格分隔）。第2n+1行为每个字符的哈夫曼编码（只输出存在的字符，格式为：字符：编码），每两组字符之间用一个空格分隔，字符按照ASCII码从小到大的顺序排列。第2n+2行为编码后的字符串，第2n+3行为解码后的字符串（与输入的字符串相同）。

样例输入1 复制

aaaaaaabbbbbccdddd
aabccc
0

样例输出1

a:7 b:5 c:2 d:4
1 7 7 0 0
2 5 6 0 0
3 2 5 0 0
4 4 5 0 0
5 6 6 3 4
6 11 7 2 5
7 18 0 1 6
a:0 b:10 c:110 d:111
00000001010101010110110111111111111
aaaaaaabbbbbccdddd
a:2 b:1 c:3
1 2 4 0 0
2 1 4 0 0
3 3 5 0 0
4 3 5 2 1
5 6 0 3 4
a:11 b:10 c:0
111110000
aabccc

代码如下：

#include<iostream>
#include<cstring>
#include<string>

using namespace std;

struct Num
{
	char ch;
	int num;
}an[7000], temp;

void sort1(Num *bn)
{
	Num te;
	int i, j;
	for (j = 0; j<26; j++)
		for (i = 0; i < 25; i++)
		{
			if (bn[i].num < bn[i + 1].num)
			{
				te = bn[i];
				bn[i] = bn[i + 1];
				bn[i + 1] = te;
			}
		}
}

void sort2(Num *bn, int n)
{
	Num te;
	int i, j;
	for (j = 0; j<26; j++)
		for (i = 0; i < 25; i++)
		{
			if (bn[i].ch > bn[i + 1].ch&&bn[i + 1].num > 0)
			{
				te = bn[i];
				bn[i] = bn[i + 1];
				bn[i + 1] = te;
			}
		}
}

typedef struct
{
	int weight;
	int parent, lchild, rchild;
	int vis;
	char ch;
	char strc[60];
	int len;
	int num;
}HTNode, *HuffmanTree;

void Select(HuffmanTree &HT, int k, int &s1, int &s2)
{
	int i;
	HTNode h1, h2;
	int t1, t2;
	t1 = t2 = 0;
	h1 = HT[1];
	h1.weight = 999999;
	for (i = 1; i <= k; i++)
	{
		if (HT[i].weight <= h1.weight &&HT[i].vis == 0)
		{
			h1 = HT[i];
			t1 = i;
		}
	}
	h2 = HT[2];
	h2.weight = 999999;
	for (i = 1; i <= k; i++)
	{
		if (HT[i].weight <= h2.weight&&t1 != i&&HT[i].vis == 0)
		{
			h2 = HT[i];
			t2 = i;
		}
	}

	if (h1.weight == h2.weight&&HT[t1].num > HT[t2].num)
	{
		int temp;
		temp = t1;
		t1 = t2;
		t2 = temp;
	}

	s1 = t1;
	s2 = t2;
	HT[s1].vis = 1;
	HT[s2].vis = 1;
	return;
}

void code(HuffmanTree &HT, int len, char str[], int n, char cc)
{
	int i;
	for (i = 0; i < len - 1; i++)
		HT[n].strc[i] = str[i];
	HT[n].strc[i] = cc;
	HT[n].strc[i + 1] = '\0';
	if (HT[n].lchild != 0)
		code(HT, len + 1, HT[n].strc, HT[n].lchild, '0');
	if (HT[n].rchild != 0)
		code(HT, len + 1, HT[n].strc, HT[n].rchild, '1');
}

void CreateHuffmanTree(HuffmanTree &HT, int n, char *str, int n1)
{
	int s1, s2;
	s1 = s2 = 0;
	if (n <= 1)
		return;
	int m = 2 * n - 1;
	HT = new HTNode[m + 1];

	for (int i = 1; i <= m; i++)
	{
		HT[i].parent = 0; HT[i].lchild = 0; HT[i].rchild = 0; HT[i].vis = 0; HT[i].len = 0; HT[i].num = i; HT[i].weight = 0;
	}
	for (int i = 1; i <= n; i++)
	{
		HT[i].weight = an[i - 1].num;
		HT[i].ch = an[i - 1].ch;
	}
	for (int i = n + 1; i <= m; i++)
	{
		Select(HT, i - 1, s1, s2);
		HT[s1].parent = i;
		HT[s2].parent = i;
		HT[i].lchild = s1;
		HT[i].rchild = s2;
		HT[i].weight = HT[s1].weight + HT[s2].weight;
	}
	int len = HT[m].len;
	HT[m].strc[len] = '1';
	if (HT[m].lchild)
		code(HT, 1, "", HT[m].lchild, '0');
	if (HT[m].rchild)
		code(HT, 1, "", HT[m].rchild, '1');
	return;
}

int main()
{
	HuffmanTree h;
	int n, n1;
	char str[700], str2[700];
	while (cin >> str)
	{
		if (str[0] == '0'&&str[1] == '\0')
			break;
		n1 = 0;
		for (int i = 0; i < 27; i++)
		{
			an[i].num = 0;
			an[i].ch = 'a' + i;
		}
		n = strlen(str);
		for (int i = 0; i < n; i++)
		{
			an[str[i] - 'a'].num++;
			str2[i] = str[i];
			str2[i + 1] = '\0';
		}
		sort1(an);
		for (int i = 0; i < 26; i++)
		{
			if (an[i].num != 0)
				n1++;
		}

		sort2(an, n1);


		CreateHuffmanTree(h, n1, str, n1);
		if (n1 == 1)
		{
			h = new HTNode[3];
			h[1].weight = n; h[1].parent = h[1].lchild = h[1].rchild = 0;
			h[1].ch = str[0]; h[1].strc[0] = '0'; h[1].strc[1] = '\0';
		}
		int f = 0;
		for (int i = 0; i < 26; i++)
			if (an[i].num > 0)
			{
				f++;
				cout << an[i].ch << ":" << an[i].num ;
				if (f == n1)
					cout << endl;
				else
					cout << " ";
			}



		for (int i = 1; i <= 2 * n1 - 1; i++)
		{
			cout << i << " " << h[i].weight << " " << h[i].parent << " " << h[i].lchild << " " << h[i].rchild << endl;
		}

		for (int i = 1; i < n1; i++)
			cout << h[i].ch << ":" << h[i].strc << " ";
		cout << h[n1].ch << ":" << h[n1].strc;
		cout << endl;

		for (int i = 0; i < n; i++)
		{
			for (int j = 1; j <= n1; j++)
				if (h[j].ch == str[i])
					cout << h[j].strc;
		}
		cout << endl;
		cout << str << endl;

		for (int i = 1; i <= 2 * n1 - 1; i++)
		{
			h[i].parent = 0; h[i].lchild = 0; h[i].rchild = 0; h[i].vis = 0; h[i].len = 0; h[i].num = i;
		}
	}

	return 0;
}