6-1 哈夫曼树及哈夫曼编码分数

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：

#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 */
void SelectTwoMin(int upbound, HuffmanTree HT, int &s1, int &s2)
{
    int min1=10000,min2=10000,num1,num2;
    for(int i=1;i<=upbound;i++)
    {
        if(HT[i].parent==0&&HT[i].weight<min1)
        {
            min2=min1;
            num2=num1;
            min1=HT[i].weight;
            num1=i;
        }
        else if (HT[i].parent==0&&HT[i].weight<min2)
        {
            min2=HT[i].weight;
            num2=i;
        }
        s1=num1;
        s2=num2;
    }


}
void HuffmanCoding(HuffmanTree &HT, HuffmanCode &HC, int *w, int n)
{
    //空间开辟
    HT=(HuffmanTree)malloc(sizeof(HTNode)*(2*n-1));
    HC=(char**)malloc(sizeof(char*)*(n+1));//注意这里只是开辟了（n+1）个数组指针，各个指针并没有开辟空间
    for(int i=0;i<(n+1);i++)//给每个字符串开辟空间
    {
        HC[i]=(char*)malloc(sizeof(char)*(n+1));
        memset(HC[i],0,sizeof(char)*(n+1));
    }

    //赋初值
    for(int i=1;i<=2*n-1;i++)
    {
        HT[i].parent=0;
        HT[i].lchild=0;
        HT[i].rchild=0;
        if(i<=n)
            HT[i].weight=w[i-1];
        else
            HT[i].weight=0;

    }

//构建哈夫曼树
    for(int i=n+1;i<=2*n-1;i++)
    {
        int s1=0,s2=0;
        SelectTwoMin(i-1,HT,s1,s2);//找出当前最小的两个数下标
        HT[i].lchild=s1;//更新
        HT[i].rchild=s2;
        HT[i].weight=HT[s1].weight+ HT[s2].weight;
        HT[s1].parent=i;
        HT[s2].parent=i;
    }
    for(int i=1;i<=n;i++)
    {
        char *a="";//从叶子开始往上找直到根节点
        a=(char*)malloc(sizeof(char)*(n+1));
        int k=i;
        int top=n-1;
        a[n]='\0';
        while(HT[k].parent)
        {
            if(HT[HT[k].parent].lchild==k)
            {
                a[top--]='0';
            }
            else if(HT[HT[k].parent].rchild==k)
            {
                a[top--]='1';
            }
        }
        strcpy(HC[k],a+top+1);//复制过去
    }

}