问题 D: Haffman编码
时间限制: 1 Sec 内存限制: 128 MB
提交: 60 解决: 46
[提交][״̬][讨论版][命题人:外部导入]
题目描述
哈弗曼编码大家一定很熟悉吧（不熟悉也没关系，自己查去。。。）。现在给你一串字符以及它们所对应的权值，让你构造哈弗曼树，从而确定每个字符的哈弗曼编码。当然，这里有一些小规定：
1.规定哈弗曼树的左子树编码为0，右子树编码为1；
2.若两个字符权值相同，则ASCII码值小的字符为左孩子，大的为右孩子；
3.创建的新节点所代表的字符与它的做孩子的字符相同；
4.所有字符为ASCII码表上32-96之间的字符（即“ ”到“`”之间的字符）。
输入
输入包含多组数据（不超过100组）
每组数据第一行一个整数n，表示字符个数。接下来n行，每行有一个字符ch和一个整数weight，表示字符ch所对应的权值，中间用空格隔开。
输入数据保证每组测试数据的字符不会重复。
输出
对于每组测试数据，按照输入顺序输出相应的字符以及它们的哈弗曼编码结果，具体格式见样例。
样例输入
3 a 10 b 5 c 8 4 a 1 b 1 c 1 d 1
样例输出
a:0 b:10 c:11 a:00 b:01 c:10 d:11
#define _CRT_SECURE_NO_WARNINGS #include <algorithm> #include <iostream> #include <cstring> using namespace std; const int MaxN = 110; int w[MaxN]; typedef struct data{ int weight; char val; }Key; typedef struct{ Key key; int parent, lchild, rchild; }HuffmanNode, *HuffmanTree; typedef char *HuffmanCode; void seletTMin(HuffmanTree HT, int n, int &m1, int &m2){ int min = INT32_MAX; for (int i = 1; i <= n; ++i){ if (HT[i].parent == 0 && min >= HT[i].key.weight){ if (min > HT[i].key.weight){ min = HT[i].key.weight; m1 = i; } else{ if (HT[m1].key.val > HT[i].key.val) m1 = i; } } } min = INT32_MAX; for (int i = 1; i <= n; ++i){ if (HT[i].parent == 0 && min >= HT[i].key.weight && i != m1){ if (min > HT[i].key.weight){ min = HT[i].key.weight; m2 = i; } else{ if (HT[m2].key.val > HT[i].key.val) m2 = i; } } } } void HuffmanCoding(HuffmanTree &HT, HuffmanCode *&HC, Key w[], int n){ if (n <= 1) return; int m = 2 * n - 1; HT = new HuffmanNode[m + 1]; for (int i = 1; i <= n; ++i){ HT[i].key = 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){ int m1, m2; seletTMin(HT, i - 1, m1, m2); HT[i].lchild = m1; HT[i].rchild = m2; HT[m1].parent = HT[m2].parent = i; HT[i].key.weight = HT[m1].key.weight + HT[m2].key.weight; } HC = new HuffmanCode[n + 1]; char * cd = new char[n]; cd[n - 1] = 0; for (int i = 1; i <= n; ++i){ int start = n - 1; for (int c = i, f = HT[i].parent; f != 0; c = f, f = HT[f].parent){ if (HT[f].lchild == c) cd[--start] = '0'; else cd[--start] = '1'; } HC[i] = new char[n - start]; strcpy(HC[i], cd + start); } delete cd; } int main(){ HuffmanTree HT; HuffmanCode *HC; int n; Key data[MaxN]; while (cin >> n){ for (int i = 1; i <= n; ++i) cin >> data[i].val >> data[i].weight; HuffmanCoding(HT, HC, data, n); for (int i = 1; i <= n; ++i) cout << data[i].val << ":" << HC[i] << endl; delete HC; delete HT; } return 0; }
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文章来源: https://blog.csdn.net/qq_36502291/article/details/88854584