In 1953, David A. Huffman published his paper "A Method for the Construction of Minimum-Redundancy Codes", and hence printed his name in the history of computer science. As a professor who gives the final exam problem on Huffman codes, I am encountering a big problem: the Huffman codes are NOT unique. For example, given a string "aaaxuaxz", we can observe that the frequencies of the characters 'a', 'x', 'u' and 'z' are 4, 2, 1 and 1, respectively. We may either encode the symbols as {'a'=0, 'x'=10, 'u'=110, 'z'=111}, or in another way as {'a'=1, 'x'=01, 'u'=001, 'z'=000}, both compress the string into 14 bits. Another set of code can be given as {'a'=0, 'x'=11, 'u'=100, 'z'=101}, but {'a'=0, 'x'=01, 'u'=011, 'z'=001} is NOT correct since "aaaxuaxz" and "aazuaxax" can both be decoded from the code 00001011001001. The students are submitting all kinds of codes, and I need a computer program to help me determine which ones are correct and which ones are not.
Input Specification:
N2≤N≤63N
c[1] f[1] c[2] f[2] ... c[N] f[N]
c[i]
f[i]
c[i]
M≤1000MN
c[i] code[i]
c[i]
i
code[i]
Output Specification:
For each test case, print in each line either "Yes" if the student's submission is correct, or "No" if not.
Note: The optimal solution is not necessarily generated by Huffman algorithm. Any prefix code with code length being optimal is considered correct.
给定一段文字,如果我们统计出字母出现的频率,是可以根据哈夫曼算法给出一套编码,使得用此编码压缩原文可以得到最短的编码总长。然而哈夫曼编码并不是唯一的。例如对字符串"aaaxuaxz",容易得到字母 'a'、'x'、'u'、'z' 的出现频率对应为 4、2、1、1。我们可以设计编码 {'a'=0, 'x'=10, 'u'=110, 'z'=111},也可以用另一套 {'a'=1, 'x'=01, 'u'=001, 'z'=000},还可以用 {'a'=0, 'x'=11, 'u'=100, 'z'=101},三套编码都可以把原文压缩到 14 个字节。但是 {'a'=0, 'x'=01, 'u'=011, 'z'=001} 就不是哈夫曼编码,因为用这套编码压缩得到 00001011001001 后,解码的结果不唯一,"aaaxuaxz" 和 "aazuaxax" 都可以对应解码的结果。本题就请你判断任一套编码是否哈夫曼编码。
输入格式:
N(2≤N≤63N
c[1] f[1] c[2] f[2] ... c[N] f[N]
其中c[i]
是集合{'0' - '9', 'a' - 'z', 'A' - 'Z', '_'}中的字符;f[i]
是c[i]
M(≤1000MN
c[i] code[i]
其中c[i]
是第i
个字符;code[i]
是不超过63个'0'和'1'的非空字符串。
输出格式:
对每套待检编码,如果是正确的哈夫曼编码,就在一行中输出"Yes",否则输出"No"。
注意:最优编码并不一定通过哈夫曼算法得到。任何能压缩到最优长度的前缀编码都应被判为正确。
输入样例:
7 A 1 B 1 C 1 D 3 E 3 F 6 G 6 4 A 00000 B 00001 C 0001 D 001 E 01 F 10 G 11 A 01010 B 01011 C 0100 D 011 E 10 F 11 G 00 A 000 B 001 C 010 D 011 E 100 F 101 G 110 A 00000 B 00001 C 0001 D 001 E 00 F 10 G 11
输出样例:
Yes Yes No No
提交结果
提交时间 | ״̬ | 分数 | 题目 | 编译器 | 耗时 | 用户 |
---|---|---|---|---|---|---|
2018/5/24 23:50:17 | 答案正确 | 30 | 7-9 | C++ (g++) | copper |
测试点 | 结果 | 耗时 | 内存 |
---|---|---|---|
0 | 答案正确 | 240KB | |
1 | 答案正确 | 240KB | |
2 | 答案正确 | 240KB | |
3 | 答案正确 | 6220KB | |
4 | 答案正确 | 236KB | |
5 | 答案正确 | 236KB | |
6 | 答案正确 | 316KB |
这道题可以说是树这块的压轴题了,无论是代码量还是思维难度都和其他题目不在一个档次。题目意思是给定一个带权的输入序列,和N个与带权输入序列元素相同的编码测试序列,若编码符合最优编码,则输出Yes,若不符合,则输出No。
这道题考察最优编码长度,实际上是在考察Huffman树与Huffman编码,出题人担心你想不到,还特地在题干开头专门介绍了David A. Huffman和他提出的"A Method for the Construction of Minimum-Redundancy Codes"(一种实现最小冗余编码结构的方法),也就是Huffman Codes(哈夫曼编码)。
对于每个给定的带权元素序列来说,其必定可以建成一棵Huffman树,尽管根据建树方法的不同,树的结构不同,但是对于这个给定的带权元素序列,其最优编码长度是固定的,即某一形式的Huffman树下的Huffman编码长度。(建立Huffman树的过程为,每次将权重最小的两个结点合成一棵二叉树,其树根结点权值为两子树权值之合。再将该二叉树当作结点进行重新合并。重复上述过程直到使用完所有结点,建立成一棵树(理论上N个结点要进行N-1次合并))但需要注意的是,Huffman编码的编码长度是最优编码长度,但是最优编码长度可以不是其Huffman编码(题目最后一句也有特地提醒,看来出题人还是比较好心的)。因此,验证输入测试序列需要靠两点:1.符合最优编码长度,2.能够无歧义解码。
那么思路就出来了。先根据输入序列建立Huffman树,并获得最优编码长度。再对提交数据进行检查:1.是否符合最优编码长度,2.是否符合无歧义解码规则(前缀码编码,数据仅存在于二叉树叶节点)。
获得最优编码长度的过程,需要先创建一棵Huffman树,又需要先将带权序列建立成最小堆,再每轮弹出2次最小堆的顶点,作为二叉树的左右子树进行合并,合并完后的二叉树进行权值更新,再继续放入最小堆进行合并……直到最小堆元素全部弹出,最后弹出一整棵Huffman树。(最小堆的建立可以参见:#数据结构与算法学习笔记#PTA14:最小堆与最大堆(C/C++))。计算每一个测试序列的编码长度,与标准Huffman编码长度比较即可。最小堆每次插入和弹出都需要对全堆某个路径(根节点到叶子结点的一条路径)进行一次调整,具体情况分析详见代码注释。
检查前缀码编码的过程,需要根据输入序列的每个元素编码,模拟其在树中的路径(相当于每次创建一个元素编码所代表一条二叉树的路径,0代表左子树,1代表右子树)。模拟过程中的两种情况可以验证不满足前缀码要求(如下图):1.后创建的分支经过或超过已经被定义的叶子结点,2.后创建分支创建结束时未达到叶子结点。具体情况分析详见代码注释。
// HuffmanCodes.cpp : 定义控制台应用程序的入口点。 // //#include <stdafx.h> #include <vector> #include <iostream> #include <string.h> using namespace std; //Huffman树结点类 class Node { public: Node() {} Node(char element, int weight) :element(element), weight(weight), left(NULL), right(NULL) {} char element; int weight; Node* left = NULL; Node* right = NULL; bool isleave = false; }; typedef Node* HFMTree; //输入测试样例结点类 class Case { public: char element; char route[1000]; int length; int getlength() { return strlen(this->route); } }; void Read(int num, vector<HFMTree>& minHeap, vector<HFMTree>& inputlist); void Insert(vector<HFMTree>& minHeap, HFMTree node); //插入数据创建最小堆 HFMTree CreateHFMT(vector<HFMTree>& minHeap); //根据最小堆创建Huffman树 HFMTree DeleteMinHeap(vector<HFMTree>& minHeap); //从最小堆中取出最小元素,删除该结点并重新调整最小堆,最后删除该结点 int getHFMLength(HFMTree hfmtree, int depth); //获得该树编码长度 void Input(vector<Case>& testcase, int num); bool isOptimalLen(vector<Case>& testcase, vector<HFMTree>& inputlist, int weight); //检查是否符合最优编码长度 bool isPrefixCode(vector<Case>& testcase); //检查是否符合前缀码编码 int main() { /*根据输入序列建立Huffman树,并获得最优编码长度*/ int num; cin >> num; vector<HFMTree> minHeap; //创建最小堆,用最小堆对序列进行存储 vector<HFMTree> inputlist; //记录输入顺序与权值大小 HFMTree flag = new Node('-', -1); minHeap.push_back(flag); Read(num, minHeap, inputlist); HFMTree hfmtree; //利用最小堆创建Huffman树 hfmtree = CreateHFMT(minHeap); int optcodelength = getHFMLength(hfmtree, 0); //通过序列创建的Huffman树获得最优编码长度 /*对提交数据进行检查:1.是否符合最优编码长度,2.是否符合无歧义解码规则(前缀码编码,数据仅存在于二叉树叶节点)*/ int count; cin >> count; for (int i = 0;i < count;i++) { vector<Case> testcase; Input(testcase, num); bool isoptimallen = isOptimalLen(testcase, inputlist, optcodelength); bool isprefixcode = isPrefixCode(testcase); if (isoptimallen && isprefixcode) { cout << "Yes" << endl; } else { cout << "No" << endl; } } system("pause"); return 0; } void Read(int num, vector<HFMTree>& minHeap, vector<HFMTree>& inputlist) { char element; int weight; for (int i = 0; i < num; i++) { cin >> element >> weight; HFMTree node = new Node(element, weight); inputlist.push_back(node); Insert(minHeap, node); } //minHeap.erase(minHeap.begin()); } void Insert(vector<HFMTree>& minHeap, HFMTree node) { int index = minHeap.size(); minHeap.push_back(node); //每次插入后自底向上进行调整 while ((*minHeap[index / 2]).weight > (*node).weight) { //此处不可单纯进行值交换,需要交换两个对象 //(*minHeap[index]).element = (*minHeap[index / 2]).element; //(*minHeap[index]).weight = (*minHeap[index / 2]).weight; minHeap[index] = minHeap[index / 2]; index /= 2; } minHeap[index] = node; } HFMTree CreateHFMT(vector<HFMTree>& minHeap) { HFMTree hfmtree = new Node(); int size = minHeap.size() - 1; //进行size-1次合并 for (int i = 1; i < size; i++) { HFMTree node = new Node(); //每次从最小堆中取出堆顶的两个结点作为该结点的左右子结点 node->left = DeleteMinHeap(minHeap); node->right = DeleteMinHeap(minHeap); node->weight = node->left->weight + node->right->weight; //将该结点作为根节点的二叉树重新加入最小堆 Insert(minHeap, node); } //从最小堆中取出建好的Huffman树 hfmtree = DeleteMinHeap(minHeap); return hfmtree; } HFMTree DeleteMinHeap(vector<HFMTree>& minHeap) { //检查是否堆空 if (minHeap.size() == 1) { return NULL; } //将该堆最大元素装入新结点并返回 HFMTree node = new Node(); node = minHeap[1]; //重新调整该堆 int size = minHeap.size(); int parent, child; //用最大堆中最后一个元素从根结点开始向上过滤下层结点 HFMTree cmp = new Node(); cmp = minHeap[size - 1]; //从根节点开始,用parent记录根结点下标,用child记录其最小子结点下标,每次循环将parent更新为上一次循环的child //当parent指向底层结点时跳出循环(会有极端情况比如偏向一边的堆使得parent最终并非指向该子树底层结点,但不影响结果) for (parent = 1; 2 * parent < size; parent = child) { child = parent * 2; //若该子结点不是堆尾结点,令child指向左右子结点中的较小者 if ((child != size - 1) && ((*minHeap[child]).weight > (*minHeap[child + 1]).weight)) { child++; } //当循环到堆尾结点值小于等于该子结点值时,可以结束(此时堆尾结点会替换parent结点而不是child结点) if (cmp->weight <= (*minHeap[child]).weight) { break; } else { minHeap[parent] = minHeap[child]; } } //将尾结点与当前父结点替换 minHeap[parent] = cmp; //删除堆尾结点 //此处不能用minHeap.erase(minHeap.end());,因为erase会返回被删除结点的下一结点,而尾结点的下一结点超限 minHeap.pop_back(); //返回该结点 return node; } int getHFMLength(HFMTree hfmtree, int depth) { //若为叶子节点,直接返回其编码长度 if (!hfmtree->left && !hfmtree->right) { return hfmtree->weight * depth; } //否则其他节点一定有两个子树,返回左右子树编码长度之合,深度相应加一 else { return getHFMLength(hfmtree->left, depth + 1) + getHFMLength(hfmtree->right, depth + 1); } } void Input(vector<Case>& testcase, int num) { for (int i = 0;i < num;i++) { Case inputcase; cin >> inputcase.element >> inputcase.route; inputcase.length = inputcase.getlength(); testcase.push_back(inputcase); } } bool isOptimalLen(vector<Case>& testcase, vector<HFMTree>& inputlist, int weight) { int testweight = 0; for (int i = 0;i < testcase.size();i++) { testweight += (testcase[i].length * (*inputlist[i]).weight); } if (testweight == weight) { return true; } else { return false; } } bool isPrefixCode(vector<Case>& testcase) { bool isprefixcode = true; HFMTree newtree = new Node(); //两种情况验证不满足前缀码要求:1.后创建的分支经过或超过已经被定义的叶子结点,2.后创建分支创建结束时未达到叶子结点 for (int i = 0;i < testcase.size();i++) { HFMTree point = newtree; if (isprefixcode == false)break; for (int j = 0;j < testcase[i].length;j++) { if (isprefixcode == false)break; if (testcase[i].route[j] == '0') { //先检查左子结点是否存在,若不存在,则创建一个左子结点 if (!point->left) { HFMTree newnode = new Node(); point->left = newnode; point = point->left; //若此时为分支的最后一环,则将该结点定义为叶子结点 if (j == testcase[i].length - 1) { point->isleave = true; } } //若左子树存在,则先将标记指针移至左子树。 else { point = point->left; //若左子树为叶子结点,则不符合要求 if (point->isleave) { isprefixcode = false; break; } //若此时为分支的最后一环且仍有叶子结点,则不符合要求 if ((j == testcase[i].length - 1) && (point->left || point->right)) { isprefixcode = false; break; } } } else if (testcase[i].route[j] == '1') { //先检查右子结点是否存在,若不存在,则创建一个右子结点 if (!point->right) { HFMTree newnode = new Node(); point->right = newnode; point = point->right; //若此时为分支的最后一环,则将该结点定义为叶子结点 if (j == testcase[i].length - 1) { point->isleave = true; } } //若左子树存在,则先将标记指针移至左子树。 else { point = point->right; //若左子树为叶子结点,则不符合要求 if (point->isleave) { isprefixcode = false; break; } //若此时为分支的最后一环且仍有叶子结点,则不符合要求 if ((j == testcase[i].length - 1) && (point->left || point->right)) { isprefixcode = false; break; } } } } } return isprefixcode; }