BJFU_数据结构习题_262基于哈夫曼树的数据压缩算法

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262基于哈夫曼树的数据压缩算法

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+4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+2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+3行为编码后的字符串，第2n+4行为解码后的字符串（与输入的字符串相同）。

以下代码粘贴于自

基于哈夫曼树的数据压缩算法

输入样例
aaaaaaabbbbbccdddd
aabccc
0
输出样例
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

#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;elsecout << " ";}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;
}