6-1 哈夫曼树及哈夫曼编码

6-1 哈夫曼树及哈夫曼编码 （10 分)

函数SelectTwoMin(int upbound, HuffmanTree HT, int &s1, int &s2)是从1到upbound中找出father为0的节点赋给s1,s2,（为了保证答案唯一，请让s1的节点编号小于s2），函数HuffmanCoding(HuffmanTree &HT, HuffmanCode &HC, int *w, int n)是构造哈夫曼树以及计算哈夫曼编码。保证输入的权重值小于1000。
函数接口定义：
void SelectTwoMin(int upbound, HuffmanTree HT, int &s1, int &s2);
void HuffmanCoding(HuffmanTree &HT, HuffmanCode &HC, int *w, int n);
其中 upbound 编号，HT是哈夫曼树，HC是哈夫曼编码，w是权值，n是叶子节点个数。
裁判测试程序样例：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
int weight;
int parent;
int lchild;
int rchild;
} HTNode, *HuffmanTree;
typedef char ** HuffmanCode;

void SelectTwoMin(int upbound, HuffmanTree HT, int &s1, int &s2);
void HuffmanCoding(HuffmanTree &HT, HuffmanCode &HC, int *w, int n);

int main() {
HuffmanTree ht;
HuffmanCode hc;

int n;
scanf("%d", &n);

int *w = (int *) malloc (n * sizeof(int));
for(int i = 0; i < n; ++ i)
    scanf("%d", &w[i]);

HuffmanCoding(ht, hc, w, n);

for (int i = 1; i <= 2 * n - 1; ++ i) {
    printf("%d %d %d %d\n", 
    ht[i].weight, ht[i].parent, ht[i].lchild, ht[i].rchild);
}

for (int i = 1; i <= n; ++ i)
    printf("%s\n", hc[i]);

free(w);
free(ht);
for (int i = 1; i <= n; ++ i)
    free(hc[i]);

return 0;

}
/* 你的代码将被嵌在这里 */
输入格式：
第一行输入一个数n，表示叶子节点的个数，接下去输入n个整数，表示每个节点的值
输出格式：
只要建树即可，输出已经确定了
输入样例：
4
1 2 3 4
输出样例：
1 5 0 0
2 5 0 0
3 6 0 0
4 7 0 0
3 6 1 2
6 7 3 5
10 0 4 6
110
111
10
0
代码如下：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    int weight;
    int parent;
    int lchild;
    int rchild;
} HTNode, *HuffmanTree;
typedef char ** HuffmanCode;

void SelectTwoMin(int upbound, HuffmanTree HT, int &s1, int &s2);
void HuffmanCoding(HuffmanTree &HT, HuffmanCode &HC, int *w, int n);

void reverse(char *CH)
{
	int n=strlen(CH);
	int i;
	for (i=0; i<n/2;i++)
	{
		char temp;
		temp = CH[i];
		CH[i] = CH[n-i-1];
		CH[n-i-1] = temp;
	}
}

int main() {
    HuffmanTree ht;
    HuffmanCode hc;

    int n;
    scanf("%d", &n);
	
    int *w = (int *) malloc (n * sizeof(int));
    for(int i = 0; i < n; ++ i)
        scanf("%d", &w[i]);

    HuffmanCoding(ht, hc, w, n);
	
    for (int i = 1; i <= 2 * n - 1; ++ i) {
        printf("%d %d %d %d\n", 
        ht[i].weight, ht[i].parent, ht[i].lchild, ht[i].rchild);
    }

    for (int i = 1; i <= n; ++ i)
        printf("%s\n", hc[i]);

    free(w);
    free(ht);
    for (int i = 1; i <= n; ++ i)
        free(hc[i]);
	
    return 0;
}
//其中 upbound 编号，HT是哈夫曼树，HC是哈夫曼编码，w是权值，n是叶子节点个数
void SelectTwoMin(int upbound, HuffmanTree HT, int &s1, int &s2)
{
    int x1=0,x2=0;
    int m1= 1000;
    int m2= 1000;
    for(int i=1; i<=upbound; i++)
    {
        if(HT[i].parent == 0&& HT[i].weight < m1)
        {
            m2= m1;
            x2 = x1;
            m1 = HT[i].weight;
            x1 = i;
        }
        else if(HT[i].parent == 0 && HT[i].weight <m2)
        {
            m2 = HT[i].weight;
            x2 = i;
        }
    }
    s1 = x1;
    s2 = x2;
}
void HuffmanCoding(HuffmanTree &HT, HuffmanCode &HC, int *w, int n)
{
    int s1=0;
    int s2=0;
    HT = (HuffmanTree)malloc(sizeof(HTNode)*(2*n));
    HC = (char **)malloc(sizeof(char *)*(n+1));
    for(int i=1;i<=n;i++)
    {
    	HC[i] = (char *)malloc(sizeof(char)*(n+1));
		memset(HC[i],0,sizeof(char)*(n+1));
	}
    for (int i = 0; i <n ; ++i) {
        HT[i+1].weight = w[i];
    }//给结构体赋值
    for(int i=1;i<=2*n-1;i++)
    {
		HT[i].parent = 0;
		HT[i].lchild = 0;
		HT[i].rchild = 0;
	}
    for (int i = 0; i <n-1 ; ++i) {
        SelectTwoMin(n+i,HT,s1,s2);
        HT[i+n+1].lchild = s1;
        HT[i+n+1].rchild = s2;
        HT[i+n+1].weight = HT[s1].weight+HT[s2].weight;
        HT[s1].parent = i+n+1;
        HT[s2].parent = i+n+1;
    }
    //
    for (int i = 1; i <= n ; ++i) {
        int c = i;
        int parent = HT[c].parent;
        while (parent!=0)
        {
            if(HT[parent].lchild==c)
            {
                strncat(HC[i],"0",1);
            }
            else
                strncat(HC[i],"1",1);
        c = parent;
        parent = HT[parent].parent;
    }
    reverse(HC[i]);
    }
}

输出样例：

该代码随难度不大到有很多细节需要注意：

1. 初始化叶节点以及非叶节点使他们的父亲结点儿子结点全部初始化为0.
2. 申请字符串数组内存注意形式。
3. 在申请过程中要保证每一个字符串都应该被初始化否则会出现乱码（亲测）。
4. 注意得到的哈夫曼编码是从叶节点到根结点，需要将它逆序。