問題 A: 算法6-12：自底向上的赫夫曼編碼

1 題目

樣例輸入 
8
5 29 7 8 14 23 3 11
樣例輸出 
0110
10
1110
1111
110
00
0111
010
      

2 代碼

#include 
#include 
#include 
#include 
#include 

using std::swap;
using  std::strcpy;

typedef char *HuffmanCode;

const int MAXN = 110;

typedef struct{
    int weight;
    int lchild, rchild, parent;
}HuffmanNode, *HuffmanTree;

void seletMin(HuffmanTree HT, int n, int &s1, int &s2){
    int min = INT32_MAX;
    for (int i = 1; i <= n; ++i) {
        if(HT[i].parent == 0 && min > HT[i].weight){
            min = HT[i].weight;
            s1 = i;
        }
    }

    min = INT32_MAX;
    for (int j = 1; j <= n; ++j) {
        if(HT[j].parent == 0 && min > HT[j].weight && j != s1){
            min = HT[j].weight;
            s2 = j;
        }
    }

    if(s1 > s2){
        swap(s1, s2);
    }

}

void HuffmanCoding(HuffmanTree &HT, HuffmanCode *&HC,int w[],int n){
    // w存放n個字元的權值（均>0）,構造哈夫曼樹HT，并求出n個字元的哈夫曼樹編碼HC
    if(n <= 1) return;
    int m = 2 * n -1;
    //0号單元未用
    HT = new HuffmanNode[m + 1];

    for (int i = 1; i <= n; ++i) {
        HT[i].weight = w[i];
        HT[i].lchild = HT[i].rchild = HT[i].parent = 0;
    }

    for (int i = n + 1; i<= m; ++i) {
        HT[i].lchild = HT[i].rchild = HT[i].parent = 0;
    }


    for (int i = n + 1; i <= m; ++i)//建立哈夫曼樹
    {
        int s1, s2;
        //在HT[1~i-1]中選擇parent為0且weight最小的兩個結點，其序号分别為s1,s2
        seletMin(HT, i - 1, s1, s2);
        HT[s1].parent = HT[s2].parent = i;
        HT[i].lchild = s1;
        HT[i].rchild = s2;
        HT[i].weight = HT[s1].weight + HT[s2].weight;

    }
    //從葉結點到根你想求每個字元的哈夫曼編碼
    HC = new HuffmanCode[n + 1];
    //配置設定n個字元編碼的頭指針向量([0]不用)
    char* cd = new char[n];//配置設定編碼工作空間

    cd[n - 1] = '\0';//編碼結束符

    for (int i = 1; i <= n; ++i)//逐個字元求哈夫曼編碼
    {
        int star = n - 1;//編碼結束位置
        for (int c = i, f = HT[i].parent;f != 0; c = f, f = HT[f].parent)
        {
            //從葉結點到根逆向求編碼
            if(HT[f].lchild == c){
                cd[--star] = '0';
            }else{
                cd[--star] = '1';
            }
        }
            //為第i個字元串編碼配置設定空間
            HC[i] = new char[n - star];
            //從cd複制編碼（串）到HC
            strcpy(HC[i], cd + star);
    }
    delete []cd;//釋放工作空間
}

int main(){
    HuffmanTree HT;
    HuffmanCode *HC;
    int n;
    int data[MAXN];

    while(scanf("%d", &n) != EOF){
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &data[i]);
        }

        HuffmanCoding(HT, HC, data, n);

        for (int j = 1; j <= n; ++j) {
            //printf("%s\n", HC[j]);
            std::cout << HC[j] << std::endl;
        }

        delete(HT);
        delete(HC);
    }

    return 0;
}
/*
Input:
8
5 29 7 8 14 23 3 11
Output:
0110
10
1110
1111
110
00
0111
010
 */