Huffman树和Huffman编码

哈夫曼树的构造(哈夫曼算法)
1.根据给定的n个权值{w1,w2,…,wn}构成二叉树集合F={T1,T2,…,Tn},其中每棵二叉树Ti中只有一个带权为wi的根结点,其左右子树为空.
2.在F中选取两棵根结点权值最小的树作为左右子树构造一棵新的二叉树,且置新的二叉树的根结点的权值为左右子树根结点的权值之和.
3.在F中删除这两棵树,同时将新的二叉树加入F中.
4.重复2、3,直到F只含有一棵树为止.(得到哈夫曼树)

 

每次选取最小的两个值可以通过优先队列实现，使用优先队列可以使得代码非常简洁。

 

建立了哈夫曼树后，可以直接在哈夫曼树上跑一遍得到哈夫曼编码

 

#include <iostream>
#include <queue>

using namespace std;
const int maxn = 1e3 + 7;
int n, code[maxn], ans[maxn][maxn];
typedef struct node {
    int weight, id;
    struct node *lson, *rson;
} *pnode;

struct cmp {
    bool operator()(pnode f1, pnode f2) // 重载括号
    {
        return (*f1).weight > (*f2).weight; // 等同于less
    }
};

priority_queue<pnode, vector<pnode>, cmp> que;

pnode get_head() {
    for (int i = 1; i <= n; ++i) {
        pnode temp = NULL;
        temp = (pnode) malloc(sizeof(node));
        temp->id = -1;
        temp->lson = que.top();
        que.pop();
        temp->rson = que.top();
        que.pop();
        temp->weight = temp->lson->weight + temp->rson->weight;
        printf("temp->lson->weight = %d temp->rson->weight = %d\n", temp->lson->weight, temp->rson->weight);
        if (que.empty()) return temp;
        que.push(temp);
    }
}

void get_ans(pnode h, int dep) {
    if (h->lson == NULL && h->rson == NULL) {
        int idx = h->id;
        for (int i = 0; i < dep; ++i)
            ans[idx][i] = code[i];
        ans[idx][maxn - 1] = dep;
        printf("dep = %d h->weight = %d\n", dep, h->weight);
        return;
    }
    printf("h->lson->weight = %d h->rson->weight = %d\n", h->lson->weight, h->rson->weight);
    code[dep] = 0;
    get_ans(h->lson, dep + 1);
    code[dep] = 1;
    get_ans(h->rson, dep + 1);
}

int main() {
    freopen("../in.txt", "r", stdin);
    while (!que.empty()) que.pop();
    for (int i = 0; i < maxn; ++i) code[i] = 1;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        pnode cnt = NULL;
        cnt = (pnode) malloc(sizeof(node));
        cnt->lson = cnt->rson = NULL;
        cnt->id = i;
        cin >> cnt->weight;
        que.push(cnt);
    }
    pnode head = get_head();
    printf("total = %d\n", head->weight);
    get_ans(head, 0);
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j < ans[i][maxn - 1]; ++j)
            printf("%d", ans[i][j]);
        printf("\n");
    }
    return 0;
}