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最新推荐文章于 2022-04-10 16:02:38 发布

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本文介绍了一种判断一组二进制编码是否为立即可解码的方法。通过构建字典树来检查任一编码是否成为其他编码的前缀，进而判断整个集合是否满足立即可解码条件。

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IMMEDIATE DECODABILITY

Description

An encoding of a set of symbols is said to be immediately decodable if no code for one symbol is the prefix of a code for another symbol. We will assume for this problem that all codes are in binary, that no two codes within a set of codes are the same, that each code has at least one bit and no more than ten bits, and that each set has at least two codes and no more than eight.

Examples: Assume an alphabet that has symbols {A, B, C, D}

The following code is immediately decodable:
A:01 B:10 C:0010 D:0000

but this one is not:
A:01 B:10 C:010 D:0000 (Note that A is a prefix of C)

Input

Write a program that accepts as input a series of groups of records from standard input. Each record in a group contains a collection of zeroes and ones representing a binary code for a different symbol. Each group is followed by a single separator record containing a single 9; the separator records are not part of the group. Each group is independent of other groups; the codes in one group are not related to codes in any other group (that is, each group is to be processed independently).

Output

For each group, your program should determine whether the codes in that group are immediately decodable, and should print a single output line giving the group number and stating whether the group is, or is not, immediately decodable.

Sample Input

Sample Output

Set 1 is immediately decodable
Set 2 is not immediately decodable
题意：其实就是简单的查找，看某个数据是不是其中某个的前缀。

字典树..

#include <iostream>
using namespace std;
char s[1111][111];
struct node
{
	node *next[26];
	int num;
	node()//构造函数
	{
		memset(next,0,sizeof(next));
		num=0;
	}
};
node *root;
void build(char *s)//建树
{
	node *p=root;
	for(int i=0;i<strlen(s);i++)
	{
		if(p->next[s[i]-'0']==NULL)
			p->next[s[i]-'0']=new node;
		p=p->next[s[i]-'0'];
		p->num++;
	}
}
int findstr(char *s)//查找
{
	node *p=root;
	for(int i=0;i<strlen(s);i++)
	{
		if(p->next[s[i]-'0']!=NULL)
			p=p->next[s[i]-'0'];
		else
			return 0;
	}
	return p->num;
}
int main()
{
	int i,j,k=0,x=1;
	while(~scanf("%s",s[k++]))
	{
		root=new node;
		build(s[k-1]);
		while(~scanf("%s",s[k++]))
		{
			if(s[k-1][0]=='9')break;
			build(s[k-1]);
		}
		bool flag=true;
		k--;
		for(i=0;i<k;i++)
		{
			if(findstr(s[i])>1)//要扣除自己本身
			{
				flag=false;
				break;
			}
		}
		if(flag==true)
			printf("Set %d is immediately decodable\n",x++);
		else
			printf("Set %d is not immediately decodable\n",x++);
		k=0;
	}
	return 0;
}