哈夫曼編碼實現

我是看着數據結構(清華大學那本)

 

這兩頁說明了編碼方式的重要性

我想說的是書上，沒說清楚，代碼有些實現的細節自己搞了

 

代碼

#include<iostream>
#include<malloc.h>
#include<stdlib.h>
#include<string.h>
#include<cstdio>
#include<string>
#include<algorithm>

using namespace std;

const int N = 1e3 + 10;
bool vis[N];
typedef struct
{
	int weight;
	int parent, lchild, rchild;
}HTNode, * HuffmanTree;

typedef char** Huffmancode;
int now = 0;
void Select(HuffmanTree& HT, int cur, int& s1, int& s2, int n)
{
	now++;
	int val = 1e6;
	int a, b;
	for (int i = 1; i <= cur; i++)
	{
		if (vis[i]) continue;//如果用過了就不可以用了
		if (HT[i].weight < val)
		{
			val = HT[i].weight;
			a = i;
		}
	}
	s1 = a;
	vis[a] = 1;

	val = 1e6;
	for (int i = 1; i <= cur; i++)
	{
		if (vis[i]) continue;
		if (HT[i].weight < val)
		{
			val = HT[i].weight;
			b = i;
		}
	}
	s2 = b;
	vis[b] = 1;
	//cout << now << " " << s1 << " " << s2 << endl;
	if (a > n&& b <= n) swap(s1, s2);//這里必須注意，必須把1到 n放在葉子結點
}

void Huffmancoding(HuffmanTree& HT, Huffmancode& HC, int* w, int n)
{
	if (n <= 1) return;
	int m = 2 * n - 1;
	HT = (HuffmanTree)malloc((m + 1) * sizeof(HTNode));//0號不用

	int i;
	for (i = 1; i <= n; ++i)
		HT[i] = { w[i],0,0,0 };

	for (; i <= m; ++i)
		HT[i] = { 0,0,0,0 };

	for (i = n + 1; i <= m; ++i)
	{
		int s1, s2;
		Select(HT, i - 1, s1, s2, n);//書上沒寫這個函數，
		HT[s1].parent = i, HT[s2].parent = i;
		//統一把小的放在左邊，書上的不一致p148那個圖有問題
		HT[i].lchild = s1, HT[i].rchild = s2;
		HT[i].weight = HT[s1].weight + HT[s2].weight;
	}
	//從葉子結點到根逆向求每個字符的哈夫曼編碼
	//分配n個字符編碼的頭指針向量
	HC = (Huffmancode)malloc((n + 1) * sizeof(char*));
	//分配求編碼的工作空間
	char* cd = (char*)malloc(n * sizeof(char));
	cd[n - 1] = '\0';

	int cur, f, start;
	for (i = 1; i <= n; i++)
	{
		int cnt = n - 1;
		//cout << i << endl;
		for (cur = i, f = HT[i].parent; f != 0; cur = f, f = HT[f].parent)
		{
			//	cout << f << " ";
			if (HT[f].lchild == cur) cd[--cnt] = '0';
			else cd[--cnt] = '1';
		}
		//puts("");
		HC[i] = (char*)malloc((n - cnt) * sizeof(char));
		strcpy(HC[i], &cd[cnt]);//從cnt這個地方開始復制
	}
	free(cd);
}

int main()
{
	int n;
	int a[100];
	memset(vis, 0, sizeof(vis));
	HuffmanTree HT;
	Huffmancode HC;
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	Huffmancoding(HT, HC, a, n);
	for (int i = 1; i <= n; i++)
	{
		cout << i << " " << a[i] << " " << HC[i] << endl;
	}
	return 0;
}
//8
//5 29 7 8 14 23 3 11