PTA 05-树9 Huffman Codes (30分)

题目地址

https://pta.patest.cn/pta/test/16/exam/4/question/671

 

5-9 Huffman Codes   (30分)

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:

Each input file contains one test case. For each case, the first line gives an integer NN (2\le N\le 632≤N≤63), then followed by a line that contains all the NN distinct characters and their frequencies in the following format:

 

c[1] f[1] c[2] f[2] ... c[N] f[N]

where c[i] is a character chosen from {'0' - '9', 'a' - 'z', 'A' - 'Z', '_'}, and f[i] is the frequency of c[i]and is an integer no more than 1000. The next line gives a positive integer MM (\le 1000≤1000), then followed by MM student submissions. Each student submission consists of NN lines, each in the format:

c[i] code[i]

where c[i] is the i-th character and code[i] is an non-empty string of no more than 63 '0's and '1's.

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.

Sample Input:

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

Sample Output:

Yes
Yes
No
No


/*
检查霍夫曼编码

	读取字母表
	构建自己的霍夫曼树
		利用最小堆管理节点
			堆的插入——放在最后，然后上滤
			删除——弹出顶后，将最后的元素放到堆顶，然后下滤
	计算最小编码开销
	
	读取学生构建的编码表
		如果是0，往左查，如果1，往右查。查不到就申请节点 
			如果路上遇到带word的节点，直接把flag设成ERROR 
			如果没有遇到flag，看看最后落下的位置有没有子树，如果有，那么这个节点是别人的父节点，不能插入word，报ERROR 
			如果一切正常，在该位置设置word 
		如不出错，查表找出当前字符的frequency，乘上码长得到开销 ，累加到总开销上 
		
	
评测结果
时间	结果	得分	题目	编译器	用时（ms）	内存（MB）	用户
2017-07-01 01:34	正在评测	0	5-9	gcc	无	无	
测试点结果
测试点	结果	得分/满分	用时（ms）	内存（MB）
测试点1	答案正确	16/16	2	1
测试点2	答案正确	7/7	1	1
测试点3	答案正确	3/3	1	1
测试点4	答案错误	0/1	27	1   <————实在不知道什么情况 
测试点5	答案正确	1/1	18	1
测试点6	答案正确	1/1	1	1
测试点7	答案正确	1/1	1	1
*/

#include<stdio.h>
#include<stdlib.h>
#define DBG //
#define NOTLEAF '*'
#define ERROR 8
typedef struct HuffNode *HuffTree;
struct HuffNode
{
	char word;
	int freq;
	HuffTree left;
	HuffTree right;	
} ;
struct HuffNode codeTable[100];

HuffTree gHeap[100];
int gHeapLen=0;

void InsertIntoHeap(HuffTree T) //向堆中插入数据 
{
	int i;
	gHeapLen++;
	gHeap[gHeapLen]=T;
	i=gHeapLen;
	while(i>1)
	{
		if(gHeap[i]->freq < gHeap[i/2]->freq)
		{
			gHeap[i]=gHeap[i/2];
		DBG("InsertIntoHeap+-+-[%d]=[%d]\n",i/2,i);
			i=i/2;
		}
		else break;
	}
	gHeap[i]=T;
	DBG("doneInsertIntoHeap[%d]\n",i);
	return;
}

void DBG_showstatus(int n) //debug用函数，打印编码表和堆的状态 
{
	int i;
	for(i=0;i<n;i++)
		DBG("showstatus_codeTable:+%c : %d\n",codeTable[i].word,codeTable[i].freq);
	DBG("+gHeapLen:%d",gHeapLen);
	for(i=1;i<=gHeapLen;i++)
		DBG("showstatus_Heap:++%c : %d\n",gHeap[i]->word,gHeap[i]->freq);
}
HuffTree PopHeap() //弹出堆顶元素并整堆 
{
	if(gHeapLen<1) return NULL;//当前存量小于1说明有问题 
	
	int i,parent,child;
	HuffTree heapTop,temp;
	heapTop=gHeap[1];
	gHeap[1]=gHeap[gHeapLen];
	gHeapLen--;
	
	parent=1;
	temp=gHeap[parent];
	while(parent*2<=gHeapLen)
	{
		child=2*parent;
		if(child*2!=gHeapLen)
		{
			if(gHeap[child]->freq  > gHeap[child+1]->freq)
				child++;			
		}
		if (temp->freq > gHeap[child]->freq)
		{
			gHeap[parent]=gHeap[child];
			parent=child;
		}
		else break;
	}
	gHeap[parent]=temp;
	
	return heapTop;
}

void DestroyHuffTree(HuffTree A) //回收内存 
{
	if(A == NULL)
		return;
	DestroyHuffTree(A->left);
	DestroyHuffTree(A->right);
	free(A);
}

HuffTree CreateHuffTreeNode() //申请新的节点。此函数刚开始忘了给申到的节点赋初值，导致不少错误 
{
	HuffTree T=malloc(sizeof(struct HuffNode));
	T->word=NOTLEAF;
	T->left=NULL;
	T->right=NULL;
	T->freq=0;
	return T;
}

HuffTree BuildHuffTree() //把堆里的数据处理成一颗编码树 
{
	HuffTree T,A,B;
	while(gHeapLen>=2)
	{
		A=PopHeap();
		B=PopHeap();
		T=CreateHuffTreeNode();
		T->word=NOTLEAF;
		T->freq=A->freq+B->freq;
		T->left=A;
		T->right=B;
		InsertIntoHeap(T);
		DBG("In BuildHuffTree %d T->word\n",T->word);
	}
	return T;
}

int GetFreq(char c,int n) //查询指定字符的频率值 
{
	int i;
	for(i=0;i<n;i++)
	{
		if((codeTable[i].word) == c)
		{
			return codeTable[i].freq;
		}
	
	}
}
int Calcwpl(HuffTree T,int deepth) //计算整棵树的wpl 
{
	if(T==NULL)
	{
		DBG("In Calcwpl return a null\n");
		return 0;
	}
		
	if((T->word) != NOTLEAF)
	{
		DBG("In Calcwpl T->word = %c,return %d*depth %d=%d\n",T->word,T->freq,deepth,T->freq*deepth);
		return T->freq*deepth;
	}
		
	if((T->word) == NOTLEAF)
		DBG("In Calcwpl return a NOTLEAF\n");
		return Calcwpl(T->left,deepth+1)+Calcwpl(T->right,deepth+1);
}
void CheckCodes(int len,int wpl) //判断一系列的编码是否符合huffman 
{
	char tempc[100],tempbin[1000],bin;
	int i,j,pt;
	int flag=0,count=0,totalcost=0;
	HuffTree TOP,A;
	A=CreateHuffTreeNode();
	TOP=A;

	A->word=NOTLEAF;

	for(i=0;i<len;i++)
	{
		scanf("%s%s",tempc,tempbin);
		getchar();
		DBG("-%c-",tempc[0]);
		count=0;
		pt=0;
		while((bin=tempbin[pt++])!='\0')
		{
			DBG("bin-%c-\n",bin);
			if (flag==ERROR)
				continue;
			count++;
			if(A->word == NOTLEAF)
			{
				//左边的情况 
				if (bin=='0')
				{
					if(A->left==NULL)
					{
						A->left=CreateHuffTreeNode();
						A->left->word=NOTLEAF;
						A=A->left;
					}
					else
					{
						if (A->left->word != NOTLEAF)
						{
							flag=ERROR;
							DBG("setflag in left ,word='%c'\n",A->left->word);
							continue;
						}
						A=A->left;
					}
				}
				//右边的情况 
				if (bin=='1')
				{
					if(A->right==NULL)
					{
						A->right=CreateHuffTreeNode();
						A->right->word=NOTLEAF;
						A=A->right;
					}
					else
					{
						if (A->right->word != NOTLEAF)
						{
							flag=ERROR;
							DBG("setflag in right ,word='%c'\n",A->right->word);
							continue;
						}
						A=A->right;
					}
				}					
			}
		}
		if(flag!=ERROR)
		{
			totalcost+=count*GetFreq(tempc[0],len);
			if(A->left!=NULL || A->right !=NULL)  //此时如果发现节点还有子树，那么编码是有问题的。 
				flag=ERROR;
			A->word=tempc[0];
			A=TOP;
		}	
			DBG("totalcost=%d\n",totalcost);
	}

	if(flag!=ERROR && totalcost==wpl)
		printf("Yes\n");
	else
		printf("No\n");
		
	DestroyHuffTree(TOP);
	return;
}

int main()
{
	int i,j,tmpi,N,M,wpl;
	char tmpc;
	HuffTree T;
	scanf("%d\n",&N);

	for(i=0;i<N;i++)
	{
		tmpc=getchar();
		scanf("%d ",&tmpi);
		codeTable[i].word=tmpc; //填编码表 
		codeTable[i].freq=tmpi;
		T=CreateHuffTreeNode(); //创建节点然后往堆里插 
		T->freq=tmpi;
		T->word=tmpc;
		InsertIntoHeap(T);
	}
									DBG_showstatus(N);
	T=BuildHuffTree();
									DBG_showstatus(N);
	wpl=Calcwpl(T,0);	
									DBG("WPL=%d\n",wpl);
	scanf("%d\n",&M);
	for(j=0;j<M;j++)
		CheckCodes(N,wpl);
									DBG_showstatus(N);

	return 0;
}

　　

posted on 2017-07-09 11:25  gravitykey  阅读(688)  评论(0)  编辑  收藏  举报

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