哈夫曼编码实现

我是看着数据结构(清华大学那本)

这两页说明了编码方式的重要性

我想说的是书上，没说清楚，代码有些实现的细节自己搞了

代码

#include<bits/stdc++.h>
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;//如果用过了就不可以用了 
		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);//这里必须注意，必须把1到 n放在叶子结点 
}

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号不用
	
	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);//书上没写这个函数， 
		HT[s1].parent = i, HT[s2].parent = i;
		//统一把小的放在左边，书上的不一致p148那个图有问题 
		HT[i].lchild = s1, HT[i].rchild = s2;
		HT[i].weight = HT[s1].weight + HT[s2].weight;
	}
	//从叶子结点到根逆向求每个字符的哈夫曼编码
	//分配n个字符编码的头指针向量
	HC = (Huffmancode)malloc((n + 1) * sizeof(char*));
	//分配求编码的工作空间
	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]);//从cnt这个地方开始复制 
	}
	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